JEE Physics Thermodynamics MCQs Set 25

Physics Objective Questions and Answers: Thermodynamics

Explore reliable objective questions for Thermodynamics tailored for JEE learners. Utilizing these Physics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Download Thermodynamics MCQs with Answers

Access the complete set of multiple-choice questions for Thermodynamics below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: A thermodynamic system is taken from an initial state \( i \) with internal energy \( U_i=100 \) J to the final state \( f \) along two different paths \( iaf \) and \( ibf \). The path \( iaf \) goes from \( i \) to \( a \) at constant volume and then from \( a \) to \( f \). The work done by the system along the paths \( af \), \( ib \) and \( bf \) are \( W_{af}=200 \) J, \( W_{ib}=50 \) J and \( W_{bf}=100 \) J. The heat supplied along \( iaf \) is \( Q_{iaf}=500 \) J and the internal energy in state \( b \) is \( U_b=200 \) J. The ratio \( Q_{bf}/Q_{ib} \) is
(a) 2 : 1
(b) 1 : 2
(c) 3 : 1
(d) 1 : 3
Show Answer & Explanation

Answer: (a) 2 : 1

Explanation:
1. \( W_{iaf}=W_{af}=200 \) J (no work from \( i \) to \( a \)), so \( \Delta U_{if}=500-200=300 \) J and \( U_f=400 \) J.
2. \( Q_{bf}=(400-200)+100=300 \) J and \( Q_{ib}=(200-100)+50=150 \) J.
3. Ratio \( =2:1 \).

Teacher's Note:
1. The change in internal energy depends only on the end states.

Question 2: One mole of a diatomic ideal gas undergoes a cyclic process \( ABC \) in which \( AB \) is at constant volume, \( BC \) is adiabatic and \( CA \) is a return to \( A \). The temperatures at \( A \), \( B \) and \( C \) are 400 K, 800 K and 600 K respectively. Choose the correct statement.
(a) The change in internal energy in the whole cyclic process is 250\( R \)
(b) The change in internal energy in the process \( CA \) is 700\( R \)
(c) The change in internal energy in the process \( AB \) is \( -350R \)
(d) The change in internal energy in the process \( BC \) is \( -500R \)
Show Answer & Explanation

Answer: (d) The change in internal energy in the process \( BC \) is \( -500R \)

Explanation:
1. \( \Delta U=\tfrac52R\,\Delta T \) for one mole of a diatomic gas.
2. \( BC \): \( \tfrac52R(600-800)=-500R \). Whole cycle: 0; \( CA \): \( -500R \); \( AB \): \( +1000R \).

Teacher's Note:
1. Internal energy is a state function: \( \Delta U=0 \) for a complete cycle.

Question 3: In a thermodynamic process, the pressure of a fixed mass of a gas is changed in such a manner that the gas releases 20 J of heat and 8 J of work is done on the gas. If the initial internal energy of the gas was 30 J, the final internal energy will be
(a) 2 J
(b) 42 J
(c) 18 J
(d) 58 J
Show Answer & Explanation

Answer: (c) 18 J

Explanation:
1. \( Q=-20 \) J and \( W=-8 \) J.
2. \( \Delta U=Q-W=-20+8=-12 \) J, so \( U_f=30-12=18 \) J.

Teacher's Note:
1. Work done on the gas is negative work by the gas.

Question 4: An ideal gas is contained in a thermally insulated and rigid container and it is heated through a resistance 100 \( \Omega \) by passing a current of 1 A for five minutes. The change in internal energy of the gas is
(a) zero
(b) 30 kJ
(c) 10 kJ
(d) 20 kJ
Show Answer & Explanation

Answer: (b) 30 kJ

Explanation:
1. Rigid, so \( W=0 \); the heating is electrical work done on the gas, \( \Delta U=I^2Rt \).
2. \( \Delta U=1\times100\times300=30 \) kJ.

Teacher's Note:
1. Joule heating supplies \( I^2Rt \) of energy.

Question 5: The \( p\)-\(V \) diagram represents the cycle of an engine operating with an ideal monoatomic gas: \( A(V_0,p_0)\to B(V_0,2p_0)\to C(2V_0,2p_0)\to D(2V_0,p_0)\to A \). The amount of heat extracted from the source in a single cycle is
(a) \( p_0V_0 \)
(b) \( \dfrac{13}2p_0V_0 \)
(c) \( \dfrac{11}2p_0V_0 \)
(d) \( 4p_0V_0 \)
Show Answer & Explanation

Answer: (b) \( \dfrac{13}2p_0V_0 \)

Explanation:
1. Heat is absorbed along \( AB \) (isochoric) and \( BC \) (isobaric).
2. \( Q_{AB}=\tfrac32(2p_0V_0-p_0V_0)=\tfrac32p_0V_0 \); \( Q_{BC}=\tfrac52(4p_0V_0-2p_0V_0)=5p_0V_0 \).
3. Total \( =\dfrac{13}2p_0V_0 \).

Teacher's Note:
1. Use \( C_V=\tfrac32R \) and \( C_P=\tfrac52R \) for the monatomic gas.

Question 6: An ideal gas is taken around the clockwise cycle \( PQRP \) with \( P(p,V) \), \( Q(3p,3V) \) and \( R(p,3V) \) in the \( p\)-\(V \) diagram. The work done during the cycle is
(a) zero
(b) \( \tfrac12pV \)
(c) \( pV \)
(d) \( 2pV \)
Show Answer & Explanation

Answer: (d) \( 2pV \)

Explanation:
1. Work equals the area of the triangle: \( \tfrac12\times(3V-V)\times(3p-p)=2pV \).

Teacher's Note:
1. Clockwise cycle on a \( p\)-\(V \) diagram means positive work.

Question 7: A sample of ideal monoatomic gas is taken round the clockwise cycle \( ABCA \) with \( A(V,p) \), \( B(3V,4p) \) and \( C(3V,p) \). The work done during the cycle is
(a) \( 3pV \)
(b) zero
(c) \( 9pV \)
(d) \( 6pV \)
Show Answer & Explanation

Answer: (a) \( 3pV \)

Explanation:
1. Area of the triangle \( =\tfrac12\times(3V-V)\times(4p-p)=3pV \).

Teacher's Note:
1. The net work is the enclosed area.

Question 8: 100 g of water is heated from 30\(^\circ\)C to 50\(^\circ\)C. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J/kg/K)
(a) 8.4 kJ
(b) 84 kJ
(c) 2.1 kJ
(d) 4.2 kJ
Show Answer & Explanation

Answer: (a) 8.4 kJ

Explanation:
1. \( \Delta U=mc\Delta T=0.1\times4184\times20=8368 \) J \( \approx8.4 \) kJ.

Teacher's Note:
1. With negligible expansion, no work is done.

Question 9: One mole of an ideal gas having initial volume \( V \), pressure \( 2p \) and temperature \( T \) undergoes the cycle \( ABCDA \): \( A(2p,T)\to B(2p,2T) \) isobaric, \( B\to C(p,2T) \) isothermal, \( C\to D(p,T) \) isobaric, \( D\to A \) isothermal. The net work done in the complete cycle is
(a) zero
(b) \( \tfrac12RT\ln2 \)
(c) \( RT\ln2 \)
(d) \( \tfrac32RT\ln2 \)
Show Answer & Explanation

Answer: (c) \( RT\ln2 \)

Explanation:
1. \( W_{AB}=R(2T-T)=RT \) and \( W_{CD}=R(T-2T)=-RT \).
2. \( W_{BC}=R(2T)\ln\dfrac{2p}{p}=2RT\ln2 \) and \( W_{DA}=RT\ln\dfrac{p}{2p}=-RT\ln2 \).
3. Net \( =RT\ln2 \).

Teacher's Note:
1. The isobaric parts cancel; the isothermal parts leave a net work.

Question 10: An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume \( V_1 \) and contains ideal gas at pressure \( p_1 \) and temperature \( T_1 \). The other chamber has volume \( V_2 \) and contains ideal gas at pressure \( p_2 \) and temperature \( T_2 \). If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be
(a) \( \dfrac{T_1T_2(p_1V_1+p_2V_2)}{p_1V_1T_2+p_2V_2T_1} \)
(b) \( \dfrac{p_1V_1T_1+p_2V_2T_2}{p_1V_1+p_2V_2} \)
(c) \( \dfrac{p_1V_1T_2+p_2V_2T_1}{p_1V_1+p_2V_2} \)
(d) \( \dfrac{T_1T_2(p_1V_1+p_2V_2)}{p_1V_1T_1+p_2V_2T_2} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{T_1T_2(p_1V_1+p_2V_2)}{p_1V_1T_2+p_2V_2T_1} \)

Explanation:
1. \( n_1=\dfrac{p_1V_1}{RT_1} \), \( n_2=\dfrac{p_2V_2}{RT_2} \); energy is conserved: \( T=\dfrac{n_1T_1+n_2T_2}{n_1+n_2} \).
2. \( T=\dfrac{p_1V_1+p_2V_2}{\frac{p_1V_1}{T_1}+\frac{p_2V_2}{T_2}}=\dfrac{T_1T_2(p_1V_1+p_2V_2)}{p_1V_1T_2+p_2V_2T_1} \).

Teacher's Note:
1. Total internal energy is the sum \( \propto nT \).

Question 11: A thermodynamic cycle \( ABCDA \) has \( A \) and \( B \) at \( 2\times10^5 \) N m\(^{-2}\) with volumes 2.0 cm\(^3\) and 4.0 cm\(^3\), and \( C \), \( D \) at \( 1\times10^5 \) N m\(^{-2}\) (\( A\to B \) isobaric, \( B\to C \) isochoric, \( C\to D \) isobaric, \( D\to A \) isochoric). Identify the incorrect statement.
(a) 0.4 J of work is done by the gas from \( A \) to \( B \)
(b) No work is done by the gas from \( B \) to \( C \)
(c) Net work done by the gas in one cycle is 0.2 J
(d) Work is done by the gas in going from \( B \) to \( C \) and on the gas from \( D \) to \( A \)
Show Answer & Explanation

Answer: (d) Work is done by the gas in going from \( B \) to \( C \) and on the gas from \( D \) to \( A \)

Explanation:
1. \( W_{AB}=2\times10^5\times2\times10^{-6}=0.4 \) J; \( W_{CD}=-0.2 \) J; net \( =0.2 \) J.
2. \( BC \) and \( DA \) are isochoric, so no work is done in them; the last statement is false.

Teacher's Note:
1. Isochoric steps do no work.

Question 12: When a system is taken from state \( i \) to state \( f \) along the path \( iaf \), it is found that \( Q=50 \) cal and \( W=20 \) cal. Along the path \( ibf \), \( Q=36 \) cal. \( W \) along the path \( ibf \) is
(a) 6 cal
(b) 16 cal
(c) 66 cal
(d) 14 cal
Show Answer & Explanation

Answer: (a) 6 cal

Explanation:
1. \( \Delta U=50-20=30 \) cal for both paths.
2. \( W_{ibf}=Q-\Delta U=36-30=6 \) cal.

Teacher's Note:
1. \( \Delta U \) is path independent.

Question 13: A thermodynamical system is changed from state \( (p_1,V_1) \) to \( (p_2,V_2) \) by two different processes. The quantity which will remain the same will be
(a) \( \Delta Q \)
(b) \( \Delta W \)
(c) \( \Delta Q+\Delta W \)
(d) \( \Delta Q-\Delta W \)
Show Answer & Explanation

Answer: (d) \( \Delta Q-\Delta W \)

Explanation:
1. \( \Delta U=\Delta Q-\Delta W \) depends only on the end states.

Teacher's Note:
1. Heat and work individually depend on the path.

Question 14: 310 J of heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from 25\(^\circ\)C to 35\(^\circ\)C. The amount of heat required to raise the temperature of the gas through the same range at constant volume is (\( R=8.31 \) J/mol K)
(a) 384 J
(b) 144 J
(c) 276 J
(d) 452 J
Show Answer & Explanation

Answer: (b) 144 J

Explanation:
1. \( nC_P\Delta T=310\Rightarrow C_P=15.5 \) J/mol K.
2. \( C_V=15.5-8.31=7.19 \); \( Q_V=2\times7.19\times10=144 \) J.

Teacher's Note:
1. \( C_P-C_V=R \).

Question 15: The \( p\)-\(V \) diagram of 2 g of helium gas for a certain process \( A\to B \) is a straight line from \( A(V_0,p_0) \) to \( B(2V_0,2p_0) \). What is the heat given to the gas during the process \( A\to B \)?
(a) \( 4p_0V_0 \)
(b) \( 6p_0V_0 \)
(c) \( 4.5p_0V_0 \)
(d) \( 2p_0V_0 \)
Show Answer & Explanation

Answer: (b) \( 6p_0V_0 \)

Explanation:
1. \( \Delta U=\tfrac32(p_BV_B-p_AV_A)=\tfrac32(4p_0V_0-p_0V_0)=4.5p_0V_0 \).
2. \( W=\) area under the line \( =\tfrac12(p_0+2p_0)V_0=1.5p_0V_0 \).
3. \( Q=\Delta U+W=6p_0V_0 \).

Teacher's Note:
1. For a monatomic gas \( \Delta U=\tfrac32\Delta(pV) \).

Question 16: 1 cm\(^3\) of water at its boiling point absorbs 540 cal of heat to become steam with a volume of 1671 cm\(^3\). If the atmospheric pressure \( =1.013\times10^5 \) N m\(^{-2}\) and the mechanical equivalent of heat \( =4.19 \) J cal\(^{-1}\), the energy spent in this process in overcoming intermolecular forces is
(a) 540 cal
(b) 40 cal
(c) zero
(d) 500 cal
Show Answer & Explanation

Answer: (d) 500 cal

Explanation:
1. \( W=p\Delta V=1.013\times10^5\times1670\times10^{-6}=169 \) J \( =40 \) cal.
2. \( \Delta U=540-40=500 \) cal, which overcomes intermolecular forces.

Teacher's Note:
1. Latent heat splits into expansion work and internal energy change.

Question 17: Two cylinders A and B fitted with pistons contain equal numbers of moles of an ideal monoatomic gas at 400 K. The piston of A is free to move while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 42 K, then the rise in temperature of the gas in B is (\( \gamma=5/3 \))
(a) 25.2 K
(b) 35 K
(c) 42 K
(d) 70 K
Show Answer & Explanation

Answer: (d) 70 K

Explanation:
1. A is at constant pressure and B at constant volume: \( nC_P(42)=nC_V\Delta T_B \), so \( \Delta T_B=\gamma\times42=70 \) K.

Teacher's Note:
1. Same heat, so \( C_P\Delta T_A=C_V\Delta T_B \).

Question 18: Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature \( T_0 \), while box B contains one mole of helium at temperature \( (7/3)T_0 \). The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature (ignore the heat capacity of boxes). Then the final temperature of the gases \( T_f \) in terms of \( T_0 \) is
(a) \( T_f=\tfrac37T_0 \)
(b) \( T_f=\tfrac73T_0 \)
(c) \( T_f=\tfrac32T_0 \)
(d) \( T_f=\tfrac52T_0 \)
Show Answer & Explanation

Answer: (c) \( T_f=\tfrac32T_0 \)

Explanation:
1. Energy conservation: \( \tfrac52RT_0+\tfrac32R\cdot\tfrac73T_0=\left(\tfrac52+\tfrac32\right)RT_f \).
2. \( 6T_0=4T_f\Rightarrow T_f=\tfrac32T_0 \).

Teacher's Note:
1. Nitrogen is diatomic (\( C_V=\tfrac52R \)); helium is monatomic (\( C_V=\tfrac32R \)).

Question 19: An ideal monoatomic gas is taken around the cycle \( ABCD \) (a rectangle with pressures \( p \) and \( 2p \) and volumes \( V \) and \( 2V \)) in the \( p\)-\(V \) diagram. The work done during the cycle is given by
(a) \( \tfrac12pV \)
(b) \( pV \)
(c) \( 2pV \)
(d) \( 4pV \)
Show Answer & Explanation

Answer: (b) \( pV \)

Explanation:
1. Work is the area of the rectangle: \( (2p-p)(2V-V)=pV \).

Teacher's Note:
1. The net work equals the enclosed area.

Question 20: A perfect gas goes from a state A to another state B by absorbing \( 8\times10^5 \) J of heat and doing \( 6.5\times10^5 \) J of external work. It is now transferred between the same two states in another process in which it absorbs \( 10^5 \) J of heat. Then
(a) work done on the gas is \( 0.5\times10^5 \) J
(b) work done by the gas is \( 0.5\times10^5 \) J
(c) work done on the gas is \( 10^5 \) J
(d) work done by the gas is \( 10^5 \) J
Show Answer & Explanation

Answer: (a) work done on the gas is \( 0.5\times10^5 \) J

Explanation:
1. \( \Delta U=8\times10^5-6.5\times10^5=1.5\times10^5 \) J.
2. Second process: \( W=Q-\Delta U=10^5-1.5\times10^5=-0.5\times10^5 \) J, i.e. work is done on the gas.

Teacher's Note:
1. Same end states give the same \( \Delta U \).

Question 21: The ratio of specific heats of a gas is \( \gamma \). The change in internal energy of one mole of the gas, when the volume changes from \( V \) to \( 2V \) at constant pressure \( p \), is
(a) \( \dfrac{\gamma-1}{pV} \)
(b) \( pV \)
(c) \( \dfrac{pV}{\gamma-1} \)
(d) \( \dfrac{pV}{\gamma} \)
Show Answer & Explanation

Answer: (c) \( \dfrac{pV}{\gamma-1} \)

Explanation:
1. \( p\Delta V=R\Delta T=pV \), so \( \Delta T=pV/R \).
2. \( \Delta U=C_V\Delta T=\dfrac{R}{\gamma-1}\cdot\dfrac{pV}R=\dfrac{pV}{\gamma-1} \).

Teacher's Note:
1. \( C_V=R/(\gamma-1) \).

Question 22: The work done by a system in going from \( A \) to \( B \) to \( C \) (\( A\to B \) at constant pressure, \( B\to C \) at constant volume) is 30 J and 40 J of heat is given to the system between \( A \) and \( C \). The change in internal energy of the system between \( A \) and \( C \) is
(a) 10 J
(b) 70 J
(c) 84 J
(d) 134 J
Show Answer & Explanation

Answer: (a) 10 J

Explanation:
1. \( \Delta U=Q-W=40-30=10 \) J.

Teacher's Note:
1. The state function \( \Delta U \) depends only on \( A \) and \( C \).

Question 23: The specific heat of solids at constant volume (\( C_V \)) increases with temperature and its value becomes \( 3R \) at large temperatures for all solids. This is a statement of
(a) Dulong and Petit law
(b) Newton's law of cooling
(c) Wien's law
(d) Stefan's law
Show Answer & Explanation

Answer: (a) Dulong and Petit law

Explanation:
1. The molar heat capacity of solids approaches \( 3R \) at high temperature (Dulong and Petit).

Teacher's Note:
1. It follows from equipartition for 3 oscillator degrees of freedom.

Question 24: A gas at state A changes to state B through path I and path II. The change in internal energy is \( \Delta U_1 \) and \( \Delta U_2 \) respectively. Then
(a) \( \Delta U_1>\Delta U_2 \)
(b) \( \Delta U_1<\Delta U_2 \)
(c) \( \Delta U_1=\Delta U_2 \)
(d) \( \Delta U_1=\Delta U_2=0 \)
Show Answer & Explanation

Answer: (c) \( \Delta U_1=\Delta U_2 \)

Explanation:
1. Internal energy is a state function, so \( \Delta U \) is the same for both paths.

Teacher's Note:
1. Only the work and heat depend on the path.

Question 25: Two moles of an ideal helium gas are in a rubber balloon at 30\(^\circ\)C. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to 35\(^\circ\)C. The amount of heat required in raising the temperature is nearly (take \( R=8.31 \) J/mol K)
(a) 62 J
(b) 104 J
(c) 124 J
(d) 208 J
Show Answer & Explanation

Answer: (d) 208 J

Explanation:
1. The balloon expands at constant (atmospheric) pressure.
2. \( Q=nC_P\Delta T=2\times\tfrac52\times8.31\times5=208 \) J.

Teacher's Note:
1. Helium is monatomic: \( C_P=\tfrac52R \).

Question 26: During an adiabatic process, the cube of the pressure is found to be inversely proportional to the fourth power of the volume. Then the ratio of specific heats is
(a) 1
(b) 1.33
(c) 1.67
(d) 1.4
Show Answer & Explanation

Answer: (b) 1.33

Explanation:
1. \( p^3\propto V^{-4}\Rightarrow pV^{4/3}=\) constant.
2. So \( \gamma=\tfrac43=1.33 \).

Teacher's Note:
1. Adiabatic: \( pV^\gamma=\) constant.

Question 27: 5.6 L of helium gas at STP is adiabatically compressed to 0.7 L. Taking the initial temperature to be \( T_1 \), the work done in the process is
(a) \( -\dfrac98RT_1 \)
(b) \( \dfrac32RT_1 \)
(c) \( \dfrac{15}8RT_1 \)
(d) \( \dfrac92RT_1 \)
Show Answer & Explanation

Answer: (a) \( -\dfrac98RT_1 \)

Explanation:
1. \( n=\dfrac{5.6}{22.4}=0.25 \) mol; \( T_2=T_1\left(\dfrac{5.6}{0.7}\right)^{2/3}=4T_1 \).
2. \( W=-nC_V\Delta T=-0.25\times\tfrac32R\times3T_1=-\dfrac98RT_1 \).

Teacher's Note:
1. Work done on the gas appears as a negative sign.

Question 28: An ideal gas is compressed isothermally until its pressure is doubled and then allowed to expand adiabatically to regain its original volume (\( \gamma=1.4 \) and \( 2^{-1.4}=0.38 \)). The ratio of the final to initial pressure is
(a) 0.76 : 1
(b) 1 : 1
(c) 0.66 : 1
(d) 0.86 : 1
Show Answer & Explanation

Answer: (a) 0.76 : 1

Explanation:
1. After the isothermal step: pressure \( 2p \), volume \( V/2 \).
2. Adiabatic back to \( V \): \( p_f=2p\left(\tfrac12\right)^{1.4}=2\times0.38p=0.76p \).

Teacher's Note:
1. Apply \( pV^\gamma=\) constant.

Question 29: A diatomic ideal gas is compressed adiabatically to \( \dfrac1{32} \) of its initial volume. If the initial temperature of the gas is \( T_i \) (in kelvin) and the final temperature is \( aT_i \), the value of \( a \) is
(a) 4
(b) 6
(c) 5
(d) 9
Show Answer & Explanation

Answer: (a) 4

Explanation:
1. \( TV^{\gamma-1} \) constant with \( \gamma=\tfrac75 \): \( a=32^{2/5}=2^2=4 \).

Teacher's Note:
1. \( 32=2^5 \).

Question 30: Which property of the system does not change in an adiabatic process?
(a) volume
(b) temperature
(c) density
(d) total amount of heat
Show Answer & Explanation

Answer: (d) total amount of heat

Explanation:
1. In an adiabatic process no heat enters or leaves, so the total heat exchanged remains zero.

Teacher's Note:
1. Volume, temperature and density can all change.

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