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PHYSICS by Pradeep Kshetrapal
2 Kinetic Theory of Gases
11.1 Introduction Introduction.. In gases the intermolecular forces are very weak and its molecule may y apart in all directions. So the gas is characterised by the following properties. (iIt has no shape and si!e and can be obtained in a vessel of any shape or si!e. (ii It e"pands inde#nitely and uniformly to #ll the available space. (iii It e"ert erts pressure on its surroundings.
11.2 Assumption of Kinetic Theory of Gases . Kineti Kinetic c theory theory of gases gases relat relates es the macro macrosco scopic pic prope properti rties es of gases gases (such (such as press pressur ure$ e$ temp temper erat atur ure e etc. etc. to the the micr micros osco copi pic c prop proper erti ties es of the the gas gas mole molecu cule les s (suc (such h as sp spee eed$ d$ momentum$ kinetic energy of molecule etc. etc. %ctually it attempts to develop a model of the molecular behaviour which w hich should result in the observed behaviour of an ideal gas. It is based on following assumptions & (' very gas consists of e"tremely small particles known as molecules. The molecules of a given gas are all identical but are di)erent than those of another gas. (* The molecules of a gas are identical$ spherical$ rigid and perfectly perf ectly elastic point masses. (+ Their si!e is negligible in comparison to intermolecular distance (', - m (/ The volume of molecules is negligible in comparison to the volume of gas. (The volume of molecules is only ,.,'/0 of the volume of the gas. (1 2olecules of a gas keep on moving randomly in all possible direction with all possible velocities. (3 The speed of gas molecules molecule s lie between !ero and in#nity (very high speed. (4 The number of molecules moving with most probable speed is ma"imum. (5 The gas molecules keep on colliding among themselves as well as with the walls of containing containing vessel. These collisions collisions are perfectly perfectly elastic. ( i.e. the i.e. the total energy before collision 6 total energy after the collision. ( 2olecules move in a straight line with constant speeds during successive collisions. (', The distance distance covered by the molecules between two successiv successive e collisions collisions is known as free path and mean of all free paths is known as mean free path. ('' The time spent M a collision between two molecules is negligible in comparison to time between two successive collisions. ('* The number of collisions per unit volume in a gas remains constant. ('+ 7o attractive or repulsive force acts between gas molecules. mol ecules.
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PHYSICS by Pradeep Kshetrapal
3 Kinetic Theory of Gases ('/ ('/ Gravit Gravitati ationa onall attrac attractio tion n among among the molecul molecules es is ine)ec ine)ectiv tive e due to e"tr e"tremel emely y small small masses and very high speed of molecules. ('1 2olecules constantly collide with the walls of container due to which their momentum changes. The change in momentum is transferred to the walls of the container. 8onse9uently pressure is e"erted by gas molecules on the walls of container. container. ('3 The density of gas is constant at all points of the container.
11.3 Pressure of an Ideal Gas. Gas . 8onsider an ideal gas (consisting of N molecules each of mass m enclosed in a cubical bo" of side L. ; It:s any molecule moves with velocity v in any direction where v v x ;i v y j; v z k
This molecule collides with the shaded wall wall ( A' with
Y
velocity v x and rebounds with velocity v x .
The change in momentum of the molecule
– v x
v x
P (mv x (mv x *mv x
X L
%s the the moment momentum um rema remains ins cons conserv erved ed in a colli collisi sion$ on$ the
Z
change in momentum of the wall A wall A' is P *mv x %fter %fter rebound this molecule travel toward opposite wall A* with velocity velocity v x $ collide to it and again rebound with velocity v x towards wall A wall A'. (' Time between two successive collision with the wall A'. t
*L =istance travelled bymolecule between twosuccessive collisio
7umber of collision per second n
' v x t *L
(* The momentum imparted per unit time to the wall by this molecule nP
v x *L
*mv x
This is also e9ual to the the force e"erted e"erted on the wall A' due to this molecule F (+ The total force on the wall A' due to all the molecules F x
m L
m * v x L
v
* x
(/ 7ow pressure is de#ned as force per unit area
F x A
m AL
P x
So
P x P y P z
v x * m V
m V
v x *
Similarly
P y
m V
* v y and
P z
m V
v z *
(v v v * x
* y
+P
m V
+P
m * (v ' v ** v ++ ...... V
v *
* z
>%s P x
P y
P z
* P and v * v x v y * v z * ?
m * v x L
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PHYSICS by Pradeep Kshetrapal
3 Kinetic Theory of Gases ('/ ('/ Gravit Gravitati ationa onall attrac attractio tion n among among the molecul molecules es is ine)ec ine)ectiv tive e due to e"tr e"tremel emely y small small masses and very high speed of molecules. ('1 2olecules constantly collide with the walls of container due to which their momentum changes. The change in momentum is transferred to the walls of the container. 8onse9uently pressure is e"erted by gas molecules on the walls of container. container. ('3 The density of gas is constant at all points of the container.
11.3 Pressure of an Ideal Gas. Gas . 8onsider an ideal gas (consisting of N molecules each of mass m enclosed in a cubical bo" of side L. ; It:s any molecule moves with velocity v in any direction where v v x ;i v y j; v z k
This molecule collides with the shaded wall wall ( A' with
Y
velocity v x and rebounds with velocity v x .
The change in momentum of the molecule
– v x
v x
P (mv x (mv x *mv x
X L
%s the the moment momentum um rema remains ins cons conserv erved ed in a colli collisi sion$ on$ the
Z
change in momentum of the wall A wall A' is P *mv x %fter %fter rebound this molecule travel toward opposite wall A* with velocity velocity v x $ collide to it and again rebound with velocity v x towards wall A wall A'. (' Time between two successive collision with the wall A'. t
*L =istance travelled bymolecule between twosuccessive collisio
7umber of collision per second n
' v x t *L
(* The momentum imparted per unit time to the wall by this molecule nP
v x *L
*mv x
This is also e9ual to the the force e"erted e"erted on the wall A' due to this molecule F (+ The total force on the wall A' due to all the molecules F x
m L
m * v x L
v
* x
(/ 7ow pressure is de#ned as force per unit area
F x A
m AL
P x
So
P x P y P z
v x * m V
m V
v x *
Similarly
P y
m V
* v y and
P z
m V
v z *
(v v v * x
* y
+P
m V
+P
m * (v ' v ** v ++ ...... V
v *
* z
>%s P x
P y
P z
* P and v * v x v y * v z * ?
m * v x L
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Kinetic Theory of Gases 4
or
v '* v ** v +* v /* .... mN +P V N
or
+P
v rms
mN * v rms V
oot mean mean s9ua s9uarre %s root
veloc elocit ity y of the the gas gas mole molecu cule le
v '* v ** v +* v /* ..... N
or
P
' mN * v rms + V
Important points (i P
' mN * v rms + V
or
(m N T V
P
* >%s v rms T ?
(a If volume and temperature of a gas are constant P mN i.e. @ressure i.e. @ressure (2ass of gas. i.e. if i.e. if mass of gas is increased$ number of molecules and hence number of collision per second increases i.e. pressure i.e. pressure will increase. (b If mass mass and tempera temperatur ture e of a gas are constan constant. t. P ('AV ('AV $ $ i.e.$ i.e.$ if volume decreases$ number of collisions collisions per second will increase increase due to lesser e)ective distance distance between the walls resulting in greater pressure. pressure. (c If mass and volume of gas are constant$ P (v rms* T i.e.$ i.e.$ if temperature increases$ the mean s9uare speed of gas molecules will increase and as gas gas molec molecul ules es are are movi moving ng fast faster er$$ they they will will colli collide de with with the the walls walls more more ofte often n with with grea greate terr momentum resulting in greater pressure. (ii P
'M * ' mN * v rms v rms + V + V
P
' * v rms +
>%s M 6 mN = T = Total otal mass of the gas? M %s V
(iii Belation between pressure and kinetic energy Kinetic energy
' M * ' ' * * M v rms Kinetic energy per unit volume (E v rms v rms C..(i * V * *
and we know P Drom (i and (ii$ we get P
' * v rms +
C..(ii
* E +
i.e. the i.e. the pressure e"erted by an ideal gas is numerically e9ual to the two third of the mean kinetic energy of translation per unit volume of the gas.
Sample Problems based on Pressure Problem 1. roblem 1. The root mean s9uare speed of hydrogen molecules of an ideal hydrogen gas kept in a gas chamber at ,EC ,EC is +'5, ms. ms. The pressure on the hydrogen gas is (=ensity of hydrogen gas is 5. ',* k!A m+ $ ' atmosphere '.,' ',1 N A m* [MP PMT 1!"
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PHYSICS by Pradeep Kshetrapal
! Kinetic Theory of Gases (a (a ,.' ,.' "tm #$%&ti$n & #$%&ti$n & (d %s P
(b '.1 '.1 "tm
(c *., "tm
(d +., "tm
' ' * (5. ', * (+'5,* +.,+ ',1 NAm* +., "t v rms + +
Problem 2. roblem 2. The temperature of a gas is raised while its volume remains constant$ the pressure e"erted by a gas on the walls of the container increases because its molecules [#$%& PMT 13"
(a Fose Fose more kinetic kinetic energy energy to the wall (b %re in contact contact with the wall wall for a shorter shorter time (c Strike the the wall more more often often with higher higher velocities (d 8ollide 8ollide with each other other less fre9uency fre9uency #$%&ti$n & (c =ue to increase increase in temperatu temperature re root mean s9uare s9uare velocity of gas molecules molecules increases. increases. So they strike the wall more often with higher velocity. ence the pressure e"erted by a gas on the walls of the container increases. roblem 3. % cylinder of capacity Problem 3. capacity *, %itres is %itres is #lled with ' * gas. The total average kinetic energy of translatory motion of its molecules is '.1 ',1 ( . The pressure of hydrogen in the cylinder is [MP P&T 13" (a * ',3 N A m*
(b + ',3 N A m*
(c / ',3 N A m*
(d 1 ',3 N A m*
#$%&ti$n & #$%&ti$n & (d (d Kine Kineti tic c ene energ rgy y E 6 '.1 ',1 ( $ volume V 6 *, %itre 6 %itre 6 *, ',+ m+
* '.1 ',1 *E 1 ',3 NAm* . @ressure + + V + *, ',
Problem 4. roblem 4. N molecules molecules each of mass mass m of gas gas A A and *N *N molecules each of mass *m * m of gas gas ) are cont contai aine ned d in the the same same vess vessel el at temp temper erat atur ure e T . The The mean mean s9ua s9uare re of the the velo veloci city ty of * molecules of gas ) is v and the mean s9uare of x of x component component of the velocity of molecules of *
gas A gas A is is * . The ratio (a '
** *
v
is
['#&(T 1)4* MP PMT 1+"
(b *
(c
#$%&ti$n & #$%&ti$n & (d 2ean s9uare s9uare velocit velocity y of molecule molecule
' +
(d
* +
+kT m
Dor gas A gas A$$ x component component of mean s9uare velocity of molecule ** * 2ean s9uare velocity +*
+kT m
Dor ) gas mean s9uare velocity v *
Drom (i and (ii
+** v *
C..(i
+kT *m
C..(ii
** * * so * . ' + v
+ + Problem !. roblem !. % ask ask contains contains ', m gas. %t a temperature$ the number of molecules of o"ygen are k! and at that temperature the +., ',** . The mass of an o"ygen molecule is 1.+ ',*3 k! and
rms velocity rms velocity of molecules is /,, mAs. The pressure in N A m* of the gas in the ask is (a 5./5 ',/
(b *.54 ',/
(c *1.// ',/
#$%&ti$n & #$%&ti$n & (a V ',+ m+ $ N +., ',** $ m 1.+ ',*3k!$ v rms /,,mAs
(d '*.4* ',/
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Kinetic Theory of Gases ,
P
' mN * ' 1.+ ',*3 +., ',** v rms (/,,* 5./5 ',/ NAm* . + + V + ',
Problem ,. % gas at a certain volume and temperature has pressure 41 cm. If the mass of the gas is doubled at the same volume and temperature$ its new pressure is (a +4.1 cm #$%&ti$n & (c
P
(b 41 cm
(c '1, cm
(d +,, cm
'M * MT v rms P + V V
%t constant volume and temperature$ if the mass of the gas is doubled then pressure will become twice.
11.4 Ideal Gas &-uation. % gas which strictly obeys the gas laws is called as perfect or an ideal gas. The si!e of the molecule of an ideal gas is !ero i.e. each molecule is a point mass with no dimension. There is no force of attraction or repulsion amongst the molecule of the gas. %ll real gases are not perfect gases. owever at e"tremely low pressure and high temperature$ the gases like hydrogen$ nitrogen$ helium etc. are nearly perfect gases. The e9uation which relates the pressure (P$ volume (V and temperature (T of the given state of an ideal gas is known as gas e9uation.
Ideal as e-uations Dor ' mole or N A molecule or M gram or **./ %itres of gas
PV 6 +T
Dor mole of gas
PV 6 +T
Dor ' molecule of gas
+ T kT N A
PV
Dor N molecules of gas
PV 6 NkT
Dor ' !m of gas
+ T rT M
PV
for n !m of gas
PV 6 nrT
(' /ni0ersal as constant R =imension >ML*T * '? +
@ressure Horkdone PV
Thus universal gas constant signi#es the work done by (or on a gas per mole per kelvin. %itre "tm ($&%e c"% '.5 ,.5**' 5.+' m$%e ke%vin m$%e ke%vin m$%e ke%vi
S.T.@. value &
(* $oltman5s constant k =imension >ML*T * '? k
+ 5.+' N 3.,*+ ',*+
'.+5 ',*+ ($&%eAke%vi
(+ %peci6c as constant r =imension >L*T * ' ? r
+ M
Jnit &
($&%e !m ke%vi
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PHYSICS by Pradeep Kshetrapal
7 Kinetic Theory of Gases Since the value of M is di)erent for di)erent gases. ence the value of r is di)erent for di)erent gases.
Sample Problems based on Ideal gas equation Problem 7. % gas at *4EC has a volume V and pressure P. n heating its pressure is doubled and volume becomes three times. The resulting temperature of the gas will be [MP P&T 2++3"
(a '5,,EC
(b '3*EC
(c '1*4EC
#$%&ti$n & (c Drom ideal gas e9uation PV +T we get
T * 3T ' 3 +,, '5,,,
Problem ). % balloon contains
(d 3,,EC
P V *P +V * * ' ' 3 T ' P' V ' P' V '
T *
'1*4 C.
1,,m+
of helium at *4EC and ' atmosphere pressure. The volume of the helium at - +EC temperature and ,.1 atmosphere pressure will be [MP PMT8P&T 1)* 9IPM&( 2++1: 2++2"
(a 1,,m+
(b 4,,m+
#$%&ti$n & (c Drom PV +T we get
(c ,,m+
(d ',,,m+
T P *4, ' * ' V * 1,, ,,m+ V ' T ' P* +,, ,.1 1 1
V *
Problem . Hhen volume of system is increased two times and temperature is decreased half of its initial temperature$ then pressure becomes [AI&&& 2++2"
(a * times
(b / times
#$%&ti$n & (c Drom PV +T we get
(c ' A / times
(d ' A * times
T V T A * V ' P * ' ' ' P* ' P' T ' V * T ' *V ' / /
P*
The e9uation of state corresponding to 5! of -* is
Problem 1+.
[#$%& PMT 14* ;PMT 2+++"
(a PV 5+T
(b PV +T A /
(c PV +T
(d PV +T A *
#$%&ti$n & (b %s +* !m -* means ' m$%e therefore 5 !m -* means ' A /m$%e i.e. So from PV +T we get PV
' /
' +T +T or PV / /
Problem 11. % ask is #lled with '+ !m of an ideal gas at *4EC and its temperature is raised to 1*EC. The mass of the gas that has to be released to maintain the temperature of the gas in the ask at 1*EC and the pressure remaining the same is [&AM#&T &n. 2+++"
(a *.1 !
(b *., !
(c '.1 !
(d '., !
#$%&ti$n & (d PV 2ass of gas Temperature In this problem pressure and volume remains constant so M 'T ' 6 M*T * 6 constant
M* M'
T ' T *
(*4 *4+ (1* *4+
+,, '* '* '* !m '*!m '+ M * M' +*1 '+ '+ '+
i.e. the mass of gas released from the ask 6 '+ !m - '* !m 6 ' !m.
Problem 12.
%ir is #lled at 3,EC in a vessel of open mouth. The vessel is heated to a temperature T so that ' A / th part of air escapes. %ssuming the volume of vessel remaining constant$ the value of T is [MP P&T 1,: " (a 5,EC (b ///EC (c +++EC (d '4'EC
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Kinetic Theory of Gases )
M +M >%s ' A /th part of air escapes? / / If pressure and volume of gas remains constant then MT 6 constant
#$%&ti$n & (d M ' M $ T ' 3, *4+ +++, $ M * M
Problem 13.
T * T '
M' M*
M / / / T * T ' +++ ///, '4'C +M A / + + +
If the intermolecular forces vanish away$ the volume occupied by the molecules contained in /.1 k! water at standard temperature and pressure will be given by [#PMT 1)" 1.3 m
+
(a #$%&ti$n & (a
(b
/.1 m
+
(c ''.* %itre
(d
''.* m
+
/.1 k! 2assof water *1,$ T 6 *4+ , and P ',1 NAm* (ST@ 2olecularwt.of water '5 ',+ k!
Drom PV +T V
+T
P
*1, 5.+ *4+ 1
',
1.33m+ .
Problem 14. The pressure P$ volume V and temperature T of a gas in the Lar A and the other gas in the Lar ) at pressure *P$ volume V A/ and temperature *T $ then the ratio of the number of molecules in the Lar A and ) will be [AIIM% 1)2" (a ' & '
(b ' & *
(c * & '
(d / & '
N +T where N 6 7umber of molecule$ N A 6 %vogadro N A
#$%&ti$n & (d Ideal gas e9uation PV +T number
Problem 1!.
P V T P V *T / ' ' * . N * P* V * T ' *P V A/ T ' N'
The e"pansion of an ideal gas of mass m at a constant pressure P is given by the straight line . Then the e"pansion of the same ideal gas of mass *m at a pressure PA * is given by the straight line
(a E
5
(b C
3 /
(c )
* '
(d A #$%&ti$n & (d Drom PV MT or V
A ) C E
Temperature
M M represents the slope of curve drawn on volume and T ere P P
temperature a"is.
M graph is (given in the problem Dor #rst condition slope P Dor second condition slope
*M M / i.e. slope becomes four time so graph A is correct in PA* P
this condition.
Problem 1,.
If the value of molar gas constant is 5.+ (Am$%eM, $ the n speci#c gas constant for hydrogen in (Am$%eM, will be (a /.'1
(b 5.+
#$%&ti$n & (a Speci#c gas constant r
(c '3.3
(d 7one of these
Jniversalgasconstant (+ 5.+ /.'1 ($&%eAm$%eM, . 2olecularweightof gas(M *
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PHYSICS by Pradeep Kshetrapal
Kinetic Theory of Gases
Problem 17.
% gas in container A is in thermal e9uilibrium with another gas in container ). both contain e9ual masses of the two gases in the respective containers. Hhich of the following can be true (a P AV A P)V )
#$%&ti$n & (b$ c i.e. A )
(b P A P)$ V A V ) (c P A P)$ V A V )
P A P ) V A V )
(d
%ccording to problem mass of gases are e9ual so number of moles will not be e9ual
Drom ideal gas e9uation PV
+T
P A V A
P) V )
A
>%s temperature of the container
)
are e9ual? Drom this relation it is clear that if P A P) then Similarly if V A V ) then
Problem 1).
P A P)
A )
V A V )
A )
' i.e. V A V )
' i.e. P A P) .
Two identical glass bulbs are interconnected by a thin glass tube. % gas is #lled in these bulbs at 7.T.@. If one bulb is placed in ice and another bulb is placed in hot bath$ then the pressure of the gas becomes '.1 times. The temperature of hot bath will be (a ',,EC (b '5*EC (c *13EC ot bath
Ice
(d 1/3EC
#$%&ti$n & (d Nuantity of gas in these bulbs is constant i.e. Initial 7o. of moles in both bulb 6 #nal number of moles ' * 'O *O PV +(*4+ T 5',
PV +(*4+
'.1 PV '.1PV +(*4+ +(T
* '.1 '.1 *4+ *4+ T
1/3C .
Problem 1.
Two containers of e9ual volume contain the same gas at pressures
P' and P*
and
absolute temperatures T ' and T * respectively. n Loining the vessels$ the gas reaches a common pressure P and common temperature T . The ratio PAT is e9ual to (a
P' P* T ' T *
(b
P'T ' P*T * (T ' T **
#$%&ti$n & (d 7umber of moles in #rst vessel '
P'V +T '
(c
P'T * P*T ' (T ' T **
'
*
P (*V P'V P* V +T +T ' +T * P P P ' * T *T ' *T *
P' P * *T ' *T *
and number of moles in second vessel *
If both vessels are Loined together then 9uantity of gas remains same i.e
(d
Initially P' T '
P* T * V
V Dinally P T
P T
V
V
P*V +T *
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Kinetic Theory of Gases 1+
Problem 2+.
%n ideal monoatomic gas is con#ned in a cylinder by a springMloaded piston if crossM section 5 ',+ m* . Initially the gas is at +,,, and occupies a volume of *./ ',+ m+ and the spring is in a rela"ed state. The gas is heated by a small heater coil '. The force constant of the spring is 5,,, NAm$ and the atmospheric pressure is '., ',1 P" . The cylinder and piston are thermally insulated. The piston and the spring are massless and there is no friction between the piston and cylinder. There is no heat loss through heater coil wire leads and thermal capacity of the heater coil is negligible. Hith all the above assumptions$ if the gas is heated by the heater until the piston moves out slowly by ,.'m$ then the #nal temperature is (a /,, ,
Gas
(b 5,, ,
'
Spring
(c '*,, , (d +,, , #$%&ti$n & (b V ' *./ ',+ m+ $
P' P, ',1
N m*
and
T ' 6 +,, ,
(given
If area of crossMsection of piston is A and it moves through distance x then increment in volume of the gas 6 Ax and if force constant of a spring is k then force F 6 kx and pressure 6
F kx A A
V * V ' Ax *./ ',+ 5 ',+ ,.' +.* ',+ and
P* P, Drom
kx 5,,, ,.' ',1 * ',1 + A 5 ',
ideal
gas
e9uation
P'V ' T '
P*V * T *
',1 *./ ',+ * ',1 +.* ',+ +,, T *
T * 5,,, Two identical containers each of volume V , are Loined by a small pipe. The containers
Problem 21.
contain identical gases at temperature T , and pressure P, . ne container is heated to temperature *T , while maintaining the other at the same temperature. The common pressure of the gas is P and n is the number of moles of gas in container at temperature *T , (a P *P, #$%&ti$n & (b$ c
(b P
/ P, +
Initially for container A
Dor container )
(c n
* P,V , + +T ,
(d n
+ P,V , * +T ,
P, V , n, +T ,
P, V , n, +T ,
n,
Initially
P, V ,
n,$ V ,
+T ,
P,$ T ,
P,$ T ,
(%
(P
Total number of moles n, n, *n,
n,$ V ,
Since even on heating the total number of moles is conserved ence
n' n* *n,
......(i
If P be the common pressure then Dor container A PV , n'+ *T , Dor container ) PV , n* +T ,
n' n*
PV , *+T , PV , +T ,
n'$ V ,
Dinally
n*$ V ,
P$ *T ,
P$ T ,
(%
(P
genius
PHYSICS by Pradeep Kshetrapal
11 Kinetic Theory of Gases Substituting the value of n, $ n' and n* in e9uation (i we get
P
PV , *+T ,
PV , +T ,
*.P, V , +T ,
/ P, + 7o. of moles in container A (at temperature *T , 6 n'
* P, V , / V , P, + +T , *+T , + *+T , PV ,
/ %s P + P,
Problem 22.
%t the top of a mountain a thermometer reads 4E C and a barometer reads 4, cm of '!. %t the bottom of the mountain these read *4EC and 43 cm of '! respectively. 8omparison of density of air at the top with that of bottom is 4oC$ 4, cm of '!
(a 41A43 (b 4,A43 (c 43A41 (d 43A4, #$%&ti$n & (a Ideal gas e9uation$ in terms of density
Top Pottom
P Top PPottom
T Pottom T Top
4, 43
+,, *5,
P' 'T '
P* *T *
*4oC$ 43 cm of '! '
constant
*
P' P*
T * T '
41 43
11.!
ero sie of molecule % certain portion of volume of a gas is covered by the molecules themselves. Therefore the space available for the free motion of molecules of gas will be slightly less than the volume V of a gas. ence the e)ective volume becomes (V - / (ii ?orce of attraction @eteen as molecules =ue to this$ molecule do not e"ert that force on the wall which they would have e"erted in the absence of intermolecular force. Therefore the observed pressure P of the gas will be less than that present in the absence of
" intermolecular force. ence the e)ective pressure becomes P * V The e9uation obtained by using above modi#cations in ideal gas e9uation is called
" P * (V / +T V
Dor m$%es of gas
* P " (V / +T V *
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Kinetic Theory of Gases 12
ere " and / are constant called "? 6 >ML1 T * ? and >/? 6 >L+? Jnits & " 6 N m/ and / 6 m+.
11., Andres #ur0es. The pressure (P versus volume (V curves for actual gases are called %ndrews curves. (' %t +1,EC$ part A) represents vapour phase of water$ ' in this part Poyle:s law is obeyed P . @art )C represents V the coMe"istence of vapour and li9uid phases. %t point C$ vapours completely change to li9uid phase. @art C is parallel to pressure a"is which shows that compressibility of the water is negligible.
1
1"s +5,2C
P
'
Fi9uid vapour region
+4/.'2C
Fi9uid
+4,2C
F C
+3,2C
E )
+1,2C
%ndrews curve for water
(* %t +3,EC portion representing the coMe"istence of li9uid vapour phase is shorter.
A
(+ %t +4,EC this portion is further decreased. (/ %t +4/.'EC$ it reduces to point (' called critical point and the temperature +4/.'EC is called critical temperature (T c of water. (1 The phase of water (at +5,EC above the critical temperature is called gaseous phase.
#ritical temperature: pressure and 0olume The point on the P0V curve at which the matter gets converted from gaseous state to li9uid state is known as critical point. %t this point the di)erence between the li9uid and vapour vanishes i.e. the densities of li9uid and vapour become e9ual. (i #ritical temperature T c The ma"imum temperature below which a gas can be li9ue#ed by pressure alone is called critical temperature and is characteristic of the gas. % gas cannot be li9ue#ed if its temperature is more than critical temperature. C-* (+,/.+ , $ -* (-''5EC$ N* (-'/4.'EC
and
'*- (+4/.'EC
(ii #ritical pressure Pc The minimum pressure necessary to li9uify a gas at critical temperature is de#ned as critical pressure. C-* (4+.54 /"r and
-* (/.4"tm
(iii #ritical 0olume V c The volume of ' mole of gas at critical pressure and critical temperature is de#ned as critical volume. C-* (1 ',-3 m+ (iv (elation @eteen
* + T c *4+ * T c " $ / and 5 Pc 3/ Pc
Sample problems based on Vander Waal gas equation
Pc V c T c
+ + 5
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PHYSICS by Pradeep Kshetrapal
13 Kinetic Theory of Gases
Problem 23.
Jnder which of the following conditions is the law PV 6 +T obeyed most closely by a real gas ['#&(T 174* MP PMT 14: 7* MP P&T 1* AM/ 2++1"
(a igh pressure and high temperature
(b Fow pressure and low temperature
(c Fow pressure and high temperature
(d igh pressure and low temperature
#$%&ti$n & (c %t low pressure and high temperature real gas obey PV 6 +T i.e. they behave as ideal gas because at high temperature we can assume that there is no force of attraction or repulsion works among the molecules and the volume occupied by the molecules is negligible in comparison to the volume occupied by the gas.
Problem 24.
"T * c V (+T / $ where "$ /$ c and + P The e9uation of state of a gas is given by V
are constants. The isotherms can be represented by P AV m )V n $ where A and ) depend only on temperature then [#$%& PMT 1!"
(a m (d
c and
n '
(b m c and n '
m c and n '
(c m
c
and
n'
"T * c * c ' * ' c c #$%&ti$n & (a P V +T / P "T V +TV /V P (+T /V ("T V V Py comparing this e9uation with given e9uation P AV m )V n we get m c and
n ' .
Problem 2!.
%n e"periment is carried on a #"ed amount of gas at di)erent temperatures and at high PV pressure such that it deviates from the ideal gas behaviour. The variation of with P is +T shown in the diagram. The correct variation will correspond to PV+T [#PMT 1))" *.,
A )
(a 8urve A (b 8urve ) (c 8urve C (d 8urve
'., C ,$ ,
*, /, 3, 5, ',, P ("tm
PV constant +T but when pressure increase$ the decrease in volume will not take place in same proportion PV so will increases. +T
#$%&ti$n & (b %t lower pressure we can assume that given gas behaves as ideal gas so
Problem 2,.
The conversion of ideal gas into solids is (a @ossible only at low pressure
(b @ossible only at low temperature
(c @ossible only at low volume
(d Impossible
#$%&ti$n & (d Pecause there is !ero attraction between the molecules of ideal gas.
11.7
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Kinetic Theory of Gases 14
(' (oot mean s-uare speed It is de#ned as the s9uare root of mean of s9uares of the speed of di)erent molecules i.e. v rms
v '* v ** v +* v /* .... N
(i Drom the e"pression for pressure of ideal gas P v rms
(ii v rms (iii v rms
+PV mN
+PV 2assof gas
++T M
+N AkT N A M
2assof gas %s V
+P
+ +T M
+PV 2assof gas
' mN * v rms + V
>%s if M is the molecular weight of gas PV 6 +T and 2ass of as 6 M ?
++T M
+kT
>%s M 6 N Am and + 6 N Ak ?
m
Boot mean s9uare velocity v rms
+P
++T M
+kT m
Important points (i Hith rise in temperature rms speed of gas molecules increases as v rms T . (ii Hith increase in molecular weight rms speed of gas molecule decreases as v rms
' M
.
e.!.$ rms speed of hydrogen molecules is four times that of o"ygen molecules at the same temperature. (iii rms speed of gas molecules is of the order of kms ++T M
e.!.$ %t 7T@ for hydrogen gas (v rms (iv rms speed of gas molecules is %s
v rms
++T and M
v s
+T
M
+
+ 5.+' *4+ * ',+
'5/,mA s .
times that of speed of sound in gas
v rms
+ v s
(v rms speed of gas molecules does not depends on the pressure of gas (if temperature remains constant because P (Poyle:s law if pressure is increased n times then density will also increases by n times but v rms remains constant. (vi 2oon has no atmosphere because v rms of gas molecules is more than escape velocity ( v eM . % planet or satellite will have atmosphere only and only if v rms v e (vii %t T 6 , v rms 6 , i.e. the rms speed of molecules of a gas is !ero at , , . This temperature is called absolute !ero. (* Most pro@a@le speed The particles of a gas have a range of speeds. This is de#ned as the speed which is possessed by ma"imum fraction of total number of molecules of the gas. e.!.$ if speeds of ', molecules of a gas are '$ *$ *$ +$ +$ +$ /$ 1$ 3$ 3 kmAs$ then the most probable speed is + kmAs$ as ma"imum fraction of total molecules possess this speed.
genius
PHYSICS by Pradeep Kshetrapal
1! Kinetic Theory of Gases 2ost probable speed v m3
*P
*+T M
*kT m
(+ A0erae speed It is the arithmetic mean of the speeds of molecules in a gas at given temperature.
v "v
v ' v * v + v / ..... N
and according to kinetic theory of gases 5P
%verage speed v "v
N$te &
5 +T M
5 kT m
v rms 4 v "v 4 v m3 (order remembering trick (B%2
v rms 5 v "v 5 v m3 =
+&
5
& * + & *.1 & *
Dor o"ygen gas molecules v rms 6 /3' ms$ v "v 6 /*/.4 ms and v rms 6 +43./ ms
Sample Problems based on Various speeds Problem 27.
%t room temperature$ the rms speed of the molecules of certain diatomic gas is found to be '+, mAs. The gas is [IIT>9&& 1)4* MP P&T 2+++* $#&#& 2++3"
(a ' *
(b F *
#$%&ti$n & (a Boot means s9uare velocity v rms
M
Problem 2).
++T *
('+,
(d C% *
(c -*
++T '+,mAs M
(given
+ 5.+' +,, * ', + k! * !m i.e. the gas is hydrogen. '+, '+,
Fet A and ) the two gases and given &
T A M A
/.
T ) M)
where T is the temperature and M
is the molecular mass. If C A and C) are the rms speed$ then the ratio to
Problem 2.
C)
will be e9ual
[$C/ 2++3"
(a * #$%&ti$n & (a %s v rms
C A
(b /
C A ++T C) M
(c '
T A A T ) M A A M )
/*
(d ,.1
T A M A %s / given M) T )
The rms speed of the molecules of a gas in a vessel is /,, ms-'. If half of the gas leaks out at constant temperature$ the rms speed of the remaining molecules will be [Kerala &n. 2++2"
(a 5,, ms-'
(b /,, * ms'
(c /,, ms-'
(d *,, ms-'
#$%&ti$n & (c Boot mean s9uare velocity does not depends upon the 9uantity of gas. Dor a given gas and at constant temperature it always remains same.
Problem 3+.
The root mean s9uare speed of hydrogen molecules at +,, , is '+, mAs. Then the root mean s9uare speed of o"ygen molecules at ,, , will be [MC #&T 2++2"
genius
Kinetic Theory of Gases 1, (a
++T M
#$%&ti$n & (b v rms
v -*
'+, /
Problem 31.
+
(b 5+3 mAs
'+, + m A s
(c 3/+ mAs
v ' *
v -*
T ' *
M '*
(d
M -*
T -*
'+, mA s +
'+, v -*
+,, +* * ,,
5+3mAs .
%t what temperature is the root mean s9uare velocity of gaseous hydrogen molecules is e9ual to that of o"ygen molecules at /4EC [#PMT 1)!* MP P&T 17* (P&T 1* AI&&& 2++2"
(a *, ,
(b 5, ,
++T -*
#$%&ti$n & (a Dor o"ygen v -*
M -*
Problem 32.
T -*
M -*
T ' * M '*
Dor hydrogen v ' *
and
++T -*
%ccording to problem
M -*
/4 *4+ +*
(c - 4+ ,
++
T ' * *
(d + ,
++
T ' * M '*
T ' * M '*
T ' *
+*, * *,, . +*
8ooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will [AI&&& 2++2"
(a Increase
(b =ecrease
(c Bemain same others
(d =ecrease for some$ while increase for
#$%&ti$n & (c If a lorry is moving with constant velocity then the v rms of gas molecule inside the container * will not change and we know that T v rms . So temperature remains same.
Problem 33.
The speeds of 1 molecules of a gas (in arbitrary units are as follows & *$ +$ /$ 1$ 3. The root mean s9uare speed for these molecules is [MP PMT 2+++"
(a *.' *
#$%&ti$n & (d v rms
v '
(b +.1*
v ** v +* v /* v 1* 1
*
*
(c /.,,
+* / * 1 * 3 * 1
',, 1
(d /.*/
*, /.*/
Problem 34. Gas at a pressure P, in contained as a vessel. If the masses of all the molecules are halved and their speeds are doubled$ the resulting pressure P will be e9ual to ['#&(T 1)4* M'( 1!* MP P&T 17* MP PMT 17* (P&T 1* /P%&AT 1: 2+++"
(a /P,
' mN * #$%&ti$n & (b P v rms + V
Problem 3!.
(c P,
(b *P,
P
* mv rms
(d *
P,
*
*
v m A * *v so * ' ' * P* *P' *P, P' m' v ' m' v ' P*
m*
Fet v $ v rms and v m3 respectively denote the mean speed$ root mean s9uare speed and most probable speed of the molecules in an ideal monoatomic gas at absolute temperature T . The mass of a molecule is m. Then
genius
PHYSICS by Pradeep Kshetrapal
17 Kinetic Theory of Gases [IIT>9&& 1)"
(a 7o molecule can have speed greater than
* v rms
(b 7o molecule can have speed less than v m3A
*
(c v m3 v v rms (d The average kinetic energy of a molecule is
v rms & v "v & v m3
v rms
and
++T $ v "v M
He know that v rms
#$%&ti$n & (c$ d
v m3
+&
*.1 &
5 +T and M
v m3
*
+T M
* so v m3 v "v v rms
+ + * * or v rms v m3 * *
+ * mv m3 /
%verage
kinetic
energy
' ' + * * mv rms m v m3 * * *
+ * mv m3 . /
Problem 3,.
The root mean s9uare speed of the molecules of a diatomic gas is v . Hhen the temperature is doubled$ the molecules dissociate into two atoms. The new root mean s9uare speed of the atom is [(oorDee 1," (a
#$%&ti$n & (c
Problem 37.
(b v
*v
(c *v
(d /v
++T . %ccording to problem T will becomes T A* and M will becomes MA* so the value M of v rms will increase by / * times i.e. new root mean s9uare velocity will be *v . v rms
The molecules of a given mass of a gas have a rms velocity of *,, mAsec at *4EC and 1 * 1 * '., ', N A m pressure. Hhen the temperature is '*4EC and pressure is ,.1 ', N A m $ the rms velocity in mAsec will be [AIIM% 1)!* MP P&T 12"
(a
',, * +
(b ',, *
(c
/,,
(d 7one of these
+
#$%&ti$n & (c 8hange in pressure will not a)ect the rms velocity of molecules. So we will calculate only the e)ect of temperature. %s
v rms
v /,,
Problem 3).
*,, * +
T
/,, +
v +,,$ v /,,$
+,, /,,
+ /
*,, v /,,
+ /
mAs .
Hhich of the following statement is true
[IIT>9&& 1)1"
(a %bsolute !ero degree temperature is not !ero energy temperature (b Two di)erent gases at the same temperature pressure have e9ual root mean s9uare velocities (c The rms speed of the molecules of di)erent ideal gases$ maintained at the same temperature are the same (d Given sample of 'cc of hydrogen and 'cc of o"ygen both at 7.T.@. o"ygen sample has a large number of molecules #$%&ti$n & (a %t absolute temperature kinetic energy of gas molecules becomes !ero but they possess potential energy so we can say that absolute !ero degree temperature is not !ero energy temperature.
genius
Kinetic Theory of Gases 1)
Problem 3.
The ratio of rms speeds of the gases in the mi"ture of nitrogen o"ygen will be (a ' & '
(b
v N* ++T v -* M
(c
+ &'
M -*
+* *5
(d
5& 4
3& 4
5 4
#$%&ti$n & (c
v rms
Problem 4+.
% vessel is partitioned in two e9ual halves by a #"ed diathermic separator. Two di)erent ideal gases are #lled in left (L and right (+ halves. The rms speed of the molecules in L part is e9ual to the mean speed of molecules in the + part. Then the ratio of the mass of a molecule in L part to that of a molecule in + part is (a
M N*
+ *
(b
A /
(c
*A +
+
L
(d + A 5
v rms
#$%&ti$n & (d Boot means s9uare velocity of molecule in left part
5 ,T m+
2ean or average speed of molecule in right part v "v
%ccording to problem
Problem 41.
+,T mL
5 ,T m+
+,T mL
+ 5 m+ mL
mL m+
+ . 5
%n ideal gas ( 6 '.1 is e"panded adiabatically. ow many times has the gas to be e"panded to reduce the root mean s9uare velocity of molecules * times (a / times
(b '3 times
(c 5 times
(d * times
#$%&ti$n & (b To reduce the rms velocity two times$ temperature should be reduced by four times (%s v rms
T
T ' T
T *
T $ V ' V /
Drom adiabatic law TV '
V * constant we get V '
'
T ' T *
/
V * V '
' '
(/
>
6
+A* given? '
V V (/ + A * ' V ' (/* '3V ' * '
V * '3 V '
11.) Kinetic &nery of Ideal Gas. 2olecules of ideal gases possess only translational motion. So they possess only translational kinetic energy.
Euantity of as Kinetic energy of a gas molecule (Em$%ec&%e
Kinetic enery
' +kT ' + * m kT mv rms * m * *
genius
PHYSICS by Pradeep Kshetrapal
1 Kinetic Theory of Gases +kT %sv rms m
Kinetic energy of ' m$%e (M !r"m gas (Em$%e
' ' ++T + * M v rms M +T * * * M
%sv rms
Kinetic energy of ' !m gas (E!r"m
+ + *M
T
++T M
+ k N A * mN A
T
+ k + T rT *m *
ere m 6 mass of each molecule$ M 6 2olecular weight of gas and N A 6 %vogadro number 6 3.,*+ ',*+
Important points (' Kinetic energy per molecule of gas does not depends upon the mass of the molecule but only depends upon the temperature of the gas. + kT or E T i.e. molecules of di)erent gases say 'e$ '* and -* etc. at same * temperature will have same translational kinetic energy though their rms speed are di)erent.
%s
v rms
E
+kT m
(* Kinetic energy per mole of gas depends only upon the temperature of gas. (+ Kinetic energy per !r"m of gas depend upon the temperature as well as molecular weight (or mass of one molecule of the gas. E !r"m
+ k T *m
E !r"m
T m
Drom the above e"pressions it is clear that higher the temperature of the gas$ more will be the average kinetic energy possessed by the gas molecules at T 6 ,$ E 6 , i.e. at absolute !ero the molecular motion stops.
Sample Problems based on Kinetic energy Problem 42.
Bead the given statements and decide which isAare correct on the basis of kinetic theory of gases [MP PMT 2++3" (I nergy of one molecule at absolute temperature is !ero (II rms speeds of di)erent gases are same at same temperature (IIIDor one gram of all ideal gas kinetic energy is same at same temperature (I<Dor one mole of all ideal gases mean kinetic energy is same at same temperature (a %ll are correct
(b I and I< are correct (c I< is correct
(d 7one of these
#$%&ti$n & (c If the gas is not ideal then its molecule will possess potential energy. ence statement (I is wrong. rms speed of di)erent gases at same temperature depends on its molecular weight
v rms
'
. ence statement (II also wrong.
M
genius
Kinetic Theory of Gases 2+
Kinetic energy of one !r"m gas depends on the molecular weight E !m
' . ence statement M
(III also wrong.
Put K.. of one mole of ideal gas does not depends on the molecular weight E
+ +T . *
ence (I< is correct.
Problem 43.
%t which of the following temperature would the molecules of a gas have twice the average kinetic energy they have at *,EC [MP P&T 12* $
(a /,EC
E T
#$%&ti$n & (c
(b 5,EC
E* E'
T * T '
*E' E'
(c +'+EC
T * (*, *4+
(d 153EC
T * *+ * 153, +'+C .
Problem 44.
% vessel contains a mi"ture of one mole of o"ygen and two moles of nitrogen at +,, , . The ratio of the average rotational kinetic energy per -* molecule to that per N* molecule is [IIT>9&& 1)* ;PMT 2+++" (a ' & ' (b ' & * (c * & ' (d =epends on the moments of inertia of the two molecules
#$%&ti$n & (a Kinetic energy per degree of freedom
' kT *
%s diatomic gas possess two degree of freedom for rotational motion therefore rotational
' kT kT *
K.. *
In the problem both gases (o"ygen and nitrogen are diatomic and have same temperature (+,, , therefore ratio of average rotational kinetic energy will be e9ual to one.
Problem 4!.
% gas mi"ture consists of molecules of type '$ * and + with molar masses m' m* m+ .
v rms and , are the rms speed and average kinetic energy of the gases. Hhich of the following is true [AM/ &n. 2+++" (a (v rms' (v rms* (v rms+ and (, ' (, * (, +
(v rms' (v rms* (v rms+ and (, '
(, *
(b
(, +
(c (v rms' (v rms* (v rms+ and (, ' (, * (, +
(d
(v rms' (v rms* (v rms+ and (, ' (, * (, + #$%&ti$n & (a The rms speed depends upon the molecular mass v rms
' M
but kinetic energy does not
depends on it E M , In the problem m' m* m+ (v rms' (v rms* (v rms+ but (, ' (, * (, +
Problem 4,.
The kinetic energy of one gram mole of a gas at normal temperature and pressure is (+ 6 5.+' (Am$%e0, [A?M# 1)* MC #&T 1* P@. PMT 2+++"
(a ,.13 ',/ ( #$%&ti$n & (d
E
(b '.+ ',* (
+ + +T 5.+' *4+ +./ ',+ ($&%e * *
(c *.4 ',* (
(d +./ ',+ (
genius
PHYSICS by Pradeep Kshetrapal
21 Kinetic Theory of Gases
Problem 47.
The average translational kinetic energy of -* (molar mass +* molecules at a particular temperature is ,.,/5 eV . The translational kinetic energy of N * (molar mass *5 molecules in eV at the same temperature is [IIT>9&& 17 (e>&Fam"
(a ,.,,'1
(b ,.,,+
(c ,.,/5
(d ,.435
#$%&ti$n & (c %verage translational kinetic energy does not depends upon the molar mass of the gas. =i)erent gases will possess same average translational kinetic energy at same temperature.
Problem 4).
The average translational energy and the rms speed of molecules in a sample of o"ygen gas at +,, , are 3.*' ',*' ( and /5/ mAs respectively. The corresponding values at 3,, , are nearly (assuming ideal gas behaviour [IIT>9&& 17 #ancelled"
(a '*./* ',*' ( $ 35m A s
(b
(c 3.*' ',*' ( $ 35m A s
(d
*'
5.45 ',
( $ 35/mA s
'*./* ',*' ( $ 35/m A s
#$%&ti$n & (d
E T but v rms T i.e. if temperature becomes twice then energy will becomes two time i.e. * 3.*' ',-*' 6
'*./* ',-*' ( Put rms speed will become
Problem 4.
* times
i.e.
/5/
*
35/mAs .
% bo" containing N molecules of a perfect gas at temperature T ' and pressure P' . The number of molecules in the bo" is doubled keeping the total kinetic energy of the gas same as before. If the new pressure is P* and temperature T * $ then [MP PMT 12"
(a P* P' $ T * T '
T *
(b P* P' $ T *
T ' *
(c P* *P' $ T * T '
(d
P* *P' $
T ' *
#$%&ti$n & (b Kinetic energy of N molecule of gas E Initially E'
+ NkT *
+ + N 'kT ' and #nally E * N *kT * * *
Put according to problem E' E* and N * *N' Since the kinetic energy constant
+ *
N'kT '
+ *
(*N'kT * T *
T ' *
+ + N 'kT ' N *kT * N 'T ' N *T * NT 6 constant * *
Drom ideal gas e9uation of N molecule PV NkT
Problem !+.
P'V ' P*V *
P' P*
>%s V ' V * and NT 6 constant?
Three closed vessels A$ ) and C are at the same temperature T and contain gases which obey the 2a"wellian distribution of velocities.
genius
Kinetic Theory of Gases 22 vessel A is V ' $ that of the N* molecules in vessel ) is V * $ the average speed of the -* molecules in vessel C is (where M is the mass of an o"ygen molecule (a (V ' V * A *
(c (V 'V *' A *
(b V '
#$%&ti$n & (b %verage speed of gas molecule v "v
[IIT>9&& 12"
(d
+kT A M
5kT . It depends on temperature and molecular m
mass. So the average speed of o"ygen will be same in vessel A and vessel C and that is e9ual to V ' .
Problem !1.
The graph which represent the variation of mean kinetic energy of molecules with temperature t EC is (a
(b E
(c E
(d E
#$%&ti$n & (c 2ean K.. of gas molecule E t
E
+ + kT k (t *4+ where T 6 temperature is in kelvin and t t t t * *
6 is in centigrade
E
+ + k t *4+k k 6 Polt!mannOs constant * *
Py comparing this e9uation with standard e9uation of straight li ne y mx c
+ + k and c *4+k . So the graph between E and t will be straight line with * * positive intercept on EMa"is and positive slope with t Ma"is. He get m
11. Gas as. (' $oyleBs la Dor a given mass of an ideal gas at constant temperature$ the volume of a gas is inversely proportional to its pressure. i.e. V
' P
or PV 6 constant m
(i PV 6 P constant
P
consta or
(ii PV 6 P
N n
P'
'
P* *
consta
or P'V '
>%s volume
P*V *
m ?
>%s m 6 constant?
>%s number of molecules per unit volume n
?
P n
consta or
>If m and T are constant?
P'
P*
n'
n*
>%s N 6 constant?
(iii %ccording to kinetic theory of gases P
' mN * v rms + V
N V V
N n
genius
PHYSICS by Pradeep Kshetrapal
23 Kinetic Theory of Gases
P
massof gas T V
>%s v rms T and mN 6 2ass of gas?
If mass and temperature of gas remain constant then P
' . This is in accordance with V
Poyle:s law. (iv Graphical representation & If m and T are constant
P
PV
PV
V
P
P
V
V
'AV
'AP
Sample Problems based on Boyle's law Problem !2.
%t constant temperature on increasing the pressure of a gas by 10 will decrease its volume by [MP P&T 2++2" (a 10
(b 1.*30
(c /.*30
(d /.430
#$%&ti$n & (d If P' P then P* P 10 of P 6 '.,1 P Drom Poyle:s law PV 6 constant
Dractional change in volume
V * V '
P' P*
P '.,1P
',, ',1
V V * V ' ',, ',1 1 V V ' ',1 ',1
@ercentage change in volume
1 V ',,0 ',,0 /.430 i.e. volume decrease by V ',1
/.430. 4 * % cylinder contained ', k! of gas at pressure ', N A m . The 9uantity of gas taken out
Problem !3.
of cylinder if #nal pressure is
3 *.1 ', N A m is
(assume the temperature of gas is constant
[&AM#&T Med. 1)"
(a Qero
(b 4.1 k!
(c *.1 k!
#$%&ti$n & (b %t constant temperature for the given volume of gas
',4 *.1 ',3
', m*
m*
*.1 ',3 ', ',4
P' P*
(d 1 k!
m' m*
*.1 k!
The 9uantity of gas taken out of the cylinder 6 ', - *.1 6 4.1 k!.
Problem !4.
If a given mass of gas occupies a volume of ', cc at ' atmospheric pressure and temperature of ',,EC (+4+.'1 , . Hhat will be its volume at / atmospheric pressure the temperature being the same ['#&(T 177" (a ',, cc
(b /,, cc
(c *.1 cc
(d ',/ cc
genius
Kinetic Theory of Gases 24
V * P ' ' ' V * ', *.1 cc V ' P* / V
#$%&ti$n & (c
P
Problem !!.
%n air bubble of volume V , is released by a #sh at a depth 6 in a lake. The bubble rises to the surface. %ssume constant temperature and standard atmospheric pressure P above the lake. The volume of the bubble Lust before touching the surface will be (density of water is
V , (b V , ( !6A P
(a V ,
(c
'
!6
(d V , '
!6
P
P
#$%&ti$n & (d %ccording to Poyle:s law multiplication of pressure and volume will remains constant at the bottom and top. If P is the atmospheric pressure at the top of the lake
P*V *
and the volume of bubble is V then from P'V ' P*V * 6
P 6 ! (P 6 !V , PV V V , P V V , '
Problem !,.
(P' V '
!6
P
The adLoining #gure shows graph of pressure and volume of a gas at two temperatures T ' and T * . Hhich of the following interferences is correct P
(a T ' T *
3
(b T ' T *
T * T '
(c T ' T *
V '
V *
V
(d 7o interference can be drawn #$%&ti$n & (c Dor a given pressure$ volume will be more if temperature is more (8harle:s law Drom the graph it is clear that V * R V '
T * R T '
(* #harle5s la (i If the pressure remains constant$ the volume of the given mass of a gas increases or ' of its volume at ,EC for each 'EC rise or fall *4+.'1 in temperature.
decreases by
V t V , '
centigrade scale.
' t . *4+.'1
This
is
8harle:s
law
V t V ,
for – *4+.'1
-
t (2C
(ii If the pressure remaining constant$ the volume of the given mass of a gas is directly proportional to its absolute temperature.
genius
PHYSICS by Pradeep Kshetrapal
2! Kinetic Theory of Gases V consta T
V T or
V '
or
V *
T '
T *
>If m and P are constant?
m V consta T T
(iii
>%s volume V or 'T ' *T *
T consta
or
>%s m 6 constant?
(iv %ccording to kinetic theory of gases P or P
m ?
' mN * v rms + V
2assof gas T V
If mass and pressure of the gas remains constant then V T . This is in accordance with 8harles law. (v Graphical representation & If m and P are constant
V
VT
7V
VT
V
T
T
Sample problems based on !arle's law Problem !7.
V or 'AV
T or 'AT 'AT
>%ll temperature T are in ke%vin?
% perfect gas at *4EC is heated at constant pressure to +*4EC. If original volume of gas at *4EC is V then volume at +*4EC is [#PMT 2++2"
(a V
(b +V
#$%&ti$n & (c Drom 8harle:s law V T
Problem !).
(c *V
V * V '
T * T '
(d V A*
+*4 *4+ 3,, * V * *V . *4 *4+ +,,
ydrogen gas is #lled in a balloon at *,EC. If temperature is made /,EC$ pressure remaining same$ what fraction of hydrogen will come out [MP PMT 2++2"
(a ,.,4 #$%&ti$n & (a %s V T
(b ,.*1
V * V '
T * T '
(d ,.41
+'+ V ' *+
V *
Draction of gas comes out
Problem !.
(c ,.1
V * V ' V '
+'+ V ' V ' *+ V '
*, . ,.,4 *+
The e"pansion of unit mass of a perfect gas at constant pressure is shown in the diagram. ere "
(a " 6 volume$ / 6 EC temperature -
/
genius
Kinetic Theory of Gases 2, (b " 6 volume$ / 6 , temperature (c " 6 EC temperature$ / 6 volume (d " 6 , temperature$ / 6 volume
#$%&ti$n & (c In the given graph line have a positive slop with X Ma"is and negative intercept on Y Ma"is. So we can write the e9uation of line y 6 mx - c %ccording to 8harle:s law V t
...... (i
V ,
t V , $ by rewriting this e9uation we get *4+
*4+ V t *4+ V ,
t
......(ii
Py comparing (i and (ii we can say that time is represented on Y Ma"is and volume in X Ma"is.
Problem ,+.
% gas is #lled in the cylinder shown in the #gure. The two pistons are Loined by a string. If the gas is heated$ the pistons will (a 2ove towards left
Gas
(b 2ove towards right (c Bemain stationary (d 7one of these #$%&ti$n & (b Hhen temperature of gas increases it e"pands. %s the crossMsectional area of right piston is more$ therefore greater force will work on it (because F 6 PA. So piston will move towards right. %n ideal gas is initially at a temperature T and volume V . Its volume is increased by V
Problem ,1.
due to an increase in temperature T $ pressure remaining constant. The 9uantity
V V T
varies with temperature as (a
(b
(c
#$%&ti$n & (c Drom ideal gas e9uation T or
T T (Temp. ,
PV
PV T 6 +T
T T +T (Temp. ,
=ividing e9uation (ii by (i we get
(d
C..(i T C..(ii
T T (Temp. ,
V ' V T V T T V T
T
T T (Temp. ,
(given
' . So the graph between and T will be rectangular hyperbola. T
(+ Gay>ussacBs la or pressure la (i The volume remaining constant$ the pressure of a given mass of a gas increases or ' of its pressure at ,EC for each 'EC rise *4+.'1 or fall in temperature.
decreases by
Pt P, (2C
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PHYSICS by Pradeep Kshetrapal
27 Kinetic Theory of Gases ' Pt P, ' t *4+.'1
This is pressure law for centigrade scale. (ii The volume remaining constant$ the pressure of a given mass of a gas is directly proportional to its absolute temperature. P T
P' P* P consta or T ' T * T
or
(iii %ccording to kinetic theory of gases P P
or
>If m and V are constant?
' mN * v rms + V
* >%s v rms T ?
massof gas T V
If mass and volume of gas remains constant then P T . This is in accordance with Gay Fussac:s law. (/ Graphical representation If m and V are constants
P
PT
T or 'AT
T
7P
PT
P or 'AP
P
T
7T
>%ll temperature T are in Sample problems based on "ay #ussac's law ke%vin?
Problem ,2.
n ,EC pressure measured by barometer is 43, mm. Hhat will be pressure on ',,EC [A?M# 2++2"
(a 43, mm #$%&ti$n & (d Drom Gay Fussac:s law
Problem ,3.
(b 4+, mm
P* P'
(c 45, mm
(d 7one of these
',, *4+ +4+ +4+ 43, ',+5mm. P* T ' , *4+ *4+ *4+
T *
If pressure of a gas contained in a closed vessel is increased by ,./0 when heated by 'EC$ the initial temperature must be ['#&(T 1)2* &AM#&T &n. 1!* (PMT 1,* MP P&T 1"
(a *1, , #$%&ti$n & (a
(b *1,EC
(c *1,, ,
P' P $ T ' 6 T $ P* P (,./0 of P P
Drom Gay FussacOs law
P' P*
T ' T *
,./
P T * T ' PP ',, *1,
P
P
(d *1EC
P *1,
T T '
>%s V 6 constant for closed
vessel? Py solving we get T 6 *1, , .
Problem ,4.
@ressure versus temperature graph of an ideal gas of e9ual number of moles of di)erent volumes are plotted as shown in #gure. 8hoose the correct alternative (a V ' V *$ V + V / and V * V +
/
P
*
+ '
genius
Kinetic Theory of Gases 2) (b V ' V *$ V + V / and V * V + (c V ' V * V + V / (d V / V + V * V '
#$%&ti$n & (a Drom ideal gas e9uation PV +T P
+
V
T
P
V 6 constant
8omparing this e9uation with y mx Slope of line tan m
+
i.e. V
V
' tan
T
It means line of smaller slope represent greater volume of gas. Dor the given problem #gure @oint ' and * are on the same line so they will represent same volume i.e. V ' V * Similarly point + and / are on the same line so they will represent same volume i.e. V + V / Put V ' V + (6 V / or V * V + (6 V /
as slope of line 'M* is less than +M/.
(1 A0oadroBs la 9ual volume of all the gases under similar conditions of temperature and pressure contain e9ual number of molecules. %ccording to kinetic theory of gases PV Dor #rst gas$ PV
' * mN v rms +
' * m' N' v rms (' +
C..(i Dor second gas$ PV
' * m* N * v rms (* +
C..(ii Drom (i and (ii
* * m' N' v rms ' m* N * v rms*
C..(iii %s
the
two
* * m' v rms ' m* v rms*
gases
are
at
the
same
temperature
' ' + * * m' v rms m* v rms kT ' * * * *
C..(iv
So from e9uation (iii we can say that N' N * . This is %vogadro:s law. (i A0oadroBs num@er $ % The number of molecules present in ' !m m$%e of a gas is de#ned as %vogadro number. N A 3.,*+ ',*+ per !m m$%e 3.,*+ ',*3 per k! m$%e.
(ii %t S.T.@. or 7.T.@. (T 6 *4+ , and P 6 ' "tm **./ %itre of each gas has 3.,*+ ',*+ molecule. (iii ne mole of any gas at S.T.@. occupy **./ %itre of volume Ex"m3%e & +* !m o"ygen$ *5 !m nitrogen and *!m hydrogen occupy the same volume at S.T.@.
genius
PHYSICS by Pradeep Kshetrapal
2 Kinetic Theory of Gases (iv Dor any gas ' m$%e 6 M !r"m 6 **./ %itre 6 3.,*+ ',*+ molecule.
Sample problems based on %&ogadro's #aw Temperature of an ideal gas is T , and average kinetic energy is E *.,4 ',*+ T ($&%eAm$%ec&%e. 7umber of molecules in ' %itre gas at S.T.@. will be
Problem ,!.
[#PMT 14"
(a *.35 ',**
(b *.35 ',*1
(c *.35 ',*5
(d '.35 ',**
#$%&ti$n & (a %s we know that at S.T.@. **./ %itre of gas contains 3.,*+ ',*+ molecules
Problem ,,.
' %itre of gas contain
The average kinetic energy per molecule of helium gas at temperature T is E and the molar gas constant is +$ then %vogadro:s number is
+T *E
(a
(b
++T E
(c
#$%&ti$n & (d %verage kinetic energy per unit molecule E Put %vagadro number N A
Problem ,7.
3.,*+ ',*+ *.35 ',** molecules. **./
E *+T
(d
++T *E
+ *E kT k * +T
+ + ++T . N A k (*E A +T *E
ne mole of a gas #lled in a container at 7.T.@.$ the number of molecules in ' cm+ of volume will be (a 3.,* ',*+ A **/, (b 3.,* ',*+
(d 3.,* ',*+ A 43
(c 'A**/,,
#$%&ti$n & (a 7umber of molecule in **./ %itre gas at 7.T.@. 3.,*+ ',*+ or number of molecule in **./ ',+ cm+ 3.,*+ ',*+
+
+
**./ ', cm
>%s
**./
%itre
?
7umber of molecules in 'cm+
3.,*+ ',*+ . **/,,
(3 GrahmBs la of diHusion Hhen two gases at the same pressure and temperature are allowed to di)use into each other$ the rate of di)usion of each gas is inversely proportional to the s9uare root of the density of the gas. He know v rms
+P
or v rms
'
and rate of di)usion of a gas is proportional to its rms velocity i.e.$ r v rms
r
r ' or r *
'
* '
(4 ;altonBs la of partial pressure The total pressure e"erted by a mi"ture of nonM reacting gases occupying a vessel is e9ual to the sum of the individual pressures which each gases e"ert if it alone occupied the same volume at a given temperature.
genius
Kinetic Theory of Gases 3+
Dor n gases P P' P * P+ .....Pn P'$ P* $ P+ $......Pn
where P 6 @ressure e"erted by mi"ture and
@artial pressure of
component gases.
Sample problems based on alton's law Problem ,).
The capacity of a vessel is + %itres. It contains 3 !m o"ygen$ 5 !m nitrogen and 1 !m C-* mi"ture at *4EC. If + 6 5.+' (Am$%e ke%vin$ then the pressure in the vessel in N A m* will be (appro".
#$%&ti$n & (a =alton:s law P P' P* P+
Problem ,.
(c ',3
(b 1 ',/
(a 1 ',1
'+T *+T + +T V
V
+T +T m' m* m+ > ' * + ? V V M' M * M +
+T +T 1 * /5 ', 1,, ', 1 ', NAm . + + ', +* *5 // +
Two gases occupy two containers A and ) the gas in A$ of volume ,.',m $ e"erts a pressure of './, MP" and that in ) of volume ,.'1m+ e"erts a pressure ,.4 MP". The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then it the temperature remains constant$ the #nal pressure in the container will be (in MP" (b ,.5
#$%&ti$n & (b %s the 9uantity of gas remains constant
P A V A P) V ) P(V A V ) +T +T +T
V
5.+' +,, 3 5 1
(a ,.4,
P
(d ',1
(c './, A
)
P
(d *.',
P A V A P) V ) V A V )
'./ ,.' ,.4 ,.'1 ,.' ,.'1
,.5 MP" .
Problem 7+.
The temperature$ pressure and volume of two gases X and Y are T $ P and V respectively. Hhen the gases are mi"ed then the volume and temperature of mi"ture become V and T respectively. The pressure and mass of the mi"ture will be (a *P and *M
(b P and M
(c P and *M
(d *P and M
#$%&ti$n & (a Drom =alton:s law$ @ressure of mi"ture P' P* P P *P Similarly mass also will become double i.e. *M.
Problem 71.
% closed vessel contains 5! of o"ygen and 4! of nitrogen. The total pressure is ', "tm at a given temperature. If now o"ygen is absorbed by introducing a suitable absorbent the pressure of the remaining gas in "tm will be (a *
(b ',
(c /
(d 1
#$%&ti$n & (d Drom =alton:s law #nal pressure of the mi"ture of nitrogen and o"ygen
P P' P*
'+T
V
* +T
V
m' +T m* +T 5 +T 4 +T +T M ' V M * V +* V *5 V *V
',
+T *V
P
+T /V
C..(i Hhen o"ygen is absorbed then for nitrogen let pressure is P
4 +T *5 V
C..(ii Drom e9uation (i and (ii we get pressure of the nitrogen P 1 "tm.
genius
PHYSICS by Pradeep Kshetrapal
31 Kinetic Theory of Gases
(5 Ideal as e-uation Drom kinetic theory of gases P
P
(massof gasT V
If mass of gas is constant then PV T
' mN * v rms + V
* >%s v rms T ?
or
PV 6 +T . This is ideal gas e9uation.
11.1+ ;eree of ?reedom. The term degree of freedom of a system refers to the possible independent motions$ systems can have. or The total number of independent modes (ways in which a system can possess energy is called the degree of freedom (8 . The independent motions can be translational$ rotational or vibrational or any combination of these. So the degree of freedom are of three types & (i Translational degree of freedom (ii Botational degree of freedom (iii
y v y v
v x
v z
x
z
(* ;iatomic as 2olecules of diatomic gas are made up of two atoms Loined rigidly to one another through a bond. This cannot only move bodily$ but also rotate about one of the three coMordinate a"es. owever its moment of inertia about the a"is Loining the two atoms is negligible compared to that about the other two a"es. ence it can have only two rotational motion. Thus a diatomic molecule has 1 degree of freedom & + translational and * rotational.
y
x z
(+ Triatomic as 'on>linear % nonMlinear molecule can rotate about any of three coMordinate a"es. ence it has 3 degrees of freedom & + translational and + rotational.
y
x z
(/ Ta@ular display of deree of freedom of diHerent ases
Atomicity of as
&Fample
%
B
( 3 % J B
?iure
genius
Kinetic Theory of Gases 32
2onoatomic
'e$ Ne$ Ar
'
,
8 6 +
=iatomic
'*$ -*
*
'
8 6 1
Triatomic linear
non
'*-
A )
A
+
+
A
8 6 3
C-*$ )eC%*
+
*
8 6 4
)
A
A A
N$te &
)
) A
Triatomic linear
A
)
)
A
The
above degrees of freedom are shown at room temperature. Durther at high temperature$ in case of diatomic or polyatomic molecules$ the atoms with in the molecule may also vibrate with respect to each other. In such cases$ the molecule will have an additional degrees of freedom$ due to vibrational motion. %n obLect which vibrates in one dimension has two additional degree of freedom. ne for the potential energy and one for the kinetic energy of vibration. % diatomic molecule that is free to vibrate (in addition to translation and rotation will have 4 (* + * degrees of freedom. %n atom in a solid though has no degree of freedom for translational and rotational motion$ due to vibration along + a"es has + * 6 3 degrees of freedom (and not like an ideal gas molecule. Hhen a diatomic or polyatomic gas dissociates into atoms it behaves as monoatomic gas whose degree of freedom are changed accordingly.
11.11 a of &-uipartition of &nery. Dor any system in thermal e9uilibrium$ the total energy is e9ually distributed among its various degree of freedom. %nd the energy associated with each molecule of the system per ' kT . degree of freedom of the system is * where k '.+5 ',*+ ( A , $ T 6 absolute temperature of the system. If the system possess degree of freedom 8 then Total energy associated with each molecule Total energy associated with N molecules Total energy associated with each mole Total energy associated with mole Total energy associated with each gram Total energy associated with M, gram
8
kT
*
N
8 kT *
8
*
+T
8
*
+T
8 rT *
M,
8 rT *
Sample problems based on #aw o( equipartition o( energy
genius
PHYSICS by Pradeep Kshetrapal
33 Kinetic Theory of Gases
Problem 72.
nergy of all molecules of a monoatomic gas having a volume V and pressure P is
+ PV . *
The total translational kinetic energy of all molecules of a diatomic gas as the same volume and pressure is [/P%&AT 2++2" (a
' PV *
(b
#$%&ti$n & (b nergy of ' mole of gas
+ PV *
(c
8 8 +T PV * *
1 PV *
(d + PV
where 8 6 =egree of freedom
2onoatomic or diatomic both gases posses e9ual degree of freedom for translational motion
E
and that is e9ual to + i.e. 8 6 +
+ PV *
Dor monoatomic gas E total
%lthough total energy will be di)erent$
Dor diatomic gas E total
Problem 73.
+ PV >%s 8 6 +? *
1 PV *
>%s 8 6 1?
The temperature of argon$ kept in a vessel is raised by 'EC at a constant volume. The total heat supplied to the gas is a combination of translational and rotational energies. Their respective shares are [$C/ 2+++" (a 3,0 and /,0
(b /,0 and 3,0
(c 1,0 and 1,0
(d ',,0 and ,0
#$%&ti$n & (d %s argon is a monoatomic gas therefore its molecule will possess only translatory kinetic energy i.e. the share of translational and rotational energies will be ',,0 and ,0 respectively.
C-*(- C - is a triatomic gas. 2ean kinetic energy of one gram gas will be (If NM
Problem 74.
%vogadroOs number$ k MPolt!mannOs constant and molecular weight of C-* // (a + A 55NkT
(b 1 A 55NkT
#$%&ti$n & (d 2ean kinetic energy for mole gas .
E
(c 3 A 55NkT 8
*
(d 4 A 55NkT
+T
' 4 m 4 4 4 NkT NkT NkT +T // * M * 55 *
>%s 8 6 4 and M 6 // for
C-* ?
Problem 7!.
%t standard temperature and pressure the density of a gas is '.+ !mA m+ and the speed of the sound in gas is ++, mAsec. Then the degree of freedom of the gas will be (a +
(b /
(c 1
m $ sec
#$%&ti$n & (c Given velocity of sound v s ++,
P '.,' ',1
N m*
(d 3
=ensity of gas '.+
k! m+
$ %tomic pressure
genius
Kinetic Theory of Gases 34 Substituting these value in v sound 7ow from '
P
we get
'./'
* * * 1. we get 8 ' './ ' 8
11.12 Mean ?ree Path. The molecules of a gas move with high speeds at a given temperature but even then a molecule of the gas takes a very long time to go from one point to another point in the container of the gas. This is due to the fact that a gas molecule su)ers a number of collisions with other gas molecules surrounding it. %s a result of these collisions$ the path followed by a gas molecule in the container of the gas is !igM!ag as shown in the #gure. =uring two successive collisions$ a molecule of a gas moves in a straight line with constant velocity and the distance travelled by a gas molecule between two successive collisions is known as free path. The distance travelled by a gas molecule between two successive collisions is not constant and hence the average distance travelled by a molecule during all collisions is to be calculated. This average distance travelled by a gas molecule is known as mean free path. Fet
! '$ ! * $ ! + $...! n
be the distance travelled by a gas molecule during n collisions
respectively$ then the mean free path of a gas molecule is given by !
(' !
' * n9*
! ' ! * ! + .... ! n n
where 9 6 =iameter of the molecule$ n 6 7umber of molecules per unit
volume (* %s
PV 6 +T 6 NkT
So
!
(+ Drom !
'
N P n 7umber of molecule per unit volume V kT
kT
* * 9 P
' *
* n9
m
* (mn9
*
m
* 9 *
>%s mn 6 2ass per unit volume 6 =ensity 6
?
(/ If average speed of molecule is v then ! v
t v T N
two collisions?
Important points
>%s N 6 7umber of collision in time t $ T 6 time interval between
genius
PHYSICS by Pradeep Kshetrapal
3! Kinetic Theory of Gases (i %s !
m *
* 9
!
'
i.e. the mean free path is inversely proportional to the density of
a gas. (ii %s ! molecules$
'
kT
* * 9 P
. Dor constant volume and hence constant number density n of gas
P is constant so that ! will not depend on P and T . Put if volume of given mass of a T
gas is allowed to change with P or T then ! T at constant pressure and !
' at constant P
temperature.
Sample Problems based on )ean (ree pat! Problem 7,.
If the mean free path of atoms is doubled then the pressure of gas will become [(PMT 2+++"
(b P A *
(a P A / #$%&ti$n & (b %s !
Problem 77.
'
kT
P
* 9 * P
' !
(c P A 5
(d P
i.e. by increasing ! two times pressure will become half.
The mean free path of nitrogen molecules at a pressure of '., "tm and temperature ,EC is ,.5 ',4 m . If the number of density of molecules is *.4 ', *1 3erm+ $ then the molecular diameter is (a
+.*nm
(c +.* m
(b +.*:
#$%&ti$n & (b 2ean free path ! ,.5 ', -4 m
(d *.+mm
number of molecules per unit volume n *.4 ',*1 per
m+ Substituting these value in !
' * n9*
we get 9
'.,/ ','
+.* ',', m
+.* :
11.13 %peci6c heat or %peci6c Ceat #apacity. It characterises the nature of the substance in response to the heat supplied to the substance. Speci#c heat can be de#ned by two following ways & Gram speci#c heat and 2olar speci#c heat. (' Gram speci6c heat Gram speci#c heat of a substance may be de#ned as the amount of heat re9uired to raise the temperature of unit mass of the substance by unit degree. Gram speci#c heat c Jnits &
; mT
c"% c"% ($&%e $ $ !m C !m ke%vi k! ke%vi
=imension & >L*T * ' ?
genius
Kinetic Theory of Gases 3,
(* Molar speci6c heat 2olar speci#c heat of a substance may be de#ned as the amount of heat re9uired to raise the temperature of one gram mole of the substance by a unit degree$ it is represented by capital (C ; T
C
Jnits &
c"%$rie m$%e C
$
c"%$rie ($&%e or m$%e ke%vi m$%e ke%vi
Important points (' C Mc
M ; m T
'
;
T
m %s M
i.e. molar speci#c heat of the substance is M times the gram speci#c heat$ where M is the molecular weight of that substance. (* Speci#c heat for hydrogen is ma"imum c +.1
c"% !m C
(+ In li9uids$ water has ma"imum speci#c heat c '
.
c"% !m C
.
(/ Speci#c heat of a substance also depends on the state of substance i.e. solid$ li9uid or gas. Ex"m3%e & cice ,.1
c"% c"% c"% $ cwater ' $ csteam ,./4 !m C !m C !m C
(1 Speci#c heat also depends on the conditions of the e"periment i.e. the way in which heat is supplied to the body. In general$ e"periments are made either at constant volume or at constant pressure. In case of solids and li9uids$ due to small thermal e"pansion$ the di)erence in measured values of speci#c heats is very small and is usually neglected. owever$ in case of gases$ speci#c heat at constant volume is 9uite di)erent from that at constant pressure.
11.14 %peci6c Ceat of Gases . In case of gases$ heat energy supplied to a gas is spent not only in raising the temperature of the gas but also in e"pansion of gas against atmospheric pressure. ence speci#c heat of a gas$ which is the amount of heat energy re9uired to raise the temperature of one gram of gas through a unit degree shall not have a single or uni9ue value. (i If the gas is compressed suddenly and no heat is supplied from outside i.e. ; 6 ,$ but the temperature of the gas raises on the account of compression.
C
; , m(T
i.e. C 6 ,
(ii If the gas is heated and allowed to e"pand at such a rate that rise in temperature due to heat supplied is e"actly e9ual to fall in temperature due to e"pansion of the gas. i.e. T 6 ,
C
; ; " m(T ,
i.e. C 6 "
genius
PHYSICS by Pradeep Kshetrapal
37 Kinetic Theory of Gases (iii If rate of e"pansion of the gas were slow$ the fall in temperature of the gas due to e"pansion would be smaller than the rise in temperature of the gas due to heat supplied. Therefore$ there will be some net rise in temperature of the gas i.e. T will be positive.
C
; m(T
positive
i.e. C 6 positive
(iv If the gas were to e"pand very fast$ fall of temperature of gas due to e"pansion would be greater than rise in temperature due to heat supplied. Therefore$ there will be some net fall in temperature of the gas i.e. T will be negative. C
; m( T
negative
i.e. C 6 negative
ence the speci#c heat of gas can have any positive value ranging from !ero to in#nity. Durther it can even be negative. The e"act value depends upon the mode of heating the gas. ut of many values of speci#c heat of a gas$ two are of special signi#cance. (' %peci6c heat of a as at constant 0olume c& The speci#c heat of a gas at constant volume is de#ned as the 9uantity of heat re9uired to raise the temperature of unit mass of gas through ' , when its volume is kept constant$ i.e.$ cv
(;v mT
If instead of unit mass$ ' mole of gas is considered$ the speci#c heat is called molar speci#c heat at constant volume and is represented by capital Cv . C v Mcv
M (;v mT
' (;v T
m %s M
(* %peci6c heat of a as at constant pressure c p The speci#c heat of a gas at constant pressure is de#ned as the 9uantity of heat re9uired to raise the temperature of unit mass of gas through ' , when its pressure is kept constant$ i.e.$ cP
(;3 mT
If instead of unit mass$ ' mole of gas is considered$ the speci#c heat is called molar speci#c heat at constant pressure and is represented by C 3. C 3 MC 3
M(; 3 mT
' (;3 T
m %s M
11.1! Mayer5s ?ormula. ut of two principle speci#c heats of a gas$ C 3 is more than Cv because in case of Cv $ volume of gas is kept constant and heat is re9uired only for raising the temperature of one gram mole of the gas through 'EC or ' , . 7o heat$ what so ever$ is spent in e"pansion of the gas. It means that heat supplied to the gas increases its internal energy only i.e.
(;v < Cv T while in case of C 3 the heat is used in two ways
C..(i
genius
Kinetic Theory of Gases 3)
(i In increasing the temperature of the gas by T (ii In doing work$ due to e"pansion at constant pressure ( So
(; 3 < = C 3 T
Drom e9uation (i and (ii
C..(ii
C 3 T Cv T =
T (C 3 Cv PV
C 3 Cv
PV T
>Dor constant P$ 6 PV ? >Drom PV 6 +T $ %t constant pressure PV 6
+T ? C 3 C v +
This relation is called 2ayer:s formula and shows that C 3 C v i.e. molar speci#c heat at constant pressure is greater than that at constant volume.
11.1, %peci6c Ceat in Terms of ;eree of ?reedom . He know that kinetic energy of one mole of the gas$ having 8 degrees of freedom can be given by E
8 +T *
C..(i
where T is the temperature of the gas but from the de#nition of Cv $ if 9E is a small amount of heat energy re9uired to raise the temperature of ' !m mole of the gas at constant volume$ through a temperature 9T then
9E Cv 9T Cv 9T or Cv
9E 9T
>%s 6 '?
@utting the value of E from e9uation (i we get Cv
Cv
C..(ii
9 8 8 +T + 9T * *
8 + *
8 8 Drom the 2ayer:s formula C 3 C v + C 3 Cv + + + ' + * *
8 ' + *
C 3
Batio of C 3 and Cv &
'
* 8
Important points
8 ' + C 3 * * ' 8 Cv 8 + *
genius
PHYSICS by Pradeep Kshetrapal
3 Kinetic Theory of Gases (i
* 8
' 8 * ' * ' 8
+
Cv * + ' ' 8 + ' + and C 3 ' + * ' '
%peci6c heat and Dinetic enery for diHerent ases
Monoatom ic
;iatomic
Triatomic non> linear
Triatomic linear
%tomicity
A
'
*
+
+
Bestriction
)
,
'
+
*
+
1
3
4
+ + *
1 + *
++
4 + *
1 + *
4 + *
/+
+ *
1T '.33 +
4T './ 1
/T '.++ +
T '.*5 4
+ +T *
1 +T *
++T
4 +T *
+ kT *
1 kT *
+kT
4 kT *
+ rT *
1 rT *
+rT
4 rT *
=egree freedom
of
8 6 + A - )
2olar speci#c heat at constant volume 2olar speci#c heat at constant pressure
Cv
8
*
+
+
'
8 + ' + ' *
C 3
Batio of C 3 and Cv
Kinetic energy of ' mole Kinetic energy of ' molecule Kinetic energy of ' !m
C 3 Cv
'
E mole
8 *
E molecule
E gram
* 8
+T 8 *
kT
8 rT *
Sample Problems based on Speci*c !eat Problem 7).
Dind the ratio of speci#c heat at constant pressure to the speci#c heat constant volume for N'+ [(PMT 2++3" (a '.++
(b './/
(c '.*5
(d '.34
#$%&ti$n & (c Dor polyatomic gas ratio of speci#c heat U '.++ Pecause we know that as the atomicity of gas increases its value of decreases.
Problem 7.
Dor a gas 2++1: 2++2"
+ Cv
,.34. This gas is made up of molecules which are
[#$%& PMT 12* 9IPM&(
genius
Kinetic Theory of Gases 4+ (a =iatomic molecules (c 2onoatomic
(b 2i"ture
of
diatomic
and
polyatomic
(d @olyatomic
#$%&ti$n & (c Py comparing with relation Cv
+
'
'
we get
,.34
or 6 '.34 i.e. the gas is
monoatomic.
Problem )+.
/, c"%$ries of heat is needed to raise the temperature of ' m$%e of an ideal monoatomic gas from *,EC to +,EC at a constant pressure. The amount of heat re9uired to raise its temperature over the same interval at a constant volume (+ *c"%$riem$%e', ' is [/P%&AT 2+++"
(a *, c"%$rie
(b /, c"%$rie
(c 3, c"%$rie
(d 5, c"%$rie
#$%&ti$n & (a %t constant pressure (; 3 C 3T ' C 3 (+, *, /, C 3 /
C v C 3 + / * *
c"%$rie m$%eke%vin
c"%$rie m$%e ke%vin
7ow (;v C v T ' * (+, *, *,c"%$ri
Problem )1.
%t constant volume the speci#c heat of a gas is
++ $ then the value of will be *
[;PMT 1"
(a
+
(b
*
1
(c
*
#$%&ti$n & (c Speci#c heat at constant volume Cv
Problem )2.
'
* +
+ '
1
(d 7one of the above
+
++ (given *
1 . +
Dor a gas the di)erence between the two speci#c heats is /'1, (Ak! , . Hhat is the speci#c heats at constant volume of gas if the ratio of speci#c heat is './ [A?M# 1)"
(a 5/41 (Ak! 0 ,
(b 1'53 (Ak! 0 ,
#$%&ti$n & (d Given c 3 cv /'1
C..(i
(c '33, (Ak! 0 , and
c 3 cv
(d ',+41 (Ak! 0 ,
'./ c 3 './cv
C..(ii
Py substituting the value of c 3 in e9uation (i we get './cv cv /'1, ,./cv /'1
Problem )3.
cv
/'1, ,./
',+41 ( Ak!M , .
Two cylinders A and ) #tted with pistons contain e9ual amounts of an ideal diatomic gas at +,,, . The piston of A is free to move while that of ) is held #"ed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is +, , $ then the rise in temperature of the gas in ) is [IIT>9&& 1)" (a +, ,
(b '5 ,
(c 1, ,
(d /* ,
#$%&ti$n & (d In both cylinders A and ) the gases are diatomic ( 6 './. @iston A is free to move i.e. it is isobaric process. @iston ) is #"ed i.e. it is isochoric process. If same amount of heat ; is given to both then
(;isobaric (;isochori C 3 (T A Cv (T )
(T )
C 3 Cv
(T A (T A './ +, /*, .
genius
PHYSICS by Pradeep Kshetrapal
41 Kinetic Theory of Gases
Problem )4.
The speci#c heat of a gas [MP P&T 1,"
(a as only two values of C 3 and Cv
(b as a uni9ue value at a given temperature
(c 8an have any value between , and "
(d =epends upon the mass of the gas
#$%&ti$n & (c Bange of speci#c heat varies from positive to negative and from !ero to in#nite. It depends upon the nature of process.
Problem )!.
The speci#c heat at constant volume for the monoatomic argon is ,.,41 kc"%Ak!M, whereas its gram molecular speci#c heat Cv *.5 c"%Am$%eA, . The mass of the argon atom is (%vogadro:s number 3.,* ',*+m$%ec&%esAm$%e [MP P&T 13"
(a 3.3, ',*+ !m
(b +.+, ',*+ !m
(c *.*, ',*+ !m (d '+.*, ',*+ !m
#$%&ti$n & (a 2olar speci#c heat 6 2olecular weight Gram speci#c heat
C v M cv
*.5
kc"% ,.,41 ',+ c"%$rie M ,.,41 M m$%e ke%vin k!M ke%vin !m ke%vi ',+ c"%$rie
molecular weight of argon M i.e.
+.4 *+
3.,*+ ',
Problem ),.
mass
of
*.5 +.4 !m ,.,41
3.,*+ ',*+ atom
6
+.4
!m
mass
of
single
atom
3.3, ', *+ !m.
Hhen an ideal diatomic gas is heated at constant pressure$ the fraction of the heat energy supplied which increases the internal energy of the gas is [IIT>9&& 1+"
(a *A1
(b +A1
(c +A4
(d 1A4
#$%&ti$n & (d Hhen a gas is heated at constant pressure then its one part goes to increase the internal energy and another part for work done against e"ternal pressure i.e. (; 3 < = C 3T Cv T PV
So fraction of energy that goes to increase the internal energy
C ' 1 < v >%s 4 (; 3 C 3 4 1
for diatomic gas?
Problem )7.
The temperature of 1 mole of a gas which was held at constant volume was changed from ',,oC to '*,oC. The change in internal energy was found to be 5, (. The total heat capacity of the gas at constant volume will be e9ual to [#PMT 1))"
(a
5 (
,
'
(b
,.5 (
,
'
(c
/ (
,
'
(d
,./ (
,
'
#$%&ti$n & (c %t constant volume total energy will be utilised in increasing the temperature of gas i.e. (;v C v T C v ('*, ',, 5,
Problem )).
Cv
5, / ($&%eAke%vin. This is the heat capacity of 1 m$%e gas. *,
% gas$ is heated at constant pressure. The fraction of heat supplied used for e"ternal work is
genius
Kinetic Theory of Gases 42 (a
'
(b '
'
(c
(d '
'
#$%&ti$n & (b He know fraction of given energy that goes to increase the internal energy
'
*
'
So we can say the fraction of given energy that supplied for e"ternal work '
Problem ).
'
.
% monoatomic gas e"pands at constant pressure on heating. The percentage of heat supplied that increases the internal energy of the gas and that is involved in the e"pansion is (a 410$ *10
#$%&ti$n & (c Draction
of
(b *10$ 410 energy
supplied
(c 3,0$ /,0 for
increment
in
(d /,0$ 3,0 internal
energy
'
+ 1
1 gas %s + for monoatomic
@ercentage energy
+, 1
3,0
1 ' ' ' + * Draction of energy supplied for e"ternal work done ' 1 1 + @ercentage energy
Problem +.
* 1
',,0 /,0.
The average degrees of freedom per molecule for a gas is 3. The gas performs *1 ( of work when it e"pands at constant pressure. The heat absorbed by gas is (a 41 (
(b ',, (
#$%&ti$n & (b %s 8 6 3 (given '
(c '1, (
(d '*1 (
* * / ' 8 3 +
Draction of energy given for e"ternal work
= ;
' '
*1 ' ' / A + ;
'
+ ' / /
; *1 / ',, ($&%e
Problem 1.
8ertain amount of an ideal gas are contained in a closed vessel. The vessel is moving with a constant velocity v . The molecular mass of gas is M. The rise in temperature of the gas when the vessel is suddenly stopped is ( C P A C V Mv * (a *+( '
(b
Mv *( '
*+
Mv * (c *+( '
#$%&ti$n & (b If m is the total mass of the gas then its kinetic energy
' mv * *
Mv * (d *+( '
genius
PHYSICS by Pradeep Kshetrapal
43 Kinetic Theory of Gases Hhen the vessel is suddenly stopped then total kinetic energy will increase the temperature m ' Cv T mv * Cv T * M
of the gas (because process will be adiabatic i.e. Cv
+
'
? m
Problem 2.
>%s
+
M '
T
' mv * *
T
Mv * ( ' . *+
The density of a polyatomic gas is standard conditions is ,.41 k!m+ . The speci#c heat of the gas at constant volume is (a
+, ( Mk!
'
*1 ( M k!
'
,
(b
'
,
'/,, ( M k!
'
,
'
P
Speci#c
m rT V
'
,
'
(d
'
#$%&ti$n & (b Ideal gas e9uation for m !r"m gas PV mrT or
''*, ( M k!
(c
heat
rT
at
r
P T
constant
>where r 6 Speci#c gas constant?
'.,'+ ',1 /33.4 ,.41 *4+
volume
cv
r '
/33.4 / ' +
( '/,, k!.ke%vi
/ omicgas + for polyat
Problem 3.
The value of C 3 Cv '.,, + for a gas in state A and C 3 C v '.,3+ in another state. If P A and P) denote the pressure and T A and T ) denote the temperatures in the two states$ then (a P A P) $ T A T ) (b P A P) $ T A T ) (c P A P) $ T A T )
(d P A P) $ T A T )
#$%&ti$n & (c Dor state A$ C 3 Cv + i.e. the gas behaves as ideal gas. Dor state )$ C 3 Cv '.,3+ ( + i.e. the gas does not behave like ideal gas. and we know that at high temperature and at low pressure nature of gas may be ideal. So we can say that P A P) and T A T )
11.17 Gaseous MiFture. If two nonMreactive gases are enclosed in a vessel of volume V . In the mi"ture ' moles of one gas are mi"ed with * moles of another gas. If N A is %vogadro:s number then 7umber of molecules of #rst gas N' ' N A and number of molecules of second gas N * * N A (i Total mole fraction
( ' * .
(ii If M ' is the molecular weight of #rst gas and M * that of second gas. Then molecular weight of mi"ture will be M
'M' *M * ' *
genius
Kinetic Theory of Gases 44
(iii Speci#c heat of the mi"ture at constant volume will be
C V mix
' *
m' A M ' m* A M * m' m* ' ' * ' M' M * +
C V mix
+ + * * + ' ' ' ' * ' * ' ' * ' ' *
'
'C V ' *C V *
(iv Speci#c heat of the mi"ture at constant pressure will be C Pmix
' * + * + ' ' + ' * ' * ' *
'
C Pmix
m'
+
CPmix
m'
m*
'
m* M' ' ' M*
M'
mi"ture
C Pmix C V mix
' '
mi"ture
'
'
' '
'
* ( 'C V ' *C V *
'C P
'
*C P*
'C V
*C V *
'
*
)# ' &# + * * +% ( ' #' ' ' * ' #$ )# + + &# % ( ' * ' #' ' ' * #$
* * *
' * ' * ' ' * '
M * * '
'
'
' *
*
( 'C P' *C P* (v
'C P' *C P*
'
* *
' '( * '( *
' * *( ' ' ' *( ' '
'
Sample problems based on )i+ture Problem 4.
If two moles of diatomic gas and one mole of monoatomic gas are mi"ed with then the ratio of speci#c heats is [MP PMT 2++3"
(a
#$%&ti$n & (c
4
(b
+
' '$ '
1 +
1 /
(for monoatomic gas and
(c
*
' '+
* $ *
(d
4 1
'1 '
(for diatomic gas
