http://www.rpmauryascienceblog.com/ HEAT AND THERMODYNAMICS PROBLEM Q.1.
In a steady state, the temperature at the end A and B of 20 cm long rod AB are 100 0C and 00C. Find the temperature of a point 9 cm from A.
Q.2.
A thermally insulated vessel containing a gas whose molar mass is M and adiabatic exponent , moves with a velocity v. Prove that rise in temperature of the gas resulting from the stoppage of the mv 2 vessel is ( 1) . 2R
Q.3.
In a cyclic process shown in the figure an ideal gas is adiabatically taken from B to A, the work done on the gas during the process B A is 30 J,when the gas is taken from A B the heat absorbed by the gas is 20J. Find out the work done by the gas in the process A B.
P
A
B V
Q.4.
The initial and final temperature of water as recorded by an observer are (40.6 0.2)C and (78.3 0.3)C. Calculate the rise in temperature with proper error limits.
Q.5.
Figure shows two paths through a gas can be taken from the state A to the state B. Calculate the work done by the gas in each path.
B
25 cc
A
10 cc
C 10 kPa
30 kPa
Q.6.
In steady state, the temperature at the end A and B of 20 cm long rod AB are 1000C and 00C. Find the temperature of a point 9 cm from A.
Q.7.
A gas is found to obey P2V = constant. The initial temperature and volume of gas are T0 & V0. If the gas expands to volume 3V0. Find the final temperature of gas.
Q.8.
A thermodynamical process is a shown in the figure with PA = 3 105 Pa, vA = 2 10-3 m3, PB = 8 104 Pa, v C = 5 10-3 m3, in the process AB and BC 600 J and 200 J heat is added to the system, respectively. Find the change in internal energy of the system in the process AC.
Q.9.
P
B
C
PB
PA A VA
VC
V
Find the specific heat capacity of mono-atomic ideal gas for thermodynamic process P = v2. Where is positive constant.
http://www.rpmauryascienceblog.com/ Q.10.
Find the molar specific heat (in terms of R) of a diatomic gas while undergoing the following 1 1 process. dQ = dU dW 2 2
Q.11. Three identical metallic rods are arranged as an equilateral triangle ABC as shown in the figure. Rod AB and BC have same thermal conductivity k. Ends A & C are maintained at constant temperature 1000C and 500C. Find the temperature of junction B and conductivity of rod AC such that heat flowing is AB is same as in AC. (provided that only conduction is takes place)
B
A
C 0 50
0
100
Q.12. An ideal gas undergoes a cyclic process ABCD as shown in the figure. Draw the corresponding P-V diagram.
C
V B D A T
Q.13. A vessel contains a mixture of one mole of CO2 and two moles of nitrogen at 300 k. Find the ratio of the average rotational kinetic energy of CO2 molecules to that of N2 molecules. Q.14. An insulated container containing monoatomic gas of molar mass m is moving with velocity v0. If the container is suddenly stopped, find the change in temperature. Q.15. Following graph shows variation of temperature with time of a cooling body. The surrounding temperature is 270 C. Find t.
0
1
50 C 0
48 C
2
Temp. 3
0
40 C
4
0
38 C t =5 min. t
time 0
50 C
Q.16. Find the temperature of junction of three rods of the same dimensions having thermal conductivity arranged as 3k, 2k, and k shown with their ends at 1000C, 500C and 00C?
2k 0
100 C
3k
k
00C
Q.17. An insulated container containing monoatomic gas of molar mass m is moving with a velocity v0. If the container is suddenly stopped, find the change in temperature.
Q.18. An ideal gas, whose adiabatic exponent is equal to is expended so that the amount of heat transferred to the gas is equal to the decrease of internal energy, find (a) the molar heat capacity of the gas in this process.
http://www.rpmauryascienceblog.com/ (b) the equation of the process in the variables T, V. Q.19. A body cools from 620C to 500 in 10 min and 420 C in the next 10 minutes. Find the temperature of the surrounding. Q.20. Two moles of Helium gases ( = 5/3) are initially at temperature 270 C and occupy a volume of 20 litres. The gas is first expanded at constant pressure until the volume doubles. Then it undergoes an adiabatic change until the temperature return to its initial value. (a) Sketch the process on P-V diagram. (b) What are the final value and pressure of the gas. (c) What is the work done by the gas. Q.21. Taking the composition of air to be 75% Nitrogen and 25% oxygen by weight, calculate the velocity of sound through air at STP. Q.22. A heater wire boils water in a given electric kettle in 3 minutes. Another heater wire boils water in the same electric kettle in 7 minutes. Find the time taken when both heater wires are connected in series in the same electric kettle across the given mains. Q.23. A circular hole of radius 2 cm is made in an iron plate at OoC. What will be its radius at 100oC? for iron = 11 10-6 /0C. P 3Po
Q.24. One mole of an ideal monoatomic gas is taken through the thermodynamic process ABC as shown in figure. Find the total heat supplied to the gas.
A
B
Po
C To
2To
T
Q.25. P-V diagram of a cyclic process is given in the figure. This is elliptical process and O is the focus of the ellipse. Find the work
N/m2 4P0
done in the diagram. O P0 v0
8v0
V (m3)
Q.26. What is the heat input needed to raise the temperature of 2 mole of helium gas from 0 0C to 1000C (take R = 8.31) (a) at constant volume ? (b) at constant pressure ? Q.27. A piece of ice of mass 100g and at temperature 00C is put in 200 g of water at 250C. Assuming that the heat is exchanged only between the ice and the water, find the final temperature of the mixture. Latent heat of fusion of ice = 80 cal/g, specific heat capacity of water = 1 cal/g 0C. Q.28. In steady state, the temperature at the end A and B of 20 cm long rod AB are 100 0C and 00C. Find the temperature of a point 9 cm from A.
http://www.rpmauryascienceblog.com/ Q.29. A cubical block of coefficient of linear expansion floats in liquid of coefficient of volume expansion . When temperature of block and liquid is raised by t0C, it is found that immersed portion of cube in liquid remains same. Find the relation between and .
x0
Q.30. A hollow spherical ball of inner radius a and outer radius 2a is made of a uniform material of constant thermal conductivity k. The temperature within the ball is maintained at 2T0 and outside the ball is T0. Find (a) the rate at which heat flows out of the ball in the steady state. 3a (b) the temperature at r = 2 Q.31. Two bodies A and B have thermal emissivity of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are same. The two bodies emit total radiant power at the same rate. The wavelength B corresponding to maximum spectral radiancy in the radiation from B is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A by 1.00 m. If the temperature of A is 5802 k, calculate (a) the temperature of B (b) wavelength B Q.32. The thickness of ice on a lake is 5.0 cm and the temperature of air is – 200C. Calculate how long will it take for the thickness of ice to be doubled. Thermal conductivity of ice = 0.005 cal/ cm sec. density of ice = 0.92 gm/cc. and latent heat of ice is 80 cal/g. Q.33. A cubical tank of water of volume 2m3 is kept at a constant temperature of 65 0C by 2kW heater. At time t = 0 the heater is switched off. Find the time taken by the tank to cool down to 50 0C given the temperature of the room is steady at 150C. density of water = 103kg/m3 and specific heat of water = 1 cal/gm-0C(Assume the tank to behave like a black body and cool according to Newton’s law of cooling) Take 1 KW = 240 Cal/sec Q.34. One mole of an ideal monatomic gas is taken round the cyclic process ABCA as shown. Calculate P 3P0
P0
B
A V0
(i) Work done by the gas (ii) Heat rejected by the gas in the path CA & heat absorbed by the gas in the path AB. (iii) Net heat absorbed by the gas in path BC (iv) Maximum temperature attained by the gas during the cycle.
C 2V0
V
http://www.rpmauryascienceblog.com/ Q.35. One mole of an ideal monoatomic gas is taken along a cycle ABCDA where AB and CD are adiabatic process. BC and DA are isochoric process. If vA/vB = 8, then find efficiency of the cycle.
C
` P
B
D
A V
Q.36. A gaseous mixture has the following volumetric composition per mole of the mixture. He = 0.2, H2 = 0.1, O2 = 0.3, N2 = 0.4 Assuming the mixture to be a perfect gas, determine (a) the apparent molecular weight of the mixture (b) c v and c p for the mixture (c) gas constant per kg of the mixture. Q.37. At 270C two moles of an ideal monoatomic gas occupy a volume V. The gas expands a adiabatically to to a volume 2V. Calculate (i) the final temperature of the gas. (ii) change in its internal energy and (iii) the work done by the gas during this process. Q.38. The speed of sound at N.T.P. in air is 332 m/sec. Calculate the speed of sound in hydrogen at (i) N.T.P. (ii) 8190C temperature and 4 Atmospheric pressure. (Air is 16 times heavier than hydrogen) Q.39. A cylinder with a piston holds a volume V1 = 1000 cm3 of air at an initial pressure p1 = 1.1 105 Pa and temperature T 1 = 300 K. Assume that air behaves like ideal gas. The sequence of changes imposed on the air in the cylinder is shown in the figure. (a) AB the air heated to 375 K at constant pressure. Calculate the new volume, V2. (b) BC the air is compressed isothermally to volume V1. Calculate the new pressure p2. (c) CA the air cools at constant volume to pressure p1. Find the net work done on the air.
C
p2
p1
B
A v1
v2
Q.40. A wall is made of 7.5 cm thick magnesia, surfaced with 0.5 cm thick steel plate on one side & 2.5 cm thick asbestos on the other side. The thermal conductivities of steel, magnesia and asbestos are 52.3, 0.075 & 0.081 W/m-K respectively. If the outer surface temperature of steel plate is 1500C and that of asbestos is 380C find (a) Rate of heat loss per square meter of surface area of wall & (b) Interface temperatures. Q.41. A hollow spherical ball of inner radius a and outer radius 2a is made of a uniform material of constant thermal conductivity K. The temperature within the ball is maintained at 2T0 and outside the ball it is T0.Find, (i) the rate at which heat flows out of the ball in the steady state, 3a (ii) the temperature at r = . Where r is radial distance from 2
T0 a 2T0 2a
K
http://www.rpmauryascienceblog.com/ the centre of shell. Assume steady state condition. Q.42. A body cools down from 500C to 450C in5 minutes and to 400C in another 8 minutes. Find the temperature of the surrounding. Q.43. At 270C two moles of an ideal monoatomic gas occupy a volume V. The gas expands adiabatically to a volume 2V. Calculate (i) the final temperature of the gas. (ii) change in its internal energy and Q.44. A hot body of mass m, specific heat s is initially at temperature T I and obeys Newton's law of cooling when placed in a surrounding of temperature T0. Find (a) the heat lost to the surroundings from t = 0 to the thermal equilibrium condition (b) the time taken for losing 60 % of the total heat lost. Q.45. The temperature of a body falls from 400C to 360C in 5 minutes when placed in a surrounding of constant temperature 160 C. Find the time taken for the temperature of the body to fall from 360C to 320 C. Q.46. A solid copper sphere (density and specific heat C) of radius r at an initial temperature 200k is suspended inside a chamber whose walls are at almost constant temperature at 0 K. Calculate the time required for the temperature of the sphere to drop to 100 K. Q.47. Hot oil is circulated through an insulated container with a wooden lid at the top whose conductivity K = 0.149 J/(m-C-sec), thickness t = 5 mm, emissivity = 0.6. Temperature of the top of the lid is maintained at T = 127. If the ambient temperature T a = 27C. Calculate (a) rate of heat loss per unit area due to radiation from the lid. (b) temperature of the oil.
T= 127C
To
Ta = 27C
17 (Given = 108) 3
Q.48. An insulated vessel fitted with a frictionless movable piston of mass ‘m’ and area of cross section ‘A’ consists of certain amount of monoatomic gas. When the temperature of the gas was T0 the piston was at a height of ‘ 0’ from the bottom of the vessel. Now the gas is heated such that piston raises to a height of 20 from the bottom then (assume that atmospheric pressure is P0) (a) Identify the process? (b) work done by the gas ? (c) final temperature ? and (d) heat absorbed by the gas ?
Hot Oil
P0
m
Piston A
T0
0
~
Q.49. An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving piston of mass M. The piston and the cylinder have equal cross-sectional area A. Atmospheric pressure is Po and the volume of the gas is V o when the piston is in equilibrium. The piston is slightly displaced from the equilibrium position. Assuming the process to be adiabatic, show that the piston executes SHM. Find the angular frequency of oscillation.
http://www.rpmauryascienceblog.com/ copper
Q.50. A copper and a tungsten plate having a thickness 2mm each are riveted together so that at 00C they form a flat bimetallic plate. Find the radius of curvature of the layer common to copper and a tungsten plates at 2000C. The coefficients of linear expansion for copper and tungsten are 1.7 10-5 k-1 and 0.4 10-5 k-1 respectively.
tungsten
Q.51. Three moles of an ideal gas (cp = 7/R) at pressure PA and temperature TA is isothermally expanded to twice is initial volume. It is then compressed at constant pressure to its original volume. Finally the gas is compressed at constant volume to its original pressure PA. (a) Sketch P –V (P on X-axis, V on y-axis) and P–T (P on X-axis, T on y-axis) diagrams for the complete process. (b) Calculate the net work done by the gas and net heat supplied to the gas during the complete process. (ln2 = 0.693). Q.52. A smooth vertical tube having two different sections is open from both ends and equipped with two pistons of different areas as shown. One mole of ideal gas is enclosed between the pistons tied with a non-stretchable thread. The cross-sectional S = 10cm area of the upper piston is 2 greater than that of the lower one. The combined mass of the two pistons is equal to m = 5.0 kg. The outside air pressure is po = 1.0 atm. By how many kelvins must the gas between the pistons be heated to shift the pistons through l = 5.0 cm. [Take g = 9.8 m/s 2]
po
4 P(N/m2)
Q.53. One mole of a monatomic gas is carried along process ABCDEA as shown in the diagram. Find the net work done by gas and heat given in the process.
po
B
C
3 2
A
1
D
O
1
E 2
3
Q.54. An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V while its temperature falls to T/2. (a) How many degrees of freedom do the gas molecules have ? (b) Obtain the work done by the gas during the expansion as a function of initial pressure and volume. Q.55. Two moles of an ideal monatomic gas, initially at pressure p1 and volume V1 undergo an adiabatic compression until its volume is V2. Then the gas is given heat Q at constant volume V2. (a) Sketch the complete process on a p-V diagram. (b) Find the total work done by the gas, the total change in its internal energy and the final change in its external energy and the final temperature of the gas. [Give your answers in terms of p, V1, V2, Q and R]
4
http://www.rpmauryascienceblog.com/ Q.56. A reversible heat engine carries 1 mole of an ideal monoatomic gas around the cycle as shown in the figure. Process 12 takes place at constant volume, process 23 is adiabatic and process 31 takes place at constant temperature. (a) If the pressure at point (1) is one atmosphere find the pressures at points (2) and (3). (b) Compute the heat exchanged, the change in internal energy and the work done for each of the three processes and for the cycle as a whole. (c) Compute the efficiency of the cycle. (Give all answers in terms of R) (ln2 = 0.693)
2
600k
P
1 3 300k V
Q.57. Heat flows radially outwards through a spherical shell of radius R2 and R1 (R2 > R1) and the temperature of inner and outer surfaces are 1 and 2 respectively. Find the radial distance from the centre of the shell at which the temperature is just halfway between 1 and 2. Q.58. Determine the work done by an ideal gas during 1 4 3 2 1. Given P1 = 105 Pa, P0 = 3 105 Pa, P3 = 4 105 Pa and v2 - v1 = 10 litres.
P 3
P3
4
P0
P1
1 v1
Q.59. An ideal monoatomic gas is used to operate an engine. P-V diagram of the cyclic used is shown in the figure. Find the efficiency of the cycle.
2 V
v2
p 3p0
p0
2
1 V0
3
4 2V0
V
Q.60. 2 moles of an ideal monatomic gas undergoes through the following changes in a cyclic process, (i) Isothermal expansion from a volume 0.04 m3 to 0.10 m3 at 870 C (ii) at constant volume, cooling to 270 C. (iii) Isothermal compression at 270 to 0.04 m3 (iv) at constant volume heating to original pressure volume and temperature Then, (a) Draw P-V diagram of the complete cycle (b) Find the heat absorbed by the gas (c) Find the work done by the gas during the complete cycle (d) Find the efficiency of the cycle. (e) Find the change in internal energy of the gas during the complete cycle. Q.61. A solid body X of heat capacity C is kept in an atmosphere whose temperature is TA = 300 k. At time t = 0 the temperature of X is T0 = 400 k. If cools according to Newton’s law of cooling. At time
http://www.rpmauryascienceblog.com/ t1, its temperature is found to be 350 k. At this time t1, the body X is connected to a large box Y at atmospheric temperature TA, through a conducting rod of length L, cross-sectional area A and thermal conductivity k. The heat capacity of Y is so large that any variation in its temperature may be neglected. The cross-sectional area A of the connecting rod is small compared to the surface area of X. Find the temperature of X at time t = 3t1. Q.62. One mole of an ideal monoatomic gas is taken along the cycle ABCDA where AB is isochoric, BC is isobaric, CD is adiabatic and DA is T isothermal. Find the efficiency of the cycle. It is given that C 4 , TA vA 1 and ln 2 = 0.693. v D 16 Q.63. The insulated box shown in figure has an insulated partition which can slide without friction along the length of the box. Initially each of the two chambers of the box has one mole of a monatomic ideal gas ( = 5/3) at a pressure Po, volume Vo and temperature To. The chamber on the left is slowly heated by an electric heater so that its gas, pushing the partition, expands until the final pressure in both the chambers becomes 243Po/32. Determine : (i) The final volume and temperature of the gas in B. (ii) The heat given to the gas in A by the heater.
P
B
C
A
D
A
V
B
Q.64. Find the molar specific heat (in terms of R) of a diatomic gas while undergoing each of the following processes : 1 1 (i) For any infinitesimal part of the first process, dQ = dU dW 2 2 (ii) For the second process : pV2T = constant. Q.65. One mole of a gas is isothermally expanded at 270 C till the volume is doubled. Then it is adiabatically compressed to its original volume. Find the total work done. ( = 1.4 and R = 8.4 J/mol -k). Q.66. An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V while its temperature falls to T/2. (a) How many degrees of freedom do the gas molecules have ? (b) Obtain the work done by the gas during the expansion as a function of initial pressure and volume. Q.67. An electrically heating coil was placed in a calorimeter containing 360 gm of water at 10 0 C. The coil consumes energy at the rate of 90 watt. The water equivalent of the calorimeter and the coil is 40 gm. Calculate what will be the temperature of the water after 10 minutes ? Q.68. A double-pane window used for insulating a room thermally from outside, consists of two glass sheets each of area 1 m2 and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steady state the room-glass interface and the glass-outdoor interface are at constant temperature of 270C and 00C respectively. Calculate the rate of heat flow through the window pane.
http://www.rpmauryascienceblog.com/ Also find the temperatures of other interfaces. Given thermal conductivities of glass and air as 0.8 and 0.08 Wm1 K1 respectively. Q.69. Three rods of material x and three rods of material y are connected a shown in figure. All the rods are of identical length and cross-sectional area. If the end A is maintained at 600 C and the junction E at 100 C, calculate the temperature of junctions B, C and D. The thermal conductivity of x is 0.92 cal/cms C0 and that of y is 0.46 cal/cm s C0. C x
x
0
100C E
60 C x A
y
B y
y D (A)
Q.70. A thermally insulated vessel is divided into two parts by a heat insulating piston which can move in the vessel without friction. The left part of the vessel contains one mole of an ideal mono-atomic gas, and the right part is empty. The piston is connected to the right wall of the vessel through a spring whose length in free state is equal to the length of the vessel. Determine the heat capacity C of the system, neglecting the heat capacities of the vessel, piston, and spring. Q.71. When a block of iron floats in mercury at 00C, a fraction K1 of its volume is submerged, while at temperature 800C fraction K2 is seen to be submerged. If the coefficient of volume expansion of iron is Fe and that of mercury is Hg. Find the ratio K1 / K2. Q.72. Two rods of different metals having the same area of cross-section A, are placed end to end between two massive walls. The first rod has a length 1, co-efficient of linear expansion 1 and Young’s modulus of elasticity Y1. The corresponding quantities for second rod are 2, 2 and Y2 respectively. The temperature of both the rods is now raised by T degrees. Find the force which the rods exert on each other at the higher temperature in terms of the given quantities. Assume, there is no change in the area of cross-section of the rods, the rods do not bend and there is no deformation of the walls. Q.73. Two moles of an ideal monatomic gas is taken through a cycle ABCA as shown in the P-T diagram. During the process AB, pressure and temperature of the gas vary such that PT = constant. If T1 = 300 K, calculate (a) The work done on the gas in the process AB (b) The heat absorbed or released by the gas in each of the processes. Give answers in terms of the gas constant R.
P 2P1
B
P1
C
A T1
2T1
T
http://www.rpmauryascienceblog.com/ Q.74. Two moles of a certain ideal gas at temperature T 0 = 300K were cooled at constant volume so that the gas pressure reduced η = 2 times. Then as a result of constant pressure process, gas expanded till its temperature got back to initial value. Find the total amount of heat absorbed by the gas in this process. Q.75. Two identical sphere with surface area A and emissivity e are connected by a metal rod of length , with high conductivity k and area of crossection a (a << A). Rod does not allow any loss of heat through It's side walls being coated with indulated cover. If initially temperatures of sphere is respectively T1 and T2 and temperature of surrounding is T 0 such that T 1 > T2 > T0 . Temperature difference between sphere and surrounding is small enough to consider Newton's laws of cooling for heat loss through radiation. Find the temperature difference between spheres as function of time. Q.76. A polyatomic gas is initially taken at a pressure P0 and temperature T0 and volume V1 = 500 ml. When the gas is compressed adiabatically to volume V2 = 100 ml the temperature increases from T0 to T1 and the pressure from P0 to P1 . Now the pressure increases from P1 to P2 isochorically and the temperature rises to T 2 . The gas is expanded adiabatically from volume V2 to V1 such that the temperature drops to T3 . Finally the gas pressure drops to P0 isochorically . Find efficiency of this cycle if = 1.33. Q.77. One mole of a mono atomic ideal gas is taken through the cycle shown in figure. A B Adiabatic expansion B C Cooling at constant volume C D Adiabatic compression. D A Heating at constant volume The pressure and temperature at A, B etc,. are denoted by PA, TA; PB, TB etc. respectively.
A B P
D
C V
Given TA = 1000K, PB = (2/3)PA and PC = (1/3)PA. Calculate (a) The work done by the gas in the process A B (b) the heat lost by the gas in the process B C and (c) temperature TD. Given (2/3)2/5 = 0.85 and R = 8.31 J/mol K. Q.78. A fixed mass of gas is taken through a process A B C A. Here A B is isobaric. B C is adiabatic and C A is isothermal. Find (a) Pressure and volume at C (b) work done in the process (take = 1.5)
10
5
B
A
2
P (N/m ) C 1
3
V(m )
4
Q.79. One mole of an ideal gas whose pressure changes with volume as P = v where is a constant, is expanded so that its volume becomes times the original. Find the change in internal energy and heat capacity of the gas.
http://www.rpmauryascienceblog.com/ Q.80. Consider the shown diagram where the two chambers separated by piston-spring arrangement contain equal amounts of certain ideal gas. Initially when the temperatures of the gas in both the chambers are kept at 300 K. The compression in the spring is 1 m. The temperature of the left and the right chambers are now raised to 400 K and 500 K respectively. If the pistons are free to slide, find the final compression in the spring.
300 K 300 K Vaccume 1m
Q.81. One mole of a gas is taken in a cylinder with a movable piston. A resistor R connected to a capacitor through a key is immersed in the gas. Initial potential difference across the plates of the capacitor is 1 equal to 213 3 V. When the key is closed for (2.5 ln 4 ) minutes , the gas expands isobarically and its temperature changes by 22 K . (a) Find the work done by the gas (b) Increment in the internal energy of the gas (c) The value of Q.82. One mole of a monoatomic ideal gas has been subjected to an isochoric-isobaric cycle ABCDA. The temperature at D is T0 = 200 K. Calculate (i) The temperatures at the points A, B, C of the cycle. (ii) The heat absorbed / released in each of the processes AB, BC, CD, DA. (iii) The net work done in the process and the efficiency of the cycle.
300 K 300 K
1m
R = 2k
C = 75mF
p 2p0
p0
A
D V0
B
C 3V0
V
Q.83. Find the number of strokes that the piston of an air pump must make in order to pump a vessel of volume Vc.c from a pressure P1 to P2 if the change in volume corresponding to one stroke is v1cc. Assume that the air in the vessel is in good thermal contact with the surroundings. Q.84. A pendulum clock consists of an iron rod connected to a small heavy bob. If it is designed to keep correct time at 200C, how fast or slow will it go in 24 hours at 400C. (iron = 12 10-6 /0C) Q.85. A vessel of volume 2000 cm3 contains 0.1 mole of oxygen and 0.2 mole of carbon dioxide. If the temperature of the mixture is 300 K, find its pressure. Q.86. At the top of a mountain a thermometer reads 70C and a barometer reads 70 cm of Hg. At the bottom of the mountain they read 270C and 76 cm of Hg respectively. Compare the density of the air at the top with that at the bottom. Q.87. A body of mass 25kg is dragged on a horizontal rough road with a constant speed of 20km/hr. It the coefficient of friction is 0.5, find the heat generated in one hr. if 50% of the heat is absorbed by the body, find the rise in temperature. Specific heat of material of body is 0.1Cal/gm-0C. (4.2 J = 1 Cal., g = 10m/s 2) Q.88. One mole of an ideal gas whose pressure changes with volume as P = v where is a constant, is expanded so that its volume becomes times the original. Find the heat supplied to the gas.
http://www.rpmauryascienceblog.com/ Q.89. A hot body placed in air cooled according to Newton’s law of cooling, the rate of decrease of temperature being k times the temperature difference from the surrounding, starting from t = 0, find the time in which the body will loose 90 % of the maximum heat it can loose.
Q.90. 1 mole of diatomic gas is kept in cylinder of volume V0 at temperature T0 and pressure P0. Now gas is taken through three processes as shown in figure. Process 1 2 follows law P = V, Process 2 3 is adiabatic while process 3 1 is isothermal. Find (a) temperature at state 2 in terms of T0 (b) efficiency of cycle.
SOLUTION Q.1.
Q.2.
dQ 1 (20 /100) = (100- 0) dt k A dQ 1 (9 / 100) (100 dt k A from (i) and (ii), = 550C 1 f nmv 2 nRT 2 2 2 f= 1 from (i) and (ii) mv 2 T = ( 1) 2R
Q.3.
From B to A, 0 = UBA + WBA UBA = + 30 From A to B 20 = UAB + WAB 20 = -30 + WAB WAB = 50
Q.4.
= 2 1 = 78.3 40.6 = 37.7C = (1 +2)
… (i) … (ii)
… (i) … (ii)
P
2
2P0
P0
1
v0
3 2v0
http://www.rpmauryascienceblog.com/ = (0.2 + 0.3) = 0.5 C = (37.7 0.5)C. Q.5.
Q.6.
WAB = Area under A B bounded by volume axis. 1 = 10 103 (25 – 10) 10-6 + (25 – 10) 10-6 20 103 2 = 0.15 + 0.15 = 3.0 J. In path A C B = 30 103 (25 – 10) 10-6 = 0.45 J.
dQ 1 (20 /100) = (100- 0) dt k A dQ 1 (9 / 100) (100 dt k A from (i) and (ii), = 550C
… (i) … (ii)
Q.7.
P2V = constant PV T2 also cons tan t 2 V cons tan t T V 2 -1 T V = constant 3V Tf2 T02 0 T f = 3 T0. V0
Q.8.
WAB = 0 QAB = vAB + WAB 600 = UAB QBC = UBC + WBC 200 = UBC + 8 104 3 10-3 UBC = - 40 UAC = 560
Q.9.
dQ dU dW ndt ndT ndT
Pdv C = Cv + ndT P = v2 PV = nRT PdV + VdP = nRdT also from P = v2 dP = 2V dV PdV + 2V2 dV = nRdT 3P dV = nRdT PdV R ndT 3 3 R 11 C = R R 2 3 6
P
B
C
PB
PA A VA
VC
V
http://www.rpmauryascienceblog.com/ 1 1 dU dW 2 2 Ist law: dQ = dU + dW 1 1 dU + dW = - dU dW dW = - 3 dU 2 2
Q.10. Process: dQ = -
dQ = dU - 3dU = -2dU C=
dQ dU 2 2C V 5R ndT ndT
Q.11. Let and a be the length and cross-section area of each rod. kA (100 T ) ka(T 50) QAB = QBC T = 750C ka(100 75) k a(100 50 ) also if QAB = QAC 25 k = 50 k’ k = k/2 Q.12.
A
D
P C
B
V
nRTfr 2 where fr = rotational degree of freedom ) 1 R T 2 Avg. rotational KE of CO2 = RT 2 2R T 2 Avg. rotational KE of N2 = 2RT 2 1 Ratio = . 2
Q.13. Average rotational K.E. =
Q.14. Applying COE 1 m3 mv 20 nC v T = RT 2 M2 Mv 02 T = 3R Q.15. Using Newton’s law of cooling k(avg surroundin g ) t for 1, 2
http://www.rpmauryascienceblog.com/
50 48 50 40 k 27 5 2
2 k( 49 27) 5 for 3, 4 40 38 40 38 k 27 t 2 2 k(39 27) t on solving (i) and (ii) we get t = 9.1 min.
… (i)
… (ii)
, R2 = 3kA 2kA R3 = kA 100 - = I1R1 , - 50 = R2I2, - 0 – R3 (I1 – I2) 200 0 = C 3
Q.16. R1 =
Q.17. Loss in K.E. of the gas
E =
1 (nm) v 02 , where n = number of moles. 2
If its temperature change by T. 3 1 RT = (nm) v 02 2 2 mv20 T = . 3R
Then n
Q.18. (a) Q = u1 – U2 = -(u2 – u1) = - u nR dQ = - du = -n CvdT = dt 1 dQ R C= dT 1 (b) not available Q.19. For first ten minutes dT 62 50 0 = - 1.2 C/min dt 10
62 50 T = T0 = (56 – T0)0C 10 -kA(56 – T0)0 = -1.2 0C / min. Similarly for next 10 minutes dT 420 500 0 = - 0.8 C/min dt 10
…(i)
http://www.rpmauryascienceblog.com/ 42 50 T = - T 0 = (46 – T0)0C 2 -0.80C / min = -kA (46 – T0)0C dividing (i) and (ii) T0 = 260
Q.20. (a) For an ideal gas nRT 2 8.3 300 P= = N/m2 V 20 105 = 2.5 105 N/m2
…(ii)
B
A
C P
(b) TA = T, TB = 2T at B, P’B = P’A =2.49 105 N/m2 vB = 2vA = 40 10-3 m3, TB = 600 k from TVr-1 = constant at B and C vc 2 1 = 23/2 vB
VA
VB V
vc = 2 2 vB = 2 1.414 40 = 113 NRTC 2 8.3 300 Pc = VC 113.13 10 3 = 0.44 105 N/m2 (c) WAB = 2.49 105 (40 – 20) 10-3 = 4980 J nR 2 8.3 W2 = [300 – 600] [T2 T1] r 1 1 (5 / 3 ) = 7470J Wnet = 4980 + 7470 = 12450 J Q.21. Molecular weight of the mixture is given by
m (m / M) M
75 25 = 28.9 75 25 28 32 of the mixture given by n1 n2 n1 n 2 m 1 1 1 2 1 m = 1.4
M=
velocity of sound v =
v2 .t R v2 R .t k.t H R1 t1 R2 t 2
Q.22. H
RT = 331.3 m/s M
VC
http://www.rpmauryascienceblog.com/ R eq R1
t eq
t eq t1
t R1 R2 .t1 1 2 t1 t1 t 2 10 min s R1 t1
Q.23. R100 R 0 (1 T)
(2 cm) 1 (11x 10 6 / o C)(100 o C) 4
(2 cm)(1 11x 10 ) = 2.0022 cm. Q.24. A B represents an isobaric process, 5 5 QAB = 1 R(2T0 T0 ) RT0 2 2 B C represents an isothermal expansion UBC = 0 3P QBC = 1.R.2T0ln 0 = 2RT0ln3 P0 Q = RT0[2.5 + 2ln3] Q.25. W = area enclosed by the ellipse = ab = (3P0) (7v0) Nm = 21P0v 0 Nm
N/m2 4P0 b
a O P0 v0
8v0
V (m3)
Q.26. (a) At constant volume Q = nCv T = 2 (3R/2)100 = 300 R = 300 8.31= 2493 Joules (b) At constant pressure Q = nCp T = 2 (5R/2) (100) = 500 8.31 = 4155 Joules Q.27. The heat required to melt the ice at 00C = 100 80 = 8000 cal. The heat given by water when it cools down from 250C to 00C = 200 1 25 = 5000 cal. Clearly, the whole of the ice can not be melted, as the required amount of heat is not provided by the water. Therefore, the final temperature of the mixture is 00C. Q.28.
dQ 1 (20 /100) = (100- 0) dt k A dQ 1 (9 / 100) (100 dt k A from (i) and (ii), = 550C
… (i) … (ii)
Q.29. Let A0 be the cross-section of cube and 0 the density of liquid before temperature rise.
http://www.rpmauryascienceblog.com/ After t0C increase in temperature, the density of liquid becomes 0 = (1 t) while new cross-sectional area of cube is, A = A0 (1 + 2t) Since Mg = A0x0 0g . . . (i) where x0 is the length of cube in liquid and M is the mass of cube. Also Mg = Ax0 g . . . (ii) x 0 0 A0 x00g = A0(1 + 2t) g (1 t) = 2.
dQ 1 dr Q.30. dT 2 dt k 4r C 1 dT 1 2 dr 4 k r Integrating, C 1 T = 1 C2 4k r At r = a, T = 2T0 and at r = 2a, T = T0 2a C2 = 0, C1 = 8akT0 T= T0 r dQ (i) 8akT0 dt 4T 3a (ii) T(r = ) 0 2 3 Q.31. (a) eA AA TA4 BA BTB4 0.01 (5802)2 = 0.81 (TB)4 TB = 1934 k. (b) A TA = BTB B - A = 1 m B = 1.5 m. Q.32.
dQ dy dm kA [0 ( 20)] L LA , dt y dt dt 10 t 2 2 10 20k 5 ydy dt , T = 2 L 0 2 5 t = 27600 sec. = 7 hrs. 40 minutes.
Q.33. We have heat supplied by heater = heat lost by tank by radiation under steady state. 2 = k (65 – 15) where k is a constant K = 2/50 = 2(240)/50 = 48/5 Cal/s-0C At any instant if the temperature of the tank be ‘T’ then dT K we have (T 15) dt m.s
http://www.rpmauryascienceblog.com/ or or or
dT K .dt T 15 m.s 50 dT K t dt 65 T 15 m.s 0 K 50 -ln T 15 65 = .t m.s K 65 15 ln = .t m.s 50 15
m.s 50 103 x 2 x 103 10 ln ln 48 K 35 7 5 = 20 hrs. (Approximately) t=
Q.34. (i) work done = area under the curve. 1 = 2v 0 v 0 3P0 P0 = P0v0 2 (ii) heat rejected in the path CA = nCP.dT (constant pressure process) 5 = 1 x R TC TA 2 5R P0 2v 0 P0 v 0 5P0 v 0 = 2 R R 2 heat absorbed in path AB = nCV.dT(constant vol process. 3 3R 3P0 v 0 Pv = 1 x x R TB TA 0 0 2 2 R R = 3 P0v0 (iii) for ABC, Q = W (cyclic process) 5 P0 v 0 3P0 v 0 QBC P0 v 0 2 5 QBC = P0v0 – 3P0v0 + P0 v 0 2 P0 v 0 = 2 Pv (iv) = constant so, when Pv is maximum, T also is maximum. T Pv is maximum for process BC. Hence temperature will be maximum between B & C. Let equation for BC be P = Kv + K1 satisfying both points B and C For B, 3P0 = Kv0 + K1 For C, P0 = K (2v0) + K1 2P0 So K= and K1 = 5P0 v0 2P0 equation for BC becomes P = v 5P0 v0 or
RT 2P0 v 0 5P0 v v0
or T =
P0 v2 5v 2 R v0
http://www.rpmauryascienceblog.com/ dT 0 dv P0 4v 5 0 R v0
for maximum
or
Q.35. =
v=
5v 0 4
P Tmax = 0 R
2 5v 5v 0 1 25 P0 v 0 0 5 = 2 8R 4 v 0 4
Wnet ( Q)sup plied
QBC QDA T TA =1- D QBC TC TB v v Let A P k vB vC
=
1
TA TD 1 TB TC k 1 =1(k ) 1 1 =1= 75 %. ( 8 )2 / 3
Q.36. Table of determination Constituent Molar frac tion
Molecular weight
Mass/mole of mixtur e
Cv
3 R 2 5 R 2 5 R 2 5 R 2
He
0.2
4
0.8
H2
0.1
2
0.2
O2
0.3
32
9.6
N2
0.4
28
11.2
(a) For this mixture mass per mole = 21.8 (b) Specific heat at constant volume for the mixture = (0.3+0.25+0.75+1)R = 2.3 R = 19.09 J mole-1k-1 and cp = cv + R = 27.39 J mol-1 k-1 R 8.3 (c) gas constant per kg = = = 0.4 Jkg-1 k-1 M 21.8
Specific heat per mole of mixture 0.3 R 0.25 R 0.75 R R
http://www.rpmauryascienceblog.com/ Q.37.
(i) TI Viy 1 = Tf Vfy 1 (ii) U = nCv T 3 where Cv = R 2
T2 = 189 k
and T = -111 k
3 (-111) = -2767.2 (J) 2 (iii) In adiabatic proces U = -W W = 2767.2 J V = 2
Q.38. (i) we have velocity of sound in a medium given by at N.T.P.
vH va
a
at N.T.P.
vH va
a = (16)1/2 = 4 H
vH = 4va = 4 (332) = 1328 m/s.
(ii) we also have v
T
v 819 273 819 =2 v0 273 v819 =2v 0 = 2 (1328) = 2656 m/s
Q.39.
(a)
V2 = V1
T2 375 1250cm3 1000 T1 300
p1V2 1.1 105 1250 = 1.375 105 Pa V1 1000 Wnet = W AB + WBC + W CA WAB = p1(V2 - V1) = 1.1 105 (1250 - 1000) 10-6 = 27.5 J 1000 V WBC = p2V1 ln 1 = (1.375 105)(1000 10-6) ln = -30.7 J V2 1250 WCA = 0 Wnet = 27.5 - 30.7 = -3.2 J, Work done on the gas = 3.2 J
(b)
p2 =
(c)
Q.40.
P
T1 T4 dQ L2 L dt L1 3 K1 K2 K3 150 38 q= = 85.58 w/m2 2 0.5 7 . 5 2 . 5 10 52.3 0.075 0.081
K1
0.5 cm
L 0.5 2 and T2 = T1 – q 1 =150 – 85.85 10 52.3 K1
T3
T2
T1
K2
7.5 cm
T4
K3
2.5 cm
http://www.rpmauryascienceblog.com/ = 149.990C L T3 = T 4 + q 3 = 64.40C K3 Q.41. In the steady state, the net outward thermal current is constant, and does not depend on the radial position. dT dQ Thermal current, C1 = K.(4r 2 ) dr dt dT C 1 1 2 dr 4K r C1 1 Integrating, T = C2 4K r At r = a, T = 2T0 and at r = 2a, T = T 0 2a C2 = 0, C1 = 8aKT 0 T= T0 r dQ (i) = 8aKT0 (ii) T(r = 3a/2) = 4T0/3 dt
Q.42.
Q.43.
50 45 k47.5 0 Where 0 is the temperature of the surrounding. 5 45 40 k 42.5 0 8 8 k[ 47.5 0 ] 5 k[ 42.5 0 ] 1.6 [ 42.5 - 0 ] = 47.5 - 0
68 - 1.6 0 = 47.5 - 0
0.6 0 = 20.5
0 = 340.
(i) Ti Vi 1 = Tf Vf 1 (ii) U = nCv T 3 where Cv = R 2
U = 2
T2 = 189 k
and T = -111 k 3 R (-111) = -2767.2 (J) 2
Q.44. Under equilibrium condition temperature of the body = T0 Heat lost = Decrease in internal energy = ms (Ti - T0) dT Newton's law of cooling k(T T0 ) where k is a constant dt
http://www.rpmauryascienceblog.com/ T T0 kt Ti T0 when 60 % of the total heat is lost ms (TI - T) = 0.6 ms (Ti - T0) substituting for T 1 t = ln 2.5. k
ln
Q.45.
d k ( 0 ) dt 36 d k (5 min.) 0 40
ln(5 / 6) 5 min. 32 d k= kt 0 36 k=-
t=
ln( 4 / 5 ) 5 min. ln( 5 / 6)
4 dT Q.46. 4r2T4 = r 3C 3 dt rc dT dt = 3 T 4 100 rc dT 7rc t= 10 6 s 3 200 T 4 72
Q.47. (a)
dQ = A[(T)4 (Ta)4], dt
Rate of heat loss per unit area = 595 watt / m2. (b) Let To be the temperature of the hot oil
KA To T 595 A t
T o 420 K or 147 0C Q.48. (a) Isobaric process
mg ) 0A A 0 A 2 0 A T0 T2
(b) W = P(v2 – v 1) = (P0 + (c)
v1 v 2 T1 T2
T2 = 2T0
mg P0 A 0 5RT0 5 Mg A (d) Q = nCp T = = 0 A 0 RT0 2 2 A
http://www.rpmauryascienceblog.com/ Q.49. For adiabatic process PV = constant. In the equilibrium state, total pressure, Mg P = Po + , and initial volume = Vo A
ΔV Thus P Vo = (P + P)(Vo + V) = (P + P) Vo 1 Vo ΔV P = P Vo Restoring force F = PA = PA
Acceleration a =
ΔV A 2 Mg = 2 Po x Vo A Vo
F A 2 Mg Po x m MV o A
Hence the motion is SHM with angular frequency =
Mg A 2 Po x MVo A
Q.50. C = 0 (1 + C ) t = 0 (1 + T ) R d / 2 1 CT (R d / 2) 1 T T R = 0.77 m
R
Q.51. (a) Using PV = nRT V B
2v A
T TA
vA
A
C
B
A
TA/2 C
PA/2
PA
(b) WAB = PdV = WBC = PdV =
P
PA/2
nRT dv 17.26TA v
PA ( v A 2v A ) 12.45TA 2
WCA = 0 Net work done = 17.26 TA – 12.45 TA = 4.81 TA As initial and final states of the gas are same U = 0 Q = U + W Q = W.
PA
P
http://www.rpmauryascienceblog.com/ PV nR V1 = A1X1 + A2X2 V2 = A1(X1 +) + A2 (X2 - )
Q.52. T =
A1 X1
= V1 + (A1 - A2) = V1 + s
X2
Net upward force due to inner and outer A2 pressure (P1 - P0) A1 - (P1 - P0)A2 = mg (P1 - P0) s = mg mg P1 = + P0 = P 2 (Pressure remains same for equilibrium) s P V P1V1 mg s T = T2 - T1 = 2 2 = P0 nR s nR mg sP0 = = 0.91 K. nR Q.53. Process ABCA is clockwise while process ADEA is anticlockwise in P-V diagram Net work done 1 1 1 Area ABCA - area ADEA = 2 1 1 1 = (J) 2 2 2 During process ABCDEA W = 1/2 J U = 0 (cyclic process) Q = U + W = 1/2 J. Q.54. (a) T1V1 1 T2 V2 1 T TV-1 = (5.66 V)-1 2 ln 2 0.693 -1= = 0.4 ln 5.66 1.7334 = 1.4 2 =1+ F=5 F (b) WA =
nR(TF Ti ) PF VF Pi Vi 1 1
from PV = nRT WA =
PF VF = R
1 PV PV = 1.25 PV. 0.4 2
T PV = 2 2
http://www.rpmauryascienceblog.com/ Q.55. (a)
A B adiabatic compression B C Heating at constant volume
(b) WAB = -
V P2 = P1 1 V2
P1
5/3
2R WAB = P1V1 P1( V1 / V2 )5 / 3 V2 (5 / 3) 1
V2
V 2 / 3 3R P1V1 1 1 2 V2 U = UAB + UBC V 2 / 3 3R =QP1V1 1 1 2 V2 For BC Q = nCv T Q Q T = = 3R 3R 2 2 For point A : P1V1 = 2RTA For Point B : P2V2 = 2RTB . For adiabatic change
V P2 = P1 1 V2 PV PV Further TB = 2 2 2 2 nR 2R V V P1 1 = 2R V2 Final temperature TC = TB + T 5/3 V V Q = 2 P1 1 2R V2 3R 5/3 2 / 3 P V V2 Q = 1 1 . 2R 3R
P1 V1 P2 V2
Q.56. (a) 12 isochoric process 600 = P2 = 2 atm 300 2 3 adiabatic process
C
P2
nR (P1V1 P2 V2 ) 1
n = 2, = 5/3,
P
P1 P2 T1 T2
1
P1- T = constant
V1
V
http://www.rpmauryascienceblog.com/ T P3 = 2 T3
/ 1
5/3
600 1 (5 / 3 ) 2 atm. P2 300 1 P3 = 2-3/2 atm = . 2 2 (b) 1 2 (isochoric process) 2 3 (adiabatic process) W12 = 0 nR W 23 = (T1 - T2) Q12 = n Cv T 1 nR 1 R(600 300) = (T2 - T 1) = r 1 5 / 3 1 R = 450 R units (600 - 300) (5 / 3 ) 1 Q = 0 U = - W = 450R units. Q = U + W U = - 450 R units Q12 = U12 = 450 R units Hence total work done in the cycle W = W12 + W23 + W31 = 138 R (c) Efficiency of the cycle total work done 138R 138 = = = 0.31 Heat absorbed 450R 450
3 1 (isothermal process) W 31 = nRT ln (P3/P1) = 1 R 300 ln 2-3/2 -312 R units U = 0 Q = W Q31 = -312 R units.
Q.57. Rate of heat flow through a concentric shell of radius x and thickness dx is d Q = k4 x2 dx dx 4 k Or d 2 x Q Integrating R 2 dx 4 k 2 d R1 x 2 Q 1
4k 1 2 R 1R 2 R 2 R1 Integrating equation (1) from R1 to r r R1 4k 1 rR 1 Q
or
Q=
. . . (1)
4k 1 R 1r . . . (2) r R1 Equating (1) and (2) and substituting = (1+2)/2, We get r = (2R1R2)/(R1+R2) Q=
Q.58. From the figure v 4 v 3 4 105 3 105 10 3 105 105 v 4 – v3 = 5 litres
http://www.rpmauryascienceblog.com/ 1 1 Now work done, W = 10 2 10 5 5 10 5 10 3 = 750 J. 2 2
Q.59. Process 1 - 2 1W2 = 0, 1Q2 = U2 - U1 = nCv(T 2 - T1) 3R v 0 = n (P2 P1 ) 2 nR = 3P0v0 Process 2 - 3 2W3 = 3P0(2v0 - v 0) = 3P0v0 15 5R 3P0 P0 v 0 ( V3 V2 ) = 2Q3 = nCp (T3 - T2) = n 2 2 nR Process 3 - 4 3W4 = 0, 3Q4 = nCv (T 4 - T3) < 0 as T4 < T3 Process 4 - 1 4W1 = P0 (v0 - 2v0) = - P0v0 as 4Q1 = nCp (T1 - T4) < 0 Net work efficiency = = Heat added =
T1 < T 4 2P0 v 0 Pv 15 0 0 3P0 v 0 2
4 19.04 %. 21
Q.60. (a) The P-V diagram will be as shown, For isothermal expansion A B v Q1 = nRT ln 2 v1
A P
0.10 = 2 8.314 360ln 0.04 = 5483.25J For isochoric process B C
Q2 = nCvT = nCv (Tc – T B) = 2
B D C V
3 R ( 300 – 360) 2
= -3 8.314 60 = -1496.5 J For isothermal compression C D v 0.04 Q3 = nRTln D = 2 8.314 300 ln 0.10 vC = -4569.4J For isochoric process D A 3 Q4 = nCvT = 2 8.314 (360 – 300) 2 = 1496.5 J (b) Heat absorbed by the gas during the cycle
http://www.rpmauryascienceblog.com/ = 5483.25 + 1496.5 = 6979.75J (c) Work done by the gas during the cycle = Q1 + Q2 + Q3 + Q4 = 5483.25 - 1496.5 – 4569.4 + 1496.5 = 913.85 J (d) Efficiency of the cycle =
Net work done Heat supplied
913.1 = 0.131. 6979.75 (e) Change in internal energy of the gas during the complete cycle = 0 =
dQ d d = -C ( heat capacity ms = C) ms dt dt dt and from Newton’s law of cooling d a( 0 ) dt d -C (a – constant , 0 - temperature of surrounding) a( 300) dt t d a dt ( 300) C0 400
Q.61. For cooling,
ln
( 300 ) a = - .t ( 400 300 ) C a .t e c
= 300 + 100 At time t = t1 ; = 350 Hence, a = C ln (2/t1) Now, when the body X is connected to body Y d d dQ + dt dt conduction dt radiation d kA ( 0 ) -C a( 0 ) dt L d kA Q ( 0 ) dt LC C F
kA ln 2 3 t1 d =- dt ( ) LC t 0 1 t! 350
kA ln 2 F 300 2t1 350 300 t1 LC kA ln 2 F = 300 + 50 exp 2t1 t1 LC
or, ln
Q.62. Taking the temperature, pressure and volume at D to be T 0 , P0 and V0 using the relations. TA = T D, PAvA = PDvB for path DA
http://www.rpmauryascienceblog.com/ vA = vB, P B = Pc
PB PA for path AB TB TA vB v c for path BC TB Tc
TD v D 1 TC v c for path CD With the given relations, we can complete the table. P V T A 16 P0 V0/16 T0 B 32P0 V0/16 2T0 C 32P0 V0/8 4T0 D P0 V0 T0 Now efficiency Wcycle Q QBC QCD QDA = = AB Q AB QBC Q AB QBC 3 Here, QAB = Cv (2T 0 – T0) = RT0 2 QBC = CP (4T0 – 2T0) = 5RT0 QCD = 0 (adiabatic) QDA = - RT0 ln 16 = - 4RT0 ln 2 putting the values (3 / 2) 5 4 ln 2 = = 0.573. ( 3 / 2) 5 (i) the final volume and temperature of the gas in B. (ii) the heat given to the gas in A by the heater. Q.63. (i) The process in B is adiabatic ( = 5/3) 243 P0 V05 / 3 = P0 ( VBf )5 / 3 , 32 8 Final volume, VBf V0 27 Pr V f 9 243 TBf B B T0 ( PBf P0 ) R 4 32 3 15 (ii) W = work done on B by A = R (TBf T0 ) = RT0 2 8 8 46 VAf 2V0 V0 = V0 27 27 243 PAf P0 32 207 TfA T0 16 573 U = RT0 . 32 633 The heat supplied by the heater = RT0. 32
http://www.rpmauryascienceblog.com/ Q.64. (i) Process : ist law :
1 1 dU dW 2 2 dQ = dU + dW 1 1 dU + dW = - dU dW 2 2 dQ = -
dW = - 3 dU
dQ = dU - 3dU = -2dU
dQ dU 2 2C V 5R dT dT
C=
(ii) or
PV2T = constant, for the process (PV / T) VT2 = const.
or
VT2 = const.
= A (say)
lnV + 2lnT = lnA dV/V + 2dT/T = 0 dV/dT = - 2V/T Now dQ = dU + PdV C = dQ/dT = CV + P(dV/dT) = CV + P (-2V/T) = CV - 2R = R/2. Q.65. In case of isothermal expansion, work done by one mole of an ideal gas = 2.303 RT log (2v/v) = 2.303 RT log (2) = 1747 J for adiabatic compression, dQ = 0 du = -dw du = nCv dT =Cv dT (n = 1) and T1 v1 1 T2v 2 1 300 (2v)0.4 = T2 (v)0.4 T2 = 300 (2)0.4 = 395.850k du = (5/2)R [ 395.85 - 300 ] = 2012.85 J and dw = - 2012.85 J total work done = 1747 - 2012.85 = - 265.85 J Q.66. (a) T1V1 1 T2 V2 1 T TV-1 = (5.66 V)-1 2 ln 2 0.693 -1= = 0.4 ln 5.66 1.7334 = 1.4 2 =1+ F=5 F
http://www.rpmauryascienceblog.com/ (b) WA =
nR( TF Ti ) PF VF Pi Vi 1 1
from PV = nRT WA =
PF VF = R
T PV = 2 2
1 PV PV = 1.25 PV. 0. 4 2
Q.67. Energy supplied by the heater to the system in 10 minutes: Q1 = P t = (90 J/s) (10 60 s) = 54 kJ i.e., Q1 = (54/4.2)kcal = 12857 cal. Now if T is the final temperature of the system energy absorbed by it to change its temperature from 100 C to T 0C Q2 = (m + W)cT = (360 + 40) 1 (T – 10) cal According to given problem Q1 = Q2 i.e., 400 (T – 10) 12857 or T = 42.140C
Q.68.
In case of thermal conduction as dQ Air G Out Room G KA side dt L R K = 0.08 K=.8 K=.8 L with R = 0.05 m KA 0.01 0 L 1 0.01 0.05 2 0.01 00C 27 C 1 Req = 2 KA A 0.80 0.08 1 5 26 and as here A = 1 m2 , Req = 40 8 40 dQ (27 0) 40 and hence = 41.5 W dt R 26 Now if 1 and 2 are the temperatures of air in contact with glass in the room and outside as shown in figure (27 1) 41.5 = 0.08 1 0.01 ( 0) and 41.5 = 0.80 1 2 0.01 solving these for 1 and 2 we get 1 = 26.480 C and 2 = 0.520 C.
Q.69. Treating the given network of rods in terms of thermal resistance Rx and Ry with L L L Rx = and Ry = as R AK A 0.92 A 0.46 so that if Rx = R, Ry = 2Rx = 2R Now as in this bridge [(P/Q) = (R/S)], so the bridge is balanced, i.e., the temperature of junctions C and D is equal and the rod CD becomes ineffective as no heat will flow through it.
http://www.rpmauryascienceblog.com/ Now as the thermal resistance of the bridge between junction B and E is 1 1 1 4 i.e., RBE = R RBE (R R) (2R 2R ) 3 The total resistance of bridge between A and E will be Req = RAB + RBE = 2R + (4/3)R = (10/3)R So the net rate of flow of heat through the bridge will be dQ (60 10) 15 dt R eq (10 / 3)R R Now if TB is the temperature at B, Q 60 TB dQ dt R 2R AB AB dQ dQ But , dt dt AB
i.e.,
60 TB 15 , i.e., TB = 300 C 2R R
Also at B 15 30 TC 30 TD dQ dQ dQ dt dt dt , i.e., R R 2R AB BC BD and as TC = T D = T, 30 = 3 (30 – T), i.e., TC = TD = 200C 3 nR(T2 T1) 2 k W = ( x 22 x12 ) 2 kx PS P= or, x = & S k nRT or, x2 = k nR W = (T2 T1) 2 Q = U + W = n2R(T 2 – T1) Q C= 2R nT
Q.70. U =
P=
nRT Sx
Q.71. Let v1 be the total volume of iron at 00C and let V1 be the volume submerged in mercury v k1 = 1 V1 at 800C k2 = v2 / V2 also v 2 = v1 { 1 + 80 fe}, V2 = V1 {1 + 80 Hg }
http://www.rpmauryascienceblog.com/
k1 1 80 Hg . k 2 1 80 Fe
Q.72. Increase in length of composite rod due to heating (L)increase = (L11 + L22)T Due to compressive forces from walls, decrease in length L L F (L)decrease = 1 2 Y1 Y2 A As the length of the composite rod remains unchanged, here F L1 L 2 L11 L 2 2 T A Y1 Y2 F=
L2
L1
1y1
2y2
A(L11 L 2 2 )T L1 L 2 Y1 Y2
P
Q.73. n = 2 moles, monatomic T1 = 300 K PT = constant for path AB Since PV = nRT (a) PT = c (say) nRc P= V
2P1
B
P1
1/ 2
A T1
VB
C
2T 1
1/ 2
T
nRc Work done on the gas = - dV V VA
= - 2 nRc VB VA = - 2nR [ TB – TA ] = 1200 R units (b) Process A B work done by the gas = -1200 R units U = 2
3R (-300) = -900R units 2
QAB = {- 900 R + (-1200R)} = -2100 R units(heat is released) Process B C (P = constant) QBC = 2 (5R/2) 300 = 1500 R units (heat is absorbed) Process C A Since UCA = 0 (T = constant) QCA = WCA = nR (2T1) ln (VA/VC) = 1200 R ln (PC/PA) = 831.77 R units. (heat is absorbed) Q.74. First process is constant volume. w = 0 and P T If pressure is reduced to η times then temperature is also reduced to η times. New temperature will be T0/ η
http://www.rpmauryascienceblog.com/ But
dT Q = n.CV.dT = n R γ 1
=
nR T0 1 η nR T0 T0 = γ 1 η ηγ 1
…(i) During second process pressure is constant, P.dv = n.R.dT nR.dT and Q = ΔU W = n.R.dT γ 1
1 γ = n.R.dT 1 = n.R.dT γ 1 γ 1 nRγ T nRγ η 1 T0 T0 0 = = … (ii) γ 1 η η γ 1 Q = Q 1 + Q2 (from (i) and (ii)) nR T0 1 η nRγ η 1 T0 = + ηγ 1 η γ 1 1 = nR.T0 1 η here n =2 R = 8.3 T0 = 300 K & η = 2 Q = 2.5 KJ Q.75. For if at any moment temperature of spheres be 1 and 2 respectively, 1 > 2 and specifice heat for spheres be C For first sphere, rate of loss of heat d1 Ka -C 4Ae T03 (1 - T 0 ) + (1 2 ) dt (Heat loss through (heat loss through radiation) conduction) For IInd sphere d k - C 2 4 AeT03 (2 T0 ) a(1 2 ) (heat gain through conduction) dt substrating equation (ii) from (I) d ) 2ka - C 1 2 4A e T03 (1 - 2) + (1 2 ) dt 2kg = (4A T03 (1 - 2) + (1 - 2) 2ka = (4A e T03 + ) (1 - 2) let 1 - 2 = and H = (4Ae T03 + 2kg/ )
. . (ii)
http://www.rpmauryascienceblog.com/ -C
d H dt
T1 T2
t
dt 0
T T C log 1 2 H -tH/C =e (T1 - T 2) .
t=
Q.76.
d H C
W C Q1 P2 Where W = work done during the complete cycle and Q1 is the heat input WBC = WDA = 0 B P1 W = WAB + WCD P0 nR nR = [ T0 T1] [T2 T3 ] 1 1 V2 nR = [ T0 T1 T2 T3 ] 1 nR R And Q1 = n Cv (T2 - T1 ) = [ T2 T1] Since Cv = 1 1 W T0 T1 T2 T3 T T0 = 1 3 Q1 T2 T1 T2 T1
Efficiency =
T0 V0 T0 V2 1 T2 V2 T3 V1 1
T1 = T 0 v1 / v 2
. . . (1)
. . . (2)
T2 = T 3 v1 / v 2
T1 - T2 = ( v1 / v 2 ) (T0 T3 ) T T2 1 (v1 / v 2 ) T0 T3 1 =1= 0.412 1 V1 V2 Q.77. (a) As for adiabatic change PV = constant
RT i.e. P cons tan t P
i.e.
T P
1
T cons tan t, so B TA 1
2 i.e. TB = TA 3
[asPV RT ]
1
P B PA
2 1000 3
1
2/5
850K
with
5 3
D
A V1
http://www.rpmauryascienceblog.com/ R[TF Ti ] 1 8.31[1000 850 ] 1 [(5 / 3) 1] i.e. WA = (3/2) 8.31 150 = 1869.75 J (b) For B C, V= constant so W = 0 So from first law of thermodynamics Q = U + W = CvT + 0 3 3 or Q 1 R TC 850 as Cv R 2 2 Now along path BC, V = constant ; P T P T (1/ 3)PA T 850 i.e. C C , TC TB B 425 PB TB (2 / 3)PA 2 2 3 So Q = 1 8.31( 425 850 ) 5297 .625 J 2 [Negative heat means, heat is lost by the system] (c) As A and D are on the same isochor PD TD T , i.e., PD PA D PA TA TA But C and D are on the same adiabatic
so WA
TD PD TC PC
1
P T A D PC TA 1
or
T
1
1
P PA T TC A , i.e. TD3 / 5 B 2 (1 / 3)PA 1000 PC TA 1 2 2. / 5 3 2 / 5 1000 i.e., TD 500K 2 3 1000
1/
D
i.e. TD3 / 5
Q.78. (a) For adiabatic process BC PB VB PC VC . . . (1) For isothermal process CA PAVA = PCVC . . . (2) From (1) and (2) 1
V 1 Vc = B = 64 m3 VA
PA VA 105 N/m2 VC 64 (b) Work done, W = WAB + WBC + WCA 1 V = P(VB - VA) + [ PVB – PCVC] + PVA n A 1 VC Putting the values W = 4.9 105 J PC =
Q.79. Say v is the initial volume of the gas. Final volume = v
2/5
(2)
http://www.rpmauryascienceblog.com/ v
v
Work done =
P.dv =
v
2
vdv = (v /2)
v v
v
2
v (2 1) . 2 As pr. varies with volume as P = . v Initial and final pressure are v and v. Change in internal energy ; du = nCv dT = Cv dT for (n = 1) P v P2 v 2 v 2 2 v 2 v 2 (2 1) And also du = 1 1 1 1 ( 1) We have Q - w = u v 2 2 v 2 2 Q=u+w = ( 1) ( 1) 1 2
=
=
2 v 2 2 ( 1) 1 2 1
=
1 v 2 2 ( 1) 2 1
P2 v 2 = 2v2/R R P1V1 V 2 Ti = initial temperature = R R 1 v 2 2 ( 1) 2 Q 1 And heat capacity = Tf Ti (v 2 / R)[ 2 1]
hence T f = final temperature =
C=
Q.80.
R 1 2 1
Let 1 and 2 be the final length of the two parts, then from gas equation P0 A 0 PA 1 PA 2 300 K 300 K 500 K . . .. (i) 400 K T0 T1 T2 Vaccume Considering the equilibrium of the piston in initial and final states, we get 1 2 P0A = k x0, PA = kx. P x . . . .(ii) P0 x 0 decrease in the length of spring = total increase in the length of the two chambers x – x0 = 1 + 2 - 20 . . .. (iii) from relation (I) P T PT 1 = 0 1 0 , 2 = 0 2 0 PT0 PT0 using (ii) x T x T 1 = 0 1 0 , 2 = 0 2 0 xT0 xT0 putting these in (iii)
http://www.rpmauryascienceblog.com/ x – x0 =
x 0 0 (T 1 + T2) – 2 0 xT0
putting values and solving, we get x =
13 1 = 1.3 m. 2
Q.81. Initial charge on the capacitor q0 = CV0 = 75 10-3 213
1 = 16C 3
The charge on the capacitor decays as q = q0 e-t/RC At t = 2.5 ln(4) minutes = 150 ln (4) sec. q = 16 e-ln(4) RC = 150s =4C q02 q2 1.6kJ 2C = heat imparted to the gas (a) Work done by the gas at constant pressure = PV = RT 0.182 kJ
Total heat dissipated in the resistor in the given time =
(b) Increment in the internal energy U = Q - W = 1.6 – 0.182 = 1.418 kJ (c) = 1 +
RT = 1.12 U
Q.82. P0V0 = RT0, T 0 = 200 K PA VA 2P0 V0 = = 400 K R R Similarly, TB = 1200 K, TC = 600 K
(i) PA VA = RTA,
TA =
(ii) QDA = Cv TDA = Cv (T A - TD) 3 = R 200 = 300 R 2 QAB = CP TAB = Cv (TB - TA) = 2000 R QBC = - 900R, QCD = -1000 R. (iii)
W = - 400 R,
=
W 4 . Q 23
Q.83. The expansion is isothermal, hence PV = constant also P.dV + V. dp = 0 dP dv or P V Let pressure decrease by P and volume increase by vI During one cycle. P1 P v v1 dP dv = P v P1 v
http://www.rpmauryascienceblog.com/ P P v v1 ln ln 1 v P1
. . . (i)
at the beginning of the second cycle, pressure becomes P1 - P and volume becomes V again. At the end of second cycle, pressure becomes P1 - 2P
P 2P v v1 = - ln ln 1 v P1 P
. . . (ii)
(ii) can also written as P 2P P1 v v1 ln 1 . ln P1 P v P1 P 2P P P v v1 ln 1 ln or ln 1 P1 v P1 P 2P P P v v1 ln ln or ln 1 P1 v P1 From (i) and (iii) P 2 P v v1 2 ln ln 1 P1 v When Process is repeated 'n' times, P nP v v1 n ln ln 1 P1 v
But P1 - n. P = P2 = final pressure P v v1 ln 2 n ln v P1 and n =
lnP1 / P2 . ln( v v 1 ) / v
Q.84. Time lost t = 43200 T = 43200 1.2 10-5 20 = 10.4 sec. O 2 RT
8.31 300 = 0.1 3 V 2 10 = 1.25 105 Pa Co 2 RT 8.31 300 PCO2 = = 0.2 3 V 2 10 = 2.5 105 Pa Total pressure = PO 2 PCO 2
Q.85. PO 2 =
= (1.25 105 + 2.5 105) Pa = 3.75 105 Pa.
. . . (iii)
http://www.rpmauryascienceblog.com/ Q.86. PV = nRT =
m RT M
P R = T M P P T Top T bottom T PT TB 70 300 75 = 0.9868. B PB TT 76 280 76
Q.87. f μN (Kinetic Friction) = mg Work done against friction = mg(s) ‘s’ is the distance moved by body. mg(s) = 0.5(25) 10 (20) (103) = (2.5) 106J 2.5 x 10 6 = 595 x 103 calories. Heat generated in calories = 4.2 Heat absorbed = 50% of (595) 103 = (297.5) 103 calories But Q = m.s. Δ t 297.5 x 103 = 25 x 0.1 x 103 x Δ t Δ t = 118.80C Q.88. Say v is the initial volume of the gas. Final volume = v v
v
Work done =
P.dv =
v
2
vdv = (v /2)
v v
v
2
v (2 1) . 2 As pr. varies with volume as P = . v Initial and final pressure are v and v. Change in internal energy ; du = nCv dT = Cv dT for (n = 1) P v P2 v 2 v 2 2 v 2 v 2 (2 1) And also du = 1 1 1 1 ( 1)
=
Q =u+w =
v 2 2 v 2 2 ( 1) ( 1) 1 2
2 v 2 2 ( 1) 1 2 1 2 1 v = (2 1) . 2 1
=
Q.89. The heat lost by body Q = ms (1 - 0) Let 2 be the temperature after loosing 90 % of Q 90 Q ms(1 2 ) 100 0.9(1 - 0) = (1 - 2) 2 = 0 = 0.1 (1 - 0)
http://www.rpmauryascienceblog.com/ 2 0 0. 1 1 0 According to Newton’s law of cooling d - 2 k( 0 ) dt ( 0 ) ln 2 kt ( 1 0 )
t=-
1 ln 10 ln(1/ 10) . k k
Q.90. For process 1 2 P=V Since PV = nRT V2 = RT T 2 = 4T0
P 2P0
2v0
P0
W 12 =
v0
Pdv
4v 20 v 20 3 2 3 v 0 RT0 2 2 2
15RT0 5 U12 = Cv T = R(3T0 ) 2 2 Q12 = 9RT0 for process 2 3 (TV-1 = constant v3 = 64 v0) Q23 = 0 U23 = Cv T = - Cv(3T0) 15 15 U23 = RT0 W23 = - U23 = RT0 2 2 for process 3 1 v U31 = 0 , Q31 = W 31 = - RT 0 ln 3 v1 Q31 = - RT0 ln 64 15 3 RT0 RT0 RT0 ln 64 work done 2 2 100 = 100 heat input 9RT0 9 6ln 2 3 2ln 2 = 100 100 = 53.8 %. 9 3
2
1
v0
3 2v0
