JEE Physics Thermodynamics MCQs Set 18

Answer: (a) 600 K

Explanation:
1. \( \eta=1-\dfrac{T_2}{T_1}=\dfrac12\Rightarrow T_2=\dfrac{T_1}2 \).
2. New sink temperature \( T_2-100 \) gives \( \dfrac{T_2-100}{T_1}=\dfrac13 \): \( \dfrac12-\dfrac{100}{T_1}=\dfrac13 \).
3. \( \dfrac{100}{T_1}=\dfrac16\Rightarrow T_1=600\ \text{K} \).

Teacher's Note:
1. A change of 100 °C is the same as a change of 100 K. Write each efficiency as one minus the temperature ratio.

Answer: (a) 6 J

Explanation:
1. \( \text{COP}=\dfrac{T_2}{T_1-T_2}=\dfrac{250}{300-250}=5 \), so 5 J is extracted from the cold space per joule of work.
2. Heat rejected = 5 + 1 = 6 J.

Teacher's Note:
1. Heat rejected equals heat removed plus the work done. Use kelvin temperatures.

Answer: (c) \( \Delta U=\dfrac{P_fV_f-P_iV_i}{1-\gamma} \)

Explanation:
1. For n moles \( \Delta U=nC_V\Delta T=\dfrac{nR\Delta T}{\gamma-1}=\dfrac{P_fV_f-P_iV_i}{\gamma-1} \).
2. Option (c) has \( 1-\gamma \) in the denominator, which gives the wrong sign. The other three are correct.

Teacher's Note:
1. Remember internal energy change as PV difference over gamma minus one. Check the sign with an adiabatic compression.

Answer: (d) Specific heat at constant volume, \( C_V=\dfrac R\gamma \)

Explanation:
1. Adiabatic: dQ = 0, so C = 0. Isothermal: dT = 0, so \( C=dQ/dT=\infty \).
2. \( C_P=\dfrac{\gamma R}{\gamma-1} \) is correct.
3. \( C_V=\dfrac{R}{\gamma-1} \), not \( R/\gamma \), so (d) is incorrect.

Teacher's Note:
1. Cv is R over gamma minus one. Cp is gamma times Cv.

Answer: (c) \( C_V+\dfrac{R}{1-x} \)

Explanation:
1. Work in a polytropic process: \( dW=\dfrac{R\,dT}{1-x} \).
2. First law: \( C=C_V+\dfrac{dW}{dT}=C_V+\dfrac{R}{1-x} \).

Teacher's Note:
1. Check the special cases. x = 0 gives Cp, x = 1 gives infinity (isothermal) and x = gamma gives zero (adiabatic).

Answer: (a) \( \dfrac{17R}{6} \)

Explanation:
1. Write \( PV^{-2}=a \), so the polytropic index is \( x=-2 \).
2. \( C=C_V+\dfrac{R}{1-x}=\dfrac{5R}{2}+\dfrac R3=\dfrac{17R}{6} \).

Teacher's Note:
1. Identify x from the given relation, including its sign. A diatomic gas has Cv equal to 5R by 2.

Answer: (a) 11 J

Explanation:
1. \( \dfrac{Q_{out}}{W}=\dfrac{T_1}{T_1-T_2}=\dfrac{300}{27}=11.1 \).
2. So about 11 J is delivered to the surroundings for each joule of work.

Teacher's Note:
1. For a reversible refrigerator the heat rejected scales as T1 over T1 minus T2. A small temperature difference means a large ratio.

Answer: (d) \( \dfrac{(\gamma-1)Mv^2}{2R} \)

Explanation:
1. Mass of the gas = 2M, so kinetic energy \( =\tfrac12(2M)v^2=Mv^2 \).
2. This becomes internal energy at constant volume: \( Mv^2=nC_V\Delta T=2\cdot\dfrac{R}{\gamma-1}\Delta T \).
3. \( \Delta T=\dfrac{(\gamma-1)Mv^2}{2R} \).

Teacher's Note:
1. The ordered kinetic energy of the whole gas turns into random molecular energy. Use the molar heat capacity at constant volume.

Answer: (a) \( C_V(T_1-T_2) \)

Explanation:
1. Internal energy of an ideal gas depends only on temperature: \( \Delta U=nC_V\Delta T \) for any process.
2. For one mole the decrease is \( C_V(T_1-T_2) \).

Teacher's Note:
1. Internal energy of an ideal gas depends only on temperature. The adiabatic condition is not needed for this result.

Answer: (c) \( 3/2 \)

Explanation:
1. \( dQ=C\,dT=4R\,dT \) and \( dW=2R\,dT \), so \( dU=dQ-dW=2R\,dT \) and \( C_V=2R \).
2. \( C_P=C_V+R=3R \), so \( \gamma=3R/2R=3/2 \).

Teacher's Note:
1. Use the first law to find Cv from the given heat capacity and work. Then add R to get Cp.

Answer: (c) \( 2RT_0 \)

Explanation:
1. \( \Delta U=\tfrac32R(2T_0-T_0)=\tfrac32RT_0 \).
2. Work \( =\int P\,dV=\dfrac K2(V_2^2-V_1^2)=\tfrac12(P_2V_2-P_1V_1)=\tfrac12R(2T_0-T_0)=\tfrac12RT_0 \).
3. \( Q=\Delta U+W=\tfrac32RT_0+\tfrac12RT_0=2RT_0 \).

Teacher's Note:
1. For P proportional to V the work is half the change in PV. Add it to the change in internal energy.

Answer: (d) 1.73 : 1

Explanation:
1. \( \eta_1=\dfrac{473-273}{473}=\dfrac{200}{473} \) and \( \eta_2=\dfrac{273-73}{273}=\dfrac{200}{273} \).
2. \( \dfrac{\eta_2}{\eta_1}=\dfrac{473}{273}=1.73 \).

Teacher's Note:
1. Convert to kelvin before taking ratios. The same temperature difference is more efficient at lower temperatures.