JEE Physics Thermodynamics MCQs Set 15

Answer: (b) impossible

Explanation:
1. Maximum efficiency (Carnot) = \( 1-\dfrac{300}{400}=25\% \).
2. No engine can exceed this, so 26 % is impossible.

Teacher's Note:
1. Use kelvin temperatures.
2. Carnot efficiency is the upper limit between two reservoirs.

Answer: (a) the second law of thermodynamics

Explanation:
1. This is the Clausius form of the second law.
2. The first law only requires energy to be conserved and would allow such a flow.

Teacher's Note:
1. The second law gives the direction of natural heat flow.
2. Refrigerators need external work to reverse it.

Answer: (d) \( \dfrac{10}{3}\times10^5 \) N/m²

Explanation:
1. Total internal energy is conserved: \( \tfrac32(P_AV_A+P_BV_B)=\tfrac32P(V_A+V_B) \).
2. \( P=\dfrac{2\times10^5\times2+4\times10^5\times4}{6}=\dfrac{2\times10^6}{6} \).
3. \( P=\dfrac{10}{3}\times10^5 \) N/m² (final temperature about 339 K).

Teacher's Note:
1. For an ideal gas \( U=\tfrac32PV \) in terms of pressure and volume.
2. The initial temperatures are not needed for the pressure.

Answer: (c) 500 J

Explanation:
1. \( W=\int_4^6(30V+100)\,dV=15(36-16)+100(2) \).
2. \( =300+200=500 \) J.

Teacher's Note:
1. Integrate the pressure over the volume change.
2. Check using the average pressure \( (220+280)/2=250 \) Pa times 2 m³.

Answer: (b) 200 J

Explanation:
1. Isobaric: \( V_2=10\times\dfrac{348}{290}=12 \) L.
2. \( W=P\Delta V=10^5\times2\times10^{-3}=200 \) J \( =2\times10^2 \) J.

Teacher's Note:
1. Use kelvin for the volume ratio.
2. 1 L = \( 10^{-3} \) m³.

Answer: (a) 1.5 J

Explanation:
1. \( P_1=\dfrac{RT}{V}=\dfrac{830}{0.83}=1000 \) Pa, so the initial spring force is \( PA=10 \) N, a compression \( x_1=0.1 \) m.
2. The spring is compressed further to \( x_2=0.2 \) m.
3. The gas work equals the spring energy gained: \( \tfrac12(100)(0.2^2-0.1^2)=1.5 \) J.

Teacher's Note:
1. With no atmosphere the gas pressure is balanced only by the spring.
2. Work done by the gas is stored as spring energy.

Answer: (d) 300 %

Explanation:
1. Adiabatic: \( T_2=T_1\left(\dfrac{V_1}{V_2}\right)^{2/3}=T_1\times8^{2/3}=4T_1 \).
2. The increase is \( (4-1)\times100=300\% \).

Teacher's Note:
1. A 400 % rise would mean a final temperature of 5T.
2. \( 12.5\%=1/8 \).

Answer: (c) 1/32

Explanation:
1. \( K_{ad}=\gamma P \) and \( P_2/P_1=8^{5/3}=32 \).
2. \( \dfrac{K_1}{K_2}=\dfrac{P_1}{P_2}=\dfrac1{32} \).

Teacher's Note:
1. \( \gamma=5/3 \) stays the same.
2. The bulk modulus is proportional to pressure.

Answer: (b) −1

Explanation:
1. Adiabatic: \( Q=0 \), so \( W=-\Delta U \).
2. The ratio is −1.

Teacher's Note:
1. Work is done on the gas, so W is negative while ΔU is positive.
2. The magnitudes are equal.

Answer: (a) \( p_0/3 \)

Explanation:
1. In a free expansion no work is done and no heat flows, so \( \Delta U=0 \) and the temperature is unchanged.
2. \( P_1V_1=P_2V_2 \), so \( P_2=p_0/3 \).

Teacher's Note:
1. Free expansion is not quasi-static, but the end states obey the ideal gas law.
2. An ideal gas has the same temperature before and after.

Answer: (d) monatomic

Explanation:
1. After free expansion the pressure is \( p_0/3 \); adiabatic compression to the original volume gives \( \dfrac{p_0}{3}\times3^{\gamma}=3^{2/3}p_0 \).
2. So \( 3^{\gamma-1}=3^{2/3} \), giving \( \gamma=\dfrac53 \), i.e. monatomic.

Teacher's Note:
1. \( \gamma=5/3 \) identifies a monatomic gas.
2. The compression ratio is 3.

Answer: (c) \( 3^{2/3} \)

Explanation:
1. The free expansion leaves T unchanged.
2. Adiabatic compression: \( \dfrac{T_f}{T_i}=3^{\gamma-1}=3^{2/3} \).
3. Average kinetic energy is proportional to T, so the ratio is \( 3^{2/3} \).

Teacher's Note:
1. Average kinetic energy per molecule is \( \tfrac32kT \).
2. Only the adiabatic step changes the temperature.

Answer: (b) Both statements are true, but Statement-2 is not a correct explanation of Statement-1

Explanation:
1. Total translational KE \( =\tfrac32nRT=\tfrac32PV \), so Statement-1 is true.
2. Collisions do change molecular velocities, so Statement-2 is true.
3. The result follows from \( PV=nRT \) and equipartition, not from the collisions, so Statement-2 does not explain Statement-1.

Teacher's Note:
1. Check the truth of each statement separately, then the explanation.
2. \( K=\tfrac32PV \) holds even if collisions are ignored.

Answer: (a) \( 3/T \)

Explanation:
1. \( V=\dfrac{nRT}{P}\propto T\cdot T^2=T^3 \).
2. \( \gamma_V=\dfrac1V\dfrac{dV}{dT}=\dfrac3T \).

Teacher's Note:
1. Use \( P\propto T^{-2} \) in the gas law.
2. For an isobaric process the coefficient would be 1/T.

Answer: (d) temperature decreases and the gas loses heat

Explanation:
1. \( T\propto PV\propto1/V \), so T decreases when V increases.
2. For \( PV^2=\text{const} \), \( C=\tfrac32R+\dfrac{R}{1-2}=\tfrac12R>0 \).
3. \( dQ=nC\,dT\lt 0 \) because dT \lt 0, so the gas loses heat.

Teacher's Note:
1. Heat exchanged is the product of the molar heat capacity and the temperature change.
2. A positive C with falling T means heat flows out.

Answer: (c) the temperature decreases and the gas gains heat

Explanation:
1. \( T\propto PV\propto V^{-1/3} \), so T decreases.
2. For \( n=\tfrac43 \): \( C=\tfrac32R+\dfrac{R}{1-4/3}=\tfrac32R-3R=-\tfrac32R \).
3. \( dQ=nC\,dT>0 \) because both C and dT are negative: heat is gained.

Teacher's Note:
1. This is a process with negative molar heat capacity (\( 1\lt n\lt \gamma \)).
2. The adiabatic exponent is 5/3, which is greater than 4/3.

Answer: (b) (2) and (4) only

Explanation:
1. \( C_p-C_v=R \) for both gases, so (1) is false.
2. Diatomic: \( C_p+C_v=6R \) against \( 4R \) for monatomic, so (2) is true.
3. \( \gamma=1.4 \) against \( 1.67 \), so (3) is false.
4. Diatomic: \( C_pC_v=\tfrac{35}{4}R^2 \) against \( \tfrac{15}{4}R^2 \), so (4) is true.

Teacher's Note:
1. Compute each quantity for both gases.
2. A larger number of degrees of freedom lowers γ.

Answer: (a) 4

Explanation:
1. \( TV^{\gamma-1}=\text{const} \) with \( \gamma=1.4 \).
2. \( a=32^{0.4}=2^{5\times0.4}=2^2=4 \).

Teacher's Note:
1. \( 32=2^5 \).
2. \( \gamma-1=2/5 \) for a diatomic gas.

Answer: (d) \( \tfrac98RT_1 \)

Explanation:
1. n = 5.6/22.4 = 0.25 mol.
2. \( T_2=T_1\times8^{2/3}=4T_1 \).
3. Work done on gas = \( \Delta U=0.25\times\tfrac32R\times3T_1=\tfrac98RT_1 \).

Teacher's Note:
1. Adiabatic: the work done on the gas goes entirely into internal energy.
2. STP molar volume is 22.4 L.

Answer: (c) 208 J

Explanation:
1. The balloon keeps the gas at constant (atmospheric) pressure, so the process is isobaric.
2. \( Q=nC_p\Delta T=2\times\tfrac52\times8.31\times5=207.75\approx208 \) J.

Teacher's Note:
1. Use \( C_p \), not \( C_v \), for a balloon.
2. The 104 J option is half of the correct value.

Answer: (b) 2

Explanation:
1. Path iaf: \( U_f=U_i+Q-W=100+500-200=400 \) J.
2. i to b: \( Q_{ib}=\Delta U+W=100+50=150 \) J.
3. b to f: \( Q_{bf}=(400-200)+100=300 \) J.
4. Ratio \( =300/150=2 \).

Teacher's Note:
1. Internal energy at f is fixed by the first path.
2. Apply the first law on each leg separately.

Answer: (a) 513 K

Explanation:
1. No work is done and the total heat is conserved: \( 2\cdot\tfrac32R(700-T)=2\cdot\tfrac52R(T-400) \).
2. \( 3(700-T)=5(T-400) \), so \( 8T=4100 \).
3. \( T=512.5\approx513 \) K.

Teacher's Note:
1. Each gas exchanges heat at constant volume.
2. The temperature is weighted by the heat capacities \( nC_v \).

Answer: (d) −100R

Explanation:
1. The total pressure is constant, so the heat exchanged is zero in terms of enthalpy: \( 2\cdot\tfrac52R(T-700)+2\cdot\tfrac72R(T-400)=0 \).
2. \( 12T=6300 \), so \( T=525 \) K.
3. \( \Delta U=2\cdot\tfrac32R(525-700)+2\cdot\tfrac52R(525-400)=-525R+625R=100R \).
4. The system is adiabatic, so \( W_{by}=-\Delta U=-100R \).

Teacher's Note:
1. At constant external pressure the heat balance uses \( C_p \).
2. Work done by the gases equals minus the rise in internal energy.

Answer: (c) (1), (2) and (4)

Explanation:
1. (1) Average energy per mole \( =\tfrac12(\tfrac52+\tfrac32)RT=2RT \): true.
2. (2) Mixture: \( C_v=2R \), \( C_p=3R \), \( \gamma=1.5 \), \( M=3 \); \( v\propto\sqrt{\gamma/M} \): \( \sqrt{1.5/3}\big/\sqrt{(5/3)/4}=\sqrt{6/5} \): true.
3. (3),(4) \( v_{rms}\propto1/\sqrt M \): \( \sqrt{2/4}=1/\sqrt2 \), so (4) is true and (3) is false.

Teacher's Note:
1. Speed of sound is \( \sqrt{\gamma RT/M} \).
2. Use molar masses 2 g and 4 g.

Answer: (b) (1), (2) and (3)

Explanation:
1. Case \( V_2=2V_1,T_2=3T_1 \): \( P_2=1.5P_1 \), spring energy \( =\tfrac12(P_2-P_1)(V_2-V_1)=\tfrac14P_1V_1 \): (1) true.
2. \( \Delta U=\tfrac32(P_2V_2-P_1V_1)=\tfrac32(3-1)P_1V_1=3P_1V_1 \): (2) true.
3. Case \( V_2=3V_1,T_2=4T_1 \): \( P_2=\tfrac43P_1 \), \( W=P_1\cdot2V_1+\tfrac12\cdot\tfrac13P_1\cdot2V_1=\tfrac73P_1V_1 \): (3) true.
4. \( \Delta U=\tfrac32\cdot3P_1V_1=\tfrac92P_1V_1 \), so \( Q=\tfrac92+\tfrac73=\tfrac{41}6P_1V_1 \), not 17/6: (4) false.

Teacher's Note:
1. The spring force is \( (P-P_1)A \), so spring energy is \( \tfrac12(P-P_1)\Delta V \).
2. The work done by the gas is area under the linear P–V line.

Answer: (a) 588 J

Explanation:
1. \( P^3V^5=\text{const} \) means \( \gamma=5/3 \).
2. Work: \( W=10^5(8-1)\times10^{-3}=700 \) J (isochoric step does none).
3. \( \Delta U=\dfrac{P_fV_f-P_iV_i}{\gamma-1}=\dfrac{25-100}{2/3}=-112.5 \) J.
4. \( Q=\Delta U+W=-112.5+700=587.5\approx588 \) J.

Teacher's Note:
1. ΔU depends only on the end states.
2. Only the isobaric step does work.

Answer: (d) 900 J

Explanation:
1. \( T_2=100\times8^{-2/3}=25 \) K.
2. \( \Delta U=\tfrac32R(T_1-T_2)=1.5\times8\times75=900 \) J.

Teacher's Note:
1. \( 8^{2/3}=4 \).
2. The internal energy falls because the gas does work.