Lecture # 1 & 2 Introduction : Matter as we know broadly exists in three states. There are always two opposite tendencies between between particles of matter matt er which determine the state of matter mat ter f orces. !nter molecular attractive forces. The molecular motion / random motion.
In this chapter the properties and behaviours of the gases will be analysed and discussed in detail. These properties are measured m easured with the help of the t he gas laws as proposed Boyle,Charles,Gay l ussac etc
Boyle’s law and measurement of pressure : Statement : For a fixed fi xed amount of gas at constant temperature, temperatur e, the volume volum e occupied occupied by the gas is inversely inve rsely proportional to the pressure applied appli ed on the gas or pressure of the t he gas. 1 V" P hence PV = const this constant will be dependent on the amount of the gas and temperature of the t he gas. P1V1 = P2V2
P P2 P1
B A V2 V1
v
Applications of Boyles Law. For the two parts A P1V1 = K & P2V2 = K ‘ ’ and ‘B’ hence it follows that P 1V1 = P2V2 .
Page # 1
Units Vol ume
Pressure :
Volume of the gas is the Volume of th the container S.I.
unit $ m3
C.G C.G.S. .S. unit unit $cm3 1 ! = 10 3 m3 1 ! = 103 cm9 –
1dm3 = 1 ! = 10 3 m3 –
Temperture
Pressure = N/m 2 = Pa $ S. S.I. unit 2 C.G.S un unit = dyne-cm
Kelv in scale $ Boiling point = 373 K ice point = 273 K
Convert 1N/m 2 into dyne/cm2
Farenheit scale $ B.P. = 212º F ice point = 32º F
1N
%
10 5 dyne
Celcius scale $ B.P. = 100ºC
1 m2 10 4 cm 2 1N/m2 = 10 dyne/cm 2
1 atm
ice point = 0ºC C&0 K & 273 F & 32 % % 100 & 0 373 & 273 212 & 32
= 1.013 × 105 N/m2
1ml = 10 3 ! = 1 cm 3
=
–
R & R(0) R(100) & R(0)
where R = Temp. Temp. on unknown scale. Atmospheric pressure : The pressure exerted by atomosphere on earth ’s surface at sea level is called 1 atm. 1 atm = 1.013 bar 1 atm = 1.013 × 105 N/m2 = 1.0 1.013 13 bar bar = 7 760 60 torr torr
Solved olved Exa Exam mples ples Example
Sol.
A rubber balloon contains some solid marbles each of volume v olume 10 ml. A gas is filled in the t he balloon at a pressure pressure of 2 atm and the total volume of the t he ba balloon lloon is 1 litre in this condition. If the external pressure is increased to 4atm the volume vol ume of Balloon Ball oon becomes 625 ml. Find the number of marbles m arbles present in the balloon. Let the no. of marbles be = n . volume of marble marbl e = 10 n ml. volume of balloon earlier = 1000 ml. later = 625 ml. Now for the gas inside the balloon temperature and amount of the gas is constant, constant, hence boyles law can be applied P1V1 = P2V2 4× (625 – 10n) = 2 × (1000 – 10n) 625 × 4 = 2000 – 20n + 40n 625 × 4 – 2000 = 20n 625 ' 4 – 2000 = n. 20
125 =n 5
n = 25
MEASUREMENT OF PRESSURE Barometer : A barometer is an instrument that t hat is used for the measurement of pressure.The construction of the barometer baromet er is as follows Perfect Vaccum Hg is filled upto brim
Mg P0
Thin nairow glass tube
P0= Patm P0
P0
Cross Cross sectional sectional view of the capillary column Page # 2
A mercury barometer is used to measure atmospheric pressure by determining the height of a mercury column supported in a selaled glass tube. The downward pressure pressure of the mercury in i n the column is exactl exactly y balanced by the outside atmospheric pressure that presses down on the mercury in the dish di sh and pushes it up the column.
A thin narrow callibrated capill ary tube is filled to the t he brim, with a liquid such as mercury, mercury, and is inverted into a trough fill ed with the same fluid.Now depending depending on the external atmospheric pressure, pressure, the level of the t he mercury inside the tube will adjust itself, the reading of which can be monitored. When the mercury column inside the capillary comes to rest, then the net forces on the column should be balanced. Applying force balance, we get, Patm × A= m×g ( A cross-sectional area of the capill ary tube) ‘ ’ is the cross-sectional If ‘(’ is the density of the fluid, then m = ( × v hence, Patm × A = ( ( × g × h) × A (v = A × h) risen in the capillary) (‘h’ is the height to which mercury has risen or, Patm = (gh
!
Normal atmospheric pressure which we call 1 atmosphere or 1 atm, is defined as the pressure
exerted by the atmosphere at at mean sea level. It comes out to be 760 mm of Hg = 76 cm of Hg. (at mean sea level lev el the reading shown by the barometer is 76 cm of Hg) 1 atm = (13.6 × 103) 9.8 × 0.76 = 1.013 × 10 5 Pas. 1 torr = 1 mm of Hg. 1 barr = 105 N/m N/m2 (Pa) Faulty Barometer : An : An ideal barometer will show a correct reading only if the space above the mercury column is vacuum, v acuum, but in case if some gas column is trapped in the space above the mercury column, colum n, then the barometer is classified classifi ed as a faulty barometer. The reading of such a barometer will be less than the true pressure. For such a faulty barometer Pgas × A
Patm × A
P0A = Mg + Pgas A P0 = (gh + Pgas
mg
or
(gh = P0 – Pgas
Solved Exampl xamples es Example
The reading of a faulty barometer is 700 mm of Hg. When W hen actual actual pressure pressure is 750 mm of Hg. The length of the air column trapped in this case is 10 cm .Find the actual v alue of the atmosph atm ospheric eric pressure pressure when reading of this barometer is 750 mm of Hg. Assume Assume that the length l ength of the Barometer tube above mercury surface in the container remains constant. Page # 3
Sol.
P0 = Pgas + 700 (g Pgas = 750 (g – 700 (g = 50 (g )
Now for the gas column in the capillary, amount and temp are constant hence P 1V2 = P2 V2
* × (50 A) (50 (g) (100 A) = Pgas )
* = 100 (g Pgas
Now, applying force balance in the new scenario.
* + 750 (g = 100 (g + 750 (g = 850 (g * = Pgas Patm Hence, the atmospheric pressure is now, 850 cm of Hg. Example
In each of the following examples, find the pressure of the trapped gas.
P0 = 75 cm of Hg
Hg 10 cm gas column Sol.
Total pressure of gas column = 75 + 10 = 85 cm of Hg.
Example
Sol.
Pgas = 65 cm of Hg.
Example
Pg = 75 + 10 sin +.
Page # 4
Sol.
From the above problem, it can be generalised that, applying force balance every single time is not necessary. If we are moving up in a fluid, then substract the vertical length, and while moving down add the vertical length.
Example
The diameter of a bubble at the surface of a lake is 4 mm and at the bottom of the lake is 1 mm. If atmospheric pressure is 1 atm and the temperature of the lake water and the atmosphere are equal. what is the depth of the lake ? (The density of the lake water and mercury are 1 g/ml and 13.6 g/ml respectively. Also neglect the contribution of the pressure due to surface tension) P1V1 = P2V2
Sol.
)
4 4 , (4 mm/2)3 = (760 mm × 13.6 × g + h × 1 × g) , (1 mm/2)3 3 3 760 × 13.6 × 64 = (760 × 13.6 + h) h = 64 × 760 × 13.6 – 760 × 13.6 h = 63 × 760 × 13.6 mm
(760 mm × 13.6 × g)
h= Example Sol.
63 ' 760 ' 13.6 km = 0.6511 km = 651.1 m Ans. 1000 ' 1000
A gas is initially at 1 atm pressure. To compress it to 1/4 th of initial volume, what will be the pressure required ? P1 = 1 atm V1 = V P2 = ?
V2 =
V 4
P1V1 = P1V2 P2 =
Example
at const. T & n
P1V1 1 atm ' V = = 4 atm Ans. V V2 4
A gas column is trapped between closed end of a tube and a mercury column of length (h) when this tube is placed with its open end upwards the length of gas column is ( !1) the length of gas column becomes (!2) when open end of tube is held downwards. Find atmospheric pressure in terms of height of Hg column. case ! P0
case !! P2
!2
Hg h
Hg
P1
P0
h
!1
Sol.
For gas
P1 = (PO + h) P2 = (PO – h) 2 V1 = ,r !1 V2 = ,r2!2 at const T. and moles. P1V1 = P2V2 ; (PO + h) ,r2!1 = (PO – h) ,r2!2 PO!2 + h!1 = PO!2 – h!2 PO!2 – Po!1 = h!, + h!2 P0 =
h(!1 - ! 2 ) cm of Hg column Ans. ( ! 2 & ! 1)
Example
If water is used in place of mercury than what should be minimum length of Barometer tube to measure normal atmospheric pressure.
Sol.
PHg =
PH2 O
= Patm.
0.76 m × 13.6 × g = h H2O × 1 × g hH2O = 0.76 × 13.6
= 10.336 m Ans. Page # 5
Example Sol.
A tube of length 50 cm is initially in open atmosphere at a pressure 75 cm of Hg. This tube is dipped in a Hg container upto half of its length. Find level of mercury column in side the tube. If after dipping the tube, the length of air column be x cm situation shown in the adjoining figure. Then by using, Pi V i = Pf V f We have. 75 cm Hg × !A = Pf × x × A ....... (1) (! = 50 cm) & also, Pf = 75 cm Hg + (x –
!
2
)
....... (2)
(2) & (1) .
) But,
)
[75 + (x – 25)] × x = 75 × 50 . x2 + 50 x – 3750 = 0 x = 41.14 or – 91.14 x can't be – ve ) x = 41.14 mercury column inside the tube = (50 – 41.14) cm = 8.86 cm Ans.
Lecture # 3 Charles law : For a fixed amount of gas at constant pressure volume occupied by the gas is directly proportional to temperature of the gas on absolute scale of temperature. V " T or V = kT = k × (t + 273) V % cons tan t T
where ‘k’ is a proportionality constant and is dependent on amount of gas and pressure.
V1 V2 % T1 T2
V = a + bt
Temperature on absolute scale, kelv in scale or ideal gas scale.
Temperature on centigrade scale.
V
T
Relation : T = t + 273 Since volume is proportional to absolute temperature, T the volume of a gas should be theoretical ly at absolute zero is zero. Infact no substance exists as gas at a temperature near absolute zero, though the straight l ine plots can be extra ploated to zero volume. Absolute zero can never be attained practically though it can be approch only. By considering –273.15°C as the l owest approachable li mit, Kelvin developed temperature scale which is known as absolute scale.
Page # 6
Solved Examples Example
If the temp. of a particular amount of gas is increased from 27ºC to 57ºC, find final volume of the gas, if inital volume = 1 lt and assume pressure is constant. V1 V2 % T1 T2
Sol.
V2 1 % (273 - 27) ( 273 - 57)
So
V2 = 1.1 lt.
Example
An open container of volume 3 litre contains air at 1 atmospheric pressure. The container is heated from initial temperature 27ºC or 300 K to tºC or (T + 273)K the amount of the gas expelled from the container measured 1.45 litre at 17 ºC and 1 atm.Find temperature t.
Sol.
) T01 = 300 K It can be assumed that the gas in the container was first heated to (T + 273), at which a v olume ‘/V’ escaped from the container hence applying charles law : 3 3 - /V = 300 T - 273
Now, this volume ‘/V’ which escapes trans the container get cooled
)
1.45 /V = T - 273 290
Solve the two equations and get the value of /V an a T determine / v & calculate t that will be the answer. Example
Sol.
(a)
An open container of volume V contains air at temperature 27ºC or 300 K.The container is heated to such a temperature so that amount of gas coming out is 2/3 of (a) amount of gas initially present in the container. (b) amount of gas finally remaining in the container. Find the temperature to which the container should be heated. Here , P & V are constant, n & T are changing. Let, initially the amount of gas present be n & temp is 27ºC or 300K. Finall y amount of gas present in container = n –
5 1 2 2 n = 3 ' n 0 & final temp. be T. 3 4 3 1
Then using n1T1 = n2T2 , we have, n × 300 =
n × T2 = T2 = 900K 3
i.e., final temp = 900K Ans. (b)
Let there be x mole of gas remaining in the container, 2 5x x) = n = = n 3 3
)
(x +
) x =
)
Using n1 T1 = n2 T2
)
T2 = 500K final temperature = 500 K Ans.
2 of x comes out 3
3n 5
n × 300 K =
3n × T2 5
Page # 7
Calculation of pay load : Pay load is defined as the maximum weight that can be lifted by a gas filled ballon. Buoyancy
balloon
M
For maximum weight that can be lifted, applying force balance Fbuoyancy = Mballoon × g + Mpay load × g . (air v.g. = (gas v.g + Mg + mg. mass of balloon = m net force on volume of balloon = v balloon = 0 density of air = (air (at equilibrium / when balloon is incoming density of gas inside the with constant speed) balloon = (gas
Solved Examples Example
A balloon of diameter 20 metre weights 100 kg. Calculate its pay-load, if it is filled with He at 1.0 atm and 27ºC. Density of air is 1.2 kg m 3. [R = 0.0082 dm 3 atm K 1 mol 1] –
–
–
Weight of balloon = 100 kg = 10 × 104 g
Sol.
4 3 4 22 5 20 2 ' 3 ' 100 0 Volume of balloon = ,r % ' 3 3 7 4 2 1
3
= 4190 × 106 cm3 = 4190 × 103 litre Weight of gas (He) in balloon =
w 2 5 3" PV % RT 0 M 1 4
PVM RT
1' 4190 ' 10 3 ' 4 = = 68.13 × 104 g 0.082 ' 300
)
Total weight of gas and balloon = 68.13 × 104 + 10 × 104 = 78.13 × 104 g Weight of air displaced =
) )
1.2 ' 4190 ' 10 6 10 3
= 502.8 × 104 g
Pay load = wt. of air displaced – (wt. of balloon + wt. of gas) Pay load = 502.8 × 104 – 78.13 × 104 = 424.67 × 104 g
Example
Find the lifting power of a 100 litre balloon filled with He at 730 mm and 25°C. (Density of air = 1.25 g /L).
Sol.
Since,
PV = nRT
730 100 ' 4 W PVM × RT W= = g ) 760 0.082 ' 298 M RT i.e., Wt. of He = 15.72 g Wt. of air displaced = 100 × 1.25 g/L = 125 g Lifting power of the balloon = 125 g – 15.72 g = 109.28 g Ans.
PV =
)
Page # 8
Gay-lussac’s law : For a fixed amount of gas at constant volume, pressure of the gas is directly proportional to temperature of the gas on absolute scale of temperature. P " T P = constant $ dependent on amount and volume of gas T P1 P2 % T1 T2
$ temp on absolute scale
originally, the law was developed on the centigrade scale, where it was found that pressure is a linear function of temperature P = P0 + bt where ‘b’ is a constant and P0 is pressure at zero degree centigrade.
P
P P0
( –273°C)
T Example
O
t
PV = K . V = K1 /p V = K 2 . V = K2 T T K1 = K2T P
K1
1 .? T &# # % # # $ where are we wrong ? This is wrong because we are varying temperature &
PT =
K2
= const . P " =
K1 = f(1) thus K1 will change according to temperature So
K1 will be a f unction of temp & not constant. K2
Solved Examples Example
Sol.
The temperature of a certain mass of a gas is doubled. If the initially the gas is at 1 atm pressure. Find the % increase in pressure ? P1 P2 = ; T1 T2
% increase = Example
Sol.
P2 1 = T 2T 1 x 100 = 100% 1
The temperature of a certain mass of a gas was increased from 27 °C to 37°C at constant volume. What will be the pressure of the gas. P1 P2 = ; T1 T2
P2 P = ; 300 310
P2 =
31 P 30
Page # 9
Lecture # 4 Avogadro’s Hypothesis :
.
For similar values of pressure & temperature equal number of molecules of different gases will occupy equal volume. 6$ V N1 6 (volume of N1 molecules at P & T of one gas) 6$ V N1 6 (volume of N1 molecules at P & T of second gas) Molar volume & volume occupied by one mole of each and every gas under similar conditions will be equal. One mole of any gas of a combination of gases occupies 22.413996 L of volume at STP. The previous standarad is still often used, and applies to all chemistry data more than decade old, in this definiti on Standard Temperature and Pressure STP denotes the same temperature of 0°C (273.15K), but a slightly higher pressure of 1 atm (101.325 kPa) . Standard Ambient Temperature and Pressure (SATP) , conditions are also used in some scientific works. SATP 5 Pa) At SATP (1 bar and 298.15 K), the molar volume of an – 1 (Ref. NCERT ) ideal gas is 24.789 L mol c o n d
i ti o n s
m
e a n
s
2 9
8 . 1 5
k
a
n d
1
b
a r
( i . e .
e x a
c t l y
1 0
Equation of State : Combining all the gas relations in a single expression which discribes relationship between pressure, volume and temperature, of a given mass of gas we get an expression known as equation of state. PV = constant (dependent on amount of the gas (n)). T
P1V1 P2 V2 = T1 T2
Ideal gas Equation : PV = constant nT
[universal constant]
=R
(ideal gas constant or universal gas constant)
experimentaly R = 8.314 J/Kmole 7 25/3 = 1.987 cal/mole 7 2 = 0.08 Latm/mole 7 1/12
Solved Examples Example
Sol.
Four one litre flasks are separately filled with the gases, O 2, F 2, CH4 and CO2 under the same conditions. The ratio of number of molecules in these gases : (A) 2 : 2 : 4 : 3 (B) 1 : 1 : 1 : 1 (C) 1 : 2 : 3 : 4 (D) 2 : 2 : 3 : 4 According to avogadro ’s hypothesis. All the flasks contains same no. of molecules Ratio of no. of molecules of O2, F2, CH4 & CO2 ) = 1 : 1 : 1 : 1 Ans (B)
Example
Sol.
Some spherical balloon each of volume 2 litre are to be filled with hydrogen gas at one atm & 27 °C from a cylinder of volume V litres. The pressure of the H 2 gas inside the cylinder is 20 atm at 127°C. Find number of balloons which can be filled using this cylinder. Assume that temp of the cylinder is 27°C. P1V1 P2 V2 = T1 T2 1' x ' 2 60 = = x 300 2 x = 30 balloons Page # 10
also cylinder is equivalent to 2 balloons gas will be transferred only till the time pressure in cylinder is > outer pressure. When pressure of cylinder = pressure of balloon. hence only = 30 – 2 = 28 balloons will be filled. Example
A weather balloon filled with hydrogen at 1 atm and 300 K has volume equal to 12000 litres. On ascending it reaches a place where temperature is 250 K and pressure is 0.5 atm. The volume of the balloon is : (A) 24000 litres (B) 20000 litres (C) 10000 litres (D) 12000 litres
Sol.
Using
P1V1 P2 V2 = ; T1 T2
0.5 atm ' V2 1 atm ' 12000 L = 300 K 250 K
)
V2 = 20,000 L Hence Ans. (B)
Daltons law of partial pressure : Partial pressure : In a mixture of non reacting gases partial pressure of any component gas is defined as pressure exerted by this component gas if whole of volum e of mixture had been occupied by this component only. Partial pressure of first component gas
P1 %
n1 RT v
P2 %
n 2 RT
P3 %
n 3 RT
v
v Total pressure = P 1 + P2 + P3.
Daltons law : For a non reacting gaseous mixture total pressure of the mixture is summation of partial pressure of the different components gases. PTotal = P1 + P2 + P 3
%
(n1 - n2 - n 3 ) RT v
P1 n % 1 % x1 (mole fraction of first component gas) PT n T P2 n % 2 % x 2 (mole fraction of second component gas) PT n T P3 PT
%
n3 nT
% x 3 (mole fraction of third component gas)
Page # 11
Solved Examples Example
The stop cock connecting the two bulbs of volume 5 litre and 10 lit re containing as ideal gas at 9 atm and 6 atm respectivel y, is opened. What is the f inal pressure if the temperature remains same.
Sol.
After the opening of the stop cock the pressure of the each bulb will remain same. At the beginning, the no. of moles of gas in A =
10 x 6 RT
At the beginning, the no. of moles of gas in B =
5x9 RT
) total no. of mole at the beginning =
105 RT
Total no. of mole of gas before opening the stop cock = total no. of moles of gas after opening stop cock =
105 RT
) pressure aft er the opening of the stop cock P=
RT 105 105 x = = 7 atm Vtotal 10 - 5 RT
Example
The partial pressure of hydrogen in a flask containing two grams of hydrogen and 32 gm of sulphur dioxide is : (A) 1/16th of the total pressure (B) 1/9th of the total pressure (C) 2/3 of the total pressure (D) 1/8th of the total pressure
Sol.
n H2 =
) PH 2
2g 32g = 1mol. n SO 2 = = 0.5 mol 2g / mol 64g / mol
nH 2
8nH2 - n 9
× P T =
SO 2
2 1 P . × P T = 3 T (1 - 0.5)
) (C) Example
Sol.
Equal volumes of two gases which do not react together are enclosed in separate vessels. Their pressures are 10 mm and 400 mm respectiv ely. If the two vessels are joined together, then what will be the pressure of the resulting mi xture (temperature remaining constant) : (A) 120 mm (B) 500 mm (C) 1000 mm (D) 205 mm Let, vol of containers be V & temp be T P1 = 10mm P2 = 400mm
) n1 =
P1V RT
) n1 + n2 =
&
n2 =
P2 V RT
(P1 - P2 ) ' V RT
After joining two containers final vol = (V+V) = 2V(for gases)
) Pfinal = =
8n1 - n2 9RT Vfinal
=
810 - 4009 mm 2
(P1 - P2 ) ' V RT
= 205 mm.
×
8P1 - P2 9 RT = 2V 2 Ans.
(D)
Page # 12
Example
A mixture of NO2 & CO having total volume of 100 ml contains 70 ml of NO2 at 1 atm, mixture is left for some time and same NO 2 get dimerised to N2O4 such that final volume of the mixture become 80 ml at 1 atm calculate mole fraction of NO 2 in final equilibrium mixture.
Sol.
Initial volume of NO2 = 70 ml Initial volume of CO = 100 – 70 = 30 ml Final volume of mixture = 80 ml Let the volume of NO 2 in final mixture be x Let ‘v ’ ml NO2 be converted to N2O4
6$ N2O4 2NO2 6 V V/2 Hence final volume = volume of CO + volume of NO2 left + volume of N2O4 formed = 30 + 70 – V + V/2 = 80 V = 40 ml Hence volume of NO2 left = 70 – V = 30 ml Now as volume : moles
)
mole fraction = volume fraction =
30 3 = 80 8
Analysis of gaseous mixture : Vapour density : Vapour density of any gas is defined as the density of any gas with respect to density of the H 2 gas under identical conditions of temperature T and pressure P. vapour density =
P=
m RT . V M
density of gas at T & P density of H2 under same P & T
.
P=(
PMgasRT
RT M
.
Mgas
vapour density = RT PM = M = H2 H2
( =
PM RT
M gas 2
Mgas = 2 × vapour density
Average molecular mass of gaseous mixture : total mass of the mixture div ided by total no. of moles in the mixture Total mass of mixture Mmix = Total no. of moles in mixture If we have ‘n ’ , ‘n ’ and ‘n ’ are moles of three different gases having of molar mass ‘M ’, ‘M ’ and ‘M ’ respectively. 1 2 3 1 2 3 n1 M1 - n 2 M2 - n 3 M3 Mmin = n1 - n 2 - n 3
Solved Examples Example
Calculate the mean molar mass of a mixture of gases having 7 g of Nitrogen, 22 g of CO2 and 5.6 litres of CO at STP.
Sol.
Moles of N2 = 7/28 = 1/4 Moles of CO2 = 22/44 = 1/2 Moles of CO = 5.6 / 22.4 = 1/4 mean molar mass =
Mmin =
n1 M1 - n 2 M2 - n3 M3 n1 - n 2 - n3
= ( 7 + 7 + 22 ) / 1 = 36
Page # 13
Lecture # 5 Graham’s Law of Diffusion/Effusion : Diffusion : Net spontaneous flow of gaseous molecule from region of high concentration (higher partial pressure) to region of lower concentration or lower partial pressure
flow will be from both sides, N2 will try to equalise its partial pressure in both the vessels, and so will O 2.
Graham’s Law : Under similar condition of pressure (partial pressure) the rate of diffusion of dif ferent gases is inversely proportional to square root of the density of different gases.” “
rate of diff usion r :
r1 = r2
d2 d1
=
M2 M1
1
V . D2 V . D1
=
r = volume flow rate = r = mole flow rate =
d = density of gas
d
V.D is vapour density
dVout dt
dn out dt
r = distance travelled by gaseous molecule per unit time =
dx dt
The general form of the grahams law of diffusion can be stated as follows, when one or all of the parameters can be varied. rate :
P TM
A
P – Pressure, A – area of hole, T – Temp. , M – mol. wt. If partial pressure of gases are not equal. Then rate of diffusion is found to be proportional to partial pressure & inversely proportional to square root of molecular mass. r : P r: r1 P1 = r2 P2
1 M M2 M1 Page # 14
Selective diffusion : If one or more than one components of a mixture are allowed to diffuse and others are not allowed then it is selective diffusion of those components.
O2 although total pressure H2 may be different both sides.
Diffusion will happen till the time partial pressure of H2 becomes same on both sides.
Diffusion
Pt membrane
!
Platinum allows only H2 gas to pass through Effusion : (forced diffusion) a gas is made to diffuse through a hole by application of external pressure.
Solved Examples Example
Sol.
In is a tube of length 5 m having 2 idendical holes at the opposite ends. H2 & O2 are made to effuse into the tube from opposite ends under identical conditions. Find the point where gases will meet for the first time.
r1 ax dt × = = r2 dt dx dx1 = 4 dx 2
x (5 & x )
m2 m1
=
.
dx1 = dx 2
32 2
dis tan ce travelled by H2 = 4 dis tan ce travelled by O 2
= 4
x = (5 – x) 4 x = 20 – 4x 5x = 20 x = 4 from H2 side Example Sol.
5 ml of H2 gas diffuses out in 1 sec from a hole. Find the volume of O2 that wil diffuse out from the same hole under identical conditions in 2 sec. 5ml Rate of diffusion of H2 = = 5ml/s = rH2 (say) 15
)
)
1 1 = 5ml/s × 4 4 Volume of O2 diffused in 2.0 seconds ro 2 = rH2 ×
=
5 × 2 ml = 2.5 ml Ans. 4
Page # 15
Example
A vessel contains H2 & O2 in the molar ratio of 8 : 1 respectively. This mixture of gases is allowed to diffuse through a hole, find composition of the mixture coming out of the hole.
Sol.
Here, n H2 : n O 2 = 8 : 1 &
.
rH2 rO 2
=
8 × 1
rH2 rO2
=
nH 2
MO2
nO 2
MH2
32 32 = 1 2
.
8no. of moles of H2 co min g out 9 / /t 8no. of moles of O2 comin g out 9 / /t
)
Required composition of H2 : O2 coming out = 32 : 1 Ans.
Example Sol.
=
32 1
The rate of diffusion of a sample of a ozonized oxygen is 0.98 times than that of oxygen. Find the % (by volume of ozone in the ozonized sample. Let, rate of dif fusion of ozonized oxygen be rg & Let, reat of diffusion of oxygen be rO 2 rg
) r = 0.98 (given) O2 rg
but r O2
......... (1)
1 / 2
5 MO 2 2 0 = 33 0 4 Mg 1
Form (1) . 0.98 =
(mean molar mass of ozonized oxygen = M g) 32 Mg
32
32
= 33.32 . (0.98)2 = M ) Mg = g 80.98 92 Let % of O3 be x ) % O2 = (100 – x) in ozonized oxygen.
)
48 x - 8100 – x 9 ' 32 = 33.32 100
. 3200 + 16x = 33.32 ) x =
132 = 8.25 % 16
i.e. % of O3 = 8.25 % Ans. Example
Sol.
Assume that you have a sample of hydrogen gas containing H2, HD and D2 that you want to separate into pure components (H = 1H and D = 2H). What are the relative rates of diffusion of the three molecules according to Graham ’s law ? Since D2 is the heaviest of the three molecules, it will diffuse most slowly, and we’ll call its relative rate 1.00. We can then compare HD and H2 with D2. Comparing HD with D 2, we have Rate of HD diffusion = Rate of D 2 diffusion
Mass of D 2 = Mass of HD
4.0 amu = 1.15 3.0 amu
Comparing H2 with D2 we have Mass of D 2 Rate of H2 diffusion = = Mass of H2 Rate of D 2 diffusion
4.0 amu = 1.41 2.0 amu
Thus, the relative rates of diffusion are H2(1.41) > HD (1.15) > D 2(1.00).
Page # 16
Lecture # 6 Kinetic Theory of Gases : Postulates / assumptions of KTG : A gas consists of tiny spherical particles called molecules of the gas which are identical in shape & size (mass) The volume occupied by the molecules is negligible in comparision to total v olume of the gas. For an ideal gas, volume of the ideal gas molecule ~ 0. Gaseous molecules are always in random motion and collide with other gaseous molecule & with walls of the container. Pressure of the gas is due to these molecular collisions with walls of the container These collisions are elastic in nature Molecular attraction forces are negligible. Infact, for an ideal gas attraction or repulsion forces are equal to zero. Newton’s laws of motion are applicable on the motion of the gaseous molecule. Effect of gravity is negligible on molecular motion. The average K.E. of gaseous molecules is proportional to absolute temp of the gas. 1 M (u 2 ) " T (bar is for average) 2 Kinetic equation of gaseous state (expression for pressure of gas)
Derivation m = mass of one molecule '
U = Ux ˆi + Uy jˆ + Uz kˆ
Consider collision with ABCD '
inital
Pi = mUx ˆi
;
final
'
Pf = – mUx ˆi
change in momentum due to collision = 2 Ux m time taken between two succussive collision with force ABCD = t = frequency of collision =
2! Ux
Ux 1 = t 2!
change in momentum in one sec. = force = 2 m
U x ' U x m U2x = 2! !
force due to all the molecules m {U2x1 - U2x 2 - ........ - Ux N2 } = !
2 average value of UN = UN2 =
Fx =
M !
Ux 12 - U2x 2 - .......... .U2xN N
{N U2x }
all the three direction are equal as the motion is totally random in all directions, hence 2 U2x = U y = U2z 2 2 2 U2 = U x - U y - U z
= 3 U2x Fx =
M !
. N
1 2 3 U Page # 17
Pressure =
)
Fx
=
!2
PV =
1 N 2 3 !3 U
1 mN U2 3
The volume of the container ‘v ’ = !3
Kinetic equation of gases
where U2 is mean square speed
5 U12 - U22 - U32 - ......UN2 2 3 0 Urms = U 2 = 3 0 N 4 1
root mean square speed =
Verification of Gaseous Laws Using Kinetic Equation : From postulates ;
PV =
1 mN U2 3
1 m U2 2
PV =
2 2 5 1 3 m U2 0 N 3 4 2 1
: T = ; T ;
Where ’;‘ is a proportionality constant
PV =
2 3
; NT
Boyle’s Law : N : constant T : constant PV = constant Charles law : N : constant P : constant v:T Kinetic energy of gaseous molecule (translation K.E.) To calculate ; we have to use ideal gas equation (experimental equation) PV = nRT kinetic equation PV = on comparing
; =
2 3
; nRT =
2 3
; (nNA) T
R 3 × NA 2
R 3 K where K = = Boltzmann constant NA 2 1 2 Average K.E. of molecules = m U = ; T 2 3 Average K.E. = K T (only dependent on temperature not on nature of the gas.) 2 5 1 3 3 2 2 Average K.E. for one mole = NA 3 m U 0 = K NA T = RT 4 2 1 2 2
; =
Root mean square speed Urms = "
U = 2
3kT m
=
3R T m NA
Where m-mass of one molecule
Dependent on nature of gas i.e mass of the gas 3R T molar mass must be in k/mole. M Average speed Uav = U1 + U2 + U3 + ............ UN
Urms =
Uav =
8RT = ,M
8KT ,m
K is Boltzmman constant
Page # 18
Most probable speed : The speed possesed by maximum number of molecules at the given temperature 2 RT = M
UMPS =
2 KT m
Solved Examples Example Sol.
In a container of capacity 1 litre there are 10 23 molecules each of mass 10 square speed is 10 5 cm/sec then calculate pressure of the gas. 1 PV = MN U2 3 P=? V = 10 3 m3 m = 10 25 kg N = 1023
–
22
gms. If root mean
–
–
5 3 U2 = 10 cm/sec = 10 m/sec
U2 = 106 m2 /sec2 –
P × 10
3
=
1 × 10 3
–
25
× 1023 × 0 × 106
1 × 10 2 × 106 × 103 3 1 × 107 pascals P= 3
P=
Example
Sol.
(i)
) )
–
(i) Calculate the pressure exerted by 10 23 gas molecules, each mass 10 22 g in a container of volume one litre. The rms velocity is 105 cm/sec. (ii) What is the total kinetic energy (in cal) of these particles ? (iii) What must be the temperature ? Here N = 1023 molecules m = 10 22g , V = 1L = 1000 cm 3 rms velocity = 105 cm/second 1 PV = Nm U2 3 1 P = Nm U2 3 –
–
=
1 3 ' 1000 cm 3
× 1023 × 10 22g × (105)2cm2 /s2 –
= 3.33 × 107 dyne/cm2 Ans. (ii)
Total K . E of the molecules 1 = Nm U2 2 1 × 1023 × 10 25 kg × (103m/s)2 = 2 10000J = = 5000 J 7 1195.0 Cal Ans. 2 –
Page # 19
(iii)
Avg K . E of one molecule =
)
T=
1 3 2 = kT U 2 2
1 mU2 × 3 k
1 10 × = 3
–
25
2 R 2 5 Kg ' 810 3 m / s 9 ' 6.023 ' 10 23 0 K 3" k % NA 1 4 8.314
= 2414.8 K Ans. Example
If f or two gases of molecular weights MA and MB at temperature TA and TB ; TAMB = TBMA, then which property has the same magnitude for both the gases. (A) Density (B) Pressure (C) KE per mol (D) RMS speed
Sol.
Given that TA MA = TB MA . 3RT M
But, r.m.s. =
3RTA MA
)
r.m.sA =
)
r.m.sA = r.m.sB
Example
Sol.
)
1.1 × 104 =
r.m.sB =
)
Ans. (D)
3RTB MB
T =
8RT ,M
8 ' 8.314 ' T
, ' 2 ' 10
81.1' 10 9
4 2
)
&
It has been considered that during the formation of earth H2 gas was available at the earth. But due to the excessive heat of the earth this had been escaped. What was the temperature of earth during its formation? (The escape velocity is 1.1 x 10 6 cm/s) Escape velocity of H2 should be equal to avg velocity of H 2. Avg velocity of H2 = 1.1 × 106 cm/s = 1.1 × 104 m/s But, avg. velocity =
.
TA TB = MA MB
–
' , ' 2 ' 10
8 ' 8.314
8M
3
H2
–
% 2g % 2 ' 10 3 kg9 –
3
K = 11430.5 K = 11157.5oC Ans.
Page # 20
Lecture # 7 Maxwell’s distributions of molecular speeds : Postulates/Assumptions of speed distributions It is based upon theory of probability. It gives the statistical averages of the speed of the whole collection of gas molecules.
Speed of gaseous molecules of may vary from 0 to <. The maxwell distribution of speed can be plotted against fraction of molecules as follows.
" " "
" " "
The area under the curve will denote f raction of molecules having speeds between zero and infinity Total area under the curve will be constant and will be unity at all temperatures. Area under the curve between zero and u1 will give fraction of molecules racing speed between 0 to u1 . This fraction is more at T1 and is less at T 2 . The peak corresponds to most probable speed. At higher temperature, fraction of molecules having speed less than a particular value decreases. For Gases with different molar masses will have following graph at a given temperature.
Real Gases : Real gases do not obey the ideal gas laws exactly under all conditions of temperature and pressure.
Real gases deviates from ideal behaviour because " Real gaseous molecules have a finite volume. {since on liquefaction real gases occupy a finite volume} " Inter molecular attraction focus between real gas molecules is not zero. {Real gases can be converted into liquid where as ideal gases cant be} Deviation of real gases from ideal behaviour can be measured by using compresibility factor : (Z) Z=
Z=
(PV )real (PV )ideal
(PV)ideal = nRT
PVm PV = nRT RT
VM is volume of one mole or molar volume.
Vm Z= V m ideal
Variation of Z with pressure at constant temperature :
Page # 21
NH3 /CO2 N2 He/H2 ideal
Z=1
P Variation of Z with pressure at different temperature (for a gas ) :
Conclusions : Z = 1 for ideal gas Z > 1 at all pressures for He/H 2 Z < 1 at low pressure (for alll other gases) Z > 1 at high pressur (for alll other gases)
Vander Waal Equation of real gases : The ideal gas equation does not considetr the effect of nattractive forces and molecular volume. vander Waal corrected the the ideal gas equation by taking the effect of (a) Molecular volume (b) Molecular attraction Volume correction : Ideal gas equation : Pi Vi = nRT ; In the equation ‘Vi’ stands for the volume which is available for free movement of the molecules. Videal = volume available for free movement of gasesous molecule hence, Vi = V – {volume not available for free movement} For an ideal gas Vi = V {V = volume of container} but for a real gas Vi = V , as all the volume is not available for free movement
Molecules have finite volume : Excluded volume per molecule =
1 C4 3@ B , (2r ) ? = Co-volume per molecule. 2 A3 >
r 1
r 2
The volume that is not available for free movement is called excluded volume. let us see, how this excluded volume is calculated. Excluded volume (not available for free momement)
For example 2, the entire shaded region is excluded, as its centre of mass cannot enter this region. If both molecules were ideal, then they would not have experienced any excluded volume but not in the case, of real gas as the centre of mass of ‘2’ cannot go further. Hence for this pair of real gas molecules. Excluded volume per molecule =
1 C 4 (2r )3 @ B , ? = 4 > 2 A3
C4 3 @ B ,r ? A3 >
C4 3 @ % excluded volume per mole of gas b = NA 4 B 3 , r ? A > for n moles excluded volume = nb Vi = V – nb
= 4 x NA x Volume of individual molecule
volume correction Page # 22
Lecture # 8 Pressure correction or effect of molecular attraction forces :
Molecule in middle of container Due to these attraction speed during collision will be reduced Momentum will be less Force applied will be less Pressure will be less. Pideal = P + {correction term} Correction term " no. of molecule attracting the colliding molecule. (n/v) Correction term " density of molecules (n/v).
5 n 2 no. of collision " density of molecules 3 0 4 v 1 5 n 2 net correction term : 3 0 4 v 1
5 n 2 an2 3 0 = 2 4 v 1 v
a’ is constant of proportionality and this is dependent on force of attraction Stronger the force of attraction greater will be ‘a’ (Constant) ‘
Pi = P +
an2
v2 Vander waal ’s equation is
5 an 2 2 3P 0 2 0 (v – nb) = nRT 3 v 4 1 VERIFICATION OF VANDER WAAL’S EQUATIONS Variation of Z with P for vander wall equation at any temp. Vander waal equation for 1mole
5 2 3P - a 0 (V – b) = RT 2 0 3 4 Vm 1 AT LOW PRESSURE (at separate temp.) At low pressure Vm will be high. a
Hence b can be neglected in comparision to V m. but
Vm2
cant be neglected as pressure is low
Thus equation would be
5 2 3P - a 0 V = RT 2 0 3 4 Vm 1 m PVm +
a = RT Vm
a PVm + = 1 Vm RT RT a Z = 1 – VmRT
Z<1
Real gas is easily compressible in comparision to an ideal gas. Page # 23
AT HIGH PRESSURE (moderate temp.) VM will be low so b cant be neglected in comparision to V m a but 2 can be neglected in comparision to much high values of P.. Vm Then vander Wall equation will be P(Vm – b) = RT PVm – Pb = RT PVm RT
Pb + 1 RT
=
Pb +1 Z>1 RT If Z > 1 then gas is more diff icult to compress in comparision to an ideal gas. Z=
At low pressure and very high temperature. Vm will be very large a hence ‘b’ can be neglected and 2 can also be neglected as V m is very large Vm PVm = RT (ideal gas condition) For H2 or He a ~ 0 because molecules are smaller in size or vander wall forces will be very weak, these are non polar so no dipole-dipole interactions are present in the actions. P(Vm – b) = RT Pb so Z=1+ RT ‘a’ factor depends on inter molecular attraction forces. a’ factor for polar molecule > ‘a’ factor for non polar molecule.
‘
"
Example
Arrange following in decreasing ’a’ factor (H2O, CO2, Ar) H2O > CO2 > Ar polar " For non polar molecules : Greater the size surface area greater will be vander wall forces so greater will be ’a’ constant.
Example
Arrange following gases according to ‘a’ He , Ar, Ne, Kr. aKr > a Ar > aNe > aHe "
More ‘a’ factor means high boiling point.
liquification pressure : Is the pressure required to convert gas into liquid. for easy liquefaction aD and PE "
When Example
Z < 1, Z < 1,
Vm < Vm, ideal Vm < Vm, ideal
. .
easily liquifiable more difficult to compress.
Arrange the following according to liquification pressure. n-pentane ; i iso pentane , neo pentane. an-pentene > aiso pentane > aneo pentane liquification pressure = LP L P n& pen tan e < L P iso pen tan e < LPneo pen tan e
b is roughly related with size of the molecule. (Thumb rule)
C4 3 @ b = NA 4 B , r ? A3 >
Page # 24
Solved Examples Example
Two vander waals gases have same value of b but different a v alues. Which of these would occupy greater volume under identical conditions ?
Sol.
If two gases have same value of b but different values of a, then the gas having a larger value of a will occupy lesser volume. This is because the gas with a larger value of a will have larger force of attraction and hence lesser distance between its molecules.
Page # 25
Lecture # 9 Virial Equation of state : It is a generalised equation of gaseous state all other equation can be written in the form of virial equation of state.
5 1 2 Z is expressed in power series expansion of P or 33 V 00 4 m 1 Z=1+
C D B + 2 + 3 + ..................... Vm Vm Vm
B – second virial coefficient , C – third virial coefficient , D – fourth virial coefficient
Vander wall equation in virial form :
5 2 3P - a 0 (V – b) = RT 2 0 3 4 Vm 1 m
Z=
PVm RT
a
RT
P=
( Vm & b ) Vm
=
( Vm & b )
1 1& x
–
–
Vm2
a 1 = Vm RT (1 & b / Vm )
–
a Vm RT
= 1 + x + x2 + x3 + ..........
5 b b 2 b3 2 3 0 1 ......... Z= 3 2 3 0 V Vm Vm m 4 1
–
1 5 b2 b3 a 2 a 3b & 0 + 2 + 3 + ................. =1+ Vm 4 RT 1 Vm Vm Vm RT
comparing with virial equation according to vander walls equation a , C = b2, P = b3 RT at low pressure : Vm will be larger
B = b –
1
hence
Vm2
,
Z=1+ If
1 Vm3
................ can be neglected
1 5 a 2 3b & 0 Vm 4 RT 1
a 2 5 3b & 0 =0 . 4 RT 1
Z=1 so at Z =
at
a Rb (ideal gas)
T=
a gas will behave an ideal gas (or follows boylesley) Rb
But at constant temperature ideal gas equation is obeying Boy ’es law as T =
a , so the temp is called Rb
Boyl ’es temp.
TB =
a Rb
a Vm RT for a single gas if we have two graphs as above we must conclude T 2 < T1 at Boyle’s temperature ‘a’ factor is compensated by factor so Z = 1.
Z = 1 –
Page # 26
Critical constant of a gas : When pressure is incerases at constant temp volume of gas decreases
AB $ gases BC $ vapour + liquid CD $ liquid critical point : At this point all the phydical properties of liquid phase will be equal to phsical properties in vapour such as density of liquid = density of vapour TC or critical temp : Temperature above which a gas can not be liquidied PC or critical pressure : minimum pressure which must be applied at critical temp to convert the gas into liquid. VC or critical volume : volume occupied by one mole of gas at T C & PC
Critical constant using vander wall equations : 5 2 3P - a 0 (V – b) = RT 2 0 3 4 Vm 1 m ( PVm2 - a ) (Vm – b) = RT Vm2 PVm3 + aVm – PbVm2 – ab – RTVm2 = 0
5 4
Vm3 + Vm2 3 b -
RT 2 0 + a Vm – ab = 0 P 1 P P
cubic can hence there will be three roots of equation at any temperature of pressure. At critical point all three roots will coincide and will give singles dx = V C at critical point Vander Waal equation will be
5 RTC 2 a ab Vm3 – Vm2 33 b - P 00 + Vm – = 0 PC PC C 1 4
...(1)
But at critical point all three roots of the equation should be equal, hence equation should be : Vm3 – 3Vm2 VC + 3Vm VC2 – VC3 = 0 ...(2) comparing with equation (1) b+
RTC PC
= 3V C
....(i)
a = 3 V C2 PC
...(ii)
ab = VC3 PC
...(iii)
PC =
a 3 VC2
substituting
PC =
a 3 (3b )
2
=
a 27b 2
Page # 27
by (i)
RTC = 3 VC – b = 9b – b = 8b PC
TC =
8a 27 Rb
At critical point, the slope of PV curve (slope of isotherm) will be zero
5 FP 2 33 00 = 0 V F 4 m 1 T
...(i)
C
at all other point slope will be negative O is the maximum value of slope.
5 FP 2 3 0 FVm 34 FVm 10 TC = 0 F
....(ii)
{Mathematically such points an known as point of inflection (where first two duivation becomes zero)} using the two TC PC and VC can be calculate
by
By any two a can be calculated but a by V C and T C and a by TC and PC may differ as these values are practical values and V C can’t be accuratly calculated so whwn we have VC TC & P C given use PC & TC to deduce ‘a’ as thet are more reliable.
Reduced Equation of state : Reduced Temp : Temperature in any state of gas with respect to critical temp of the gas Tr =
T TC
Reduced pressure :
P r =
P PC
Reduced volume :
Vr =
Vm VC
Vander wall equation
5 2 3 P - a 0 (V – b) = RT 2 0 3 4 Vm 1 m Substitute values :
5 2 3Pr PC - a 0 (V V – b) = R T T 3 r C r C Vr2 VC2 10 4 Substiture the value of P C TC and VC
5 2 3 Pr a 0 3 27 b 2 0 4
5 2 a 8a 3 Pr a 0 3 27 b 2 V 2 (3b)2 0 (3b Vr – b) = RTr 27 Rb r 4 1
5 Pr 1 2 33 - 00 (3 V – 1) = 8 R Tr r 3 4 3 Vr
5 2 3Pr - 3 0 (3V – 1) = 8 T 3 r r Vr2 10 4
Reduced equation of state
Equation is independent from a, b and R so will be followed by each and every gas independent of its nature. Page # 28
Solved Examples Example
Sol.
The vander waals constant for HCI are a = 371.843 KPa and b = 40.8 cm 3 mol 1 find the critical constant of this substance. –
The critical pressure, P C =
a 27b 2
=
371.843 '10 3 27 ' ( 40.8 )2 '10 &6
=
371.843 '10 9 27 ' ( 40.8)2
= 8.273 x 106
Pa = 8.273 MPa 8a 27Rb
The critical pressure, T C = –
–
R = 8.314 KPa dm 3 K 1 mol TC =
1
8 ' 371.843 8a = = 324.79 = 324.8 K 27Rb 8.314 ' 27 ' 40.8 ' 10 &3
The critical volume, V C = 3b = 3 x 40.8 = 122.4 cm 3 Example
The vander waals constant for gases A, B and C are as follows : Gas a/dm6 KPa mol 2 b/dm3 mol 1 A 405.3 0.027 B 1215.9 0.030 C 607.95 0.032 Which gas has (i) Highest critical temperature (ii) The largest molecular volume (iii) Most ideal behaviour around STP ? –
Sol.
TC =
–
8a Since, R is constant, higher the value of a/b higher will be criti cal temperature. 27Rb
VC = 3b and VC : Vm (for a particular gas) therefore higher the value of V C higher will be molar volume of the gas. If the critical temperature close to 273 K, gas will behave ideally around the STP. Let us illustrate the result in a tabular form. Gas a/dm6 KPa mol 2 b/dm3 mol 1 TC VC a/ b A 405.3 0.027 534.97 K 0.081 1.501 x 104 B 1215.9 0.030 1.444.42 K 0.09 4.053 x 104 C 607.95 0.032 677.07 K 0.096 1.89 x 10 4 –
–
(i) B gas has the largest criti cal temperature. (ii) C gas has the largest molecular v olume. (iii) A gas has the m ost ideal behaviour around STP.
Page # 29
Lecture # 10 Vapour pressure of a liquid (aqueous Tension of water) :
Vapour pressure depends on (a) Temperature (T D . VP D ) (b) Nature of the liquid Vapour pressure is independent of amount of liquid & surface area of liquid. Vapour pressure of the liquid is independent of pressure of any gas in the container, Ptotal = Pgas + Pwater vapour
Solved Examples Example
In a container of capacity 1 litre, air and some liquid water is present in equilibrium at total pressure of 200 mm of Hg this container is connected to another one litre evacuated container find total pressure inside the container when equilibrium is again stablised (aqueous tension or vapour pressure at this temp. is 96 mm Hg).
Sol.
Total pressure = 100 mm of Hg = Pgas + Pvapour water Pgas + 96 = 200 . Pgas = 107 mm of Hg Initially when second coataitive connected P1 = 107 mm of Hg P2 = ? V1 = 1 V2 = 2 litre P1 V1 = P2 V2 107 × 1 = P2 × 2 53.5 = P2 After equilibrium is established Ptotal = 53.5 + 93 = Pgas + Pwater = 146.5 mm of Hg at equilibrium
Eudiometry : The analysis of gaseous mixtures is called eudiometry. The gases are identified by absorbing them in specified and specific reagents.
Some Common Facts : Liquids and solutions can absorb gases. If a hydrocarbon is burnt gases liberated will be CO 2 & H2O. [H2O is seperated out by cooling the mixture & CO2 by absorption by aqueous KOH] If organic compound contains S or P then these are converted into SO 2 & P4O10 by burning the organic compound. If nitrogen is present then it is converted into N2. [The only exception : if organic compound contains – NO2 group then NO2 is liberated] If mixture contains N2 gas & this is exploded with O2 gas, do not assume any oxide formation unless specified. Ozone is absorbed in turpentine oil and oxygen in alkaline pyragallol.
Page # 30
Solved Examples Example
Carbon dioxide gas (CO2) measuring 1 litre is passed over heated coke the total volume of the gases coming out becomes 1.6 litre. Find % conversion of CO2 into carbon monoxide.
Sol.
6$ 2CO CO2 + C 6 CO2 CO 1 0 at time t 1 – x 2x Initial volume = 1 litre final volume = 1.6 litre no of litres = (1 + x) litres 1 + x = 1.6 x = 0.6 x = 0.6 0.6 × 10 = 60% of CO 2 will be converted into CO 1
Example
100 ml of hydrocarbon is mixed with excess of oxygen and exploded on cooling the mixture was reported to have a contraction of 250 ml the remaining gas when passed through a solution of aqueous KOH the mixture shows a further contraction of 300 ml. Find molecular formula of the hydrocarbon. 5 Y 2 y H2O 6$ x CO2 + Cx Hy + 3 x - 0 O2 6 2 4 4 1 y 100 ml 100. x 100 " 2
Sol.
Example.
Sol.
100 ml of an hydrocarbon is burnt in excess of oxygen in conditions so that water formed gets condensed out the total contraction in volume of reaction mixture was found to be 250 ml when the reaction mixture is further exposed to aqueous KOH a further contraction of 300 ml is observed find molecular formula of hydrocarbon. Cx Hy + O2 6 6$ CO2 + H2O 100 ml excess 300 ml By POAC on ‘C’ atoms x × 100 = 300 x=3 POAC on ‘H’ atoms y × 100 = 2 × moles of H 2O POAC on O atoms 2 × v = 2 × 300 + 1 × H2O {v = volume of O2 consumed} 2 × v = 600 + 50 y 600 - 50 y v= volume of O2 consumed 2 The total volume contraction is 250 ml. Hence, 100 + v – 300 = 250 – 200 + v = 250 = 450 2 × 450 – 600 = 50 y 300 =y=6 50 Hydro carbon will be C 3 H6
Alternative : Using balanced chemical equation Cx Hy t=0
100 ml
t
0
5 x - y 2 0 O 6 6$ xCO2 4 4 1 2
+ 3
v ml
+
0
5 x - y 2 0 4 4 1
v – 100 3
100 x ml
y HO 2 2 0 100 y 2
volume remained
Page # 31
250 = (volume of reactants) – (volume products – volume of unused reactant) = (100 + v) – {100 x +
100 y + v – 100 {x + y/4}} 2
...(1)
volume of CO2 100 x = 300 x=3 . The above two equations can be solved to get the required answer. Example.
A sample of water gas has a composition by volume of 50% H2, 45% CO and 5% CO 2. Calculate the volume in litre at STP at water gas which on treatment with excess of steam will produce 5 litre H 2. The equation for the reaction is :
6$ CO2 + H2 CO + H2O 6 If x L CO in needed then
Sol.
1 2 x 5 x 2 5 x ' 50% 0 L = 3 ' 0L % L 0.9 4 0.45 1 4 0.45 2 1
volume of H2 in water gas = 3
)
But, from equation : CO + H2O 6 6$ CO2 + H2 & Gay-Lussac’s law, we get, that the v olume of H2 produced = volume of CO taken. Volume of H2 due to reaction = x L
)
Total volume of H2 = 3
. ) )
5 x 2 - x 0 L = 5 L 4 0.9 1
1 .9 x =5L 0 .9 0. 9 ' 5 x= 1 .9
Volume of water gas =
x 0 .9 ' 5 L% L = 5.263 L Ans. 0.45 1.9 ' 0.45
Example.
One litre of oxygen gas is passed through a ozonizer & the final volume of the mixture becomes 820 ml. If this mixture is passed through oil of turpentine. Find final volume of gas remaining.
Sol.
6$ 2O3. The reaction that takes place in ozonizer as 3O2 6 If Vml of O2 out of 1000 ml is ozonized then vol of O3 obtained =
)
. . ) Example.
Sol.
2 V 3
Final vol of ozonized oxygen 2 = (1000 – V + V) mL = 820 mL 3 1 1000 – V = 820 3 V = 180 ) V = 540 3 O2 remaining = (1000 – 540) mL ( " O3 is absorbed in oil of turpentine) = 460 mL Ans.
A gaseous mixture containing CO, methane CH4 & N2 gas has total volume of 40 ml. This mixture is exploded with excess of oxygen on cooling this mixture a contraction of 30 ml is observed & when this mixture is exposed to aqueous KOH a further contraction of 30 ml is observed .find the composition of the mixture. Let vol of CO be x mL vol of CH4 be y mL vol of N2 be z mL On explosion with excess of oxygen the following reactions takes place
Page # 32
CO(g) +
1 O (g) 2 2
$ CO2 (g)
(By Gaylussac`s law of combing volume)
x mL x mL 6$ CO2(g) + 2H2O(g) CH4(g) + O2(g) 6
)
) )
y mL y mL 2 ymL N2 remains unreacted On cooling H2O (g) liquifies hence volume reduction of 30 mL is observed 2y = 30 ) y = 15 But, vol of CO2 obtained = (x + y) mL This is absorbed in KOH & vol reduction of 30 mL is observed. x + y = 30 . x = 30 – y = (30 – 15) = 15 and, x + y + z = 40 . z = 40 – x – y = 40 – 15 – 15 = 10 Composition of mixture is vol of CO = 15 mL Ans. vol of CH4 = 15 mL Ans. vol of N2 = 10 mL Ans.
Page # 33
