JEE Physics Thermodynamics MCQs Set 14

Answer: (b) 66.6 %

Explanation:
1. Temperatures: 900 K and 300 K.
2. \( \eta=1-\dfrac{300}{900}=\dfrac23=66.6\% \).

Teacher's Note:
1. Use absolute temperatures.
2. The efficiency is always less than 100 %.

Answer: (a) \( 2.8\times10^6 \) J

Explanation:
1. \( Q_1=10^6\times4.2=4.2\times10^6 \) J.
2. \( W=\eta Q_1=\tfrac23\times4.2\times10^6=2.8\times10^6 \) J.

Teacher's Note:
1. Convert kilocalories to joules.
2. Work = efficiency × heat absorbed.

Answer: (d) 373.3 K

Explanation:
1. Sink = 280 K. For 50 %: \( T_1=\dfrac{280}{0.5}=560 \) K.
2. For 70 %: \( T_1'=\dfrac{280}{0.3}=933.3 \) K.
3. Increase = 933.3 − 560 = 373.3 K.

Teacher's Note:
1. \( T_1=T_2/(1-\eta) \).
2. Do the arithmetic in kelvin.

Answer: (c) 79 kcal

Explanation:
1. Heat extracted from the water = 10000 g × 80 cal/g = 800 kcal.
2. \( \dfrac{Q_L}{W}=\dfrac{T_L}{T_H-T_L}=\dfrac{273}{27} \).
3. \( W=800\times\dfrac{27}{273}\approx79 \) kcal.

Teacher's Note:
1. The heat released into the room is the 800 kcal plus the work, about 879 kcal.
2. The coefficient of performance of this refrigerator is about 10.

Answer: (b) 900 calories

Explanation:
1. \( \dfrac{Q_H}{Q_L}=\dfrac{T_H}{T_L}=\dfrac{300}{250} \).
2. \( Q_H=750\times1.2=900 \) calories.

Teacher's Note:
1. For a reversible cycle, heats are proportional to temperatures.
2. The extra 150 calories is the work supplied.

Answer: (a) 24 J

Explanation:
1. Path acb: \( \Delta U=200-80=120 \) J.
2. Path adb has the same ΔU: \( W=144-120=24 \) J.

Teacher's Note:
1. Internal energy change depends only on the end states.
2. Apply Q = ΔU + W on each path.

Answer: (d) −172 J

Explanation:
1. Return path: \( \Delta U=-120 \) J.
2. Work done by the system = −52 J.
3. \( Q=\Delta U+W=-120-52=-172 \) J (heat is released).

Teacher's Note:
1. Work done on the system is negative work by the system.
2. The sign of Q shows heat leaves the system.

Answer: (c) 160 J

Explanation:
1. \( U_b-U_a=120 \) J (from the first path).
2. \( U_b=40+120=160 \) J.

Teacher's Note:
1. Add the change to the initial value.
2. The change from a to b is +120 J.

Answer: (b) 72 J

Explanation:
1. \( U_b-U_d=160-88=72 \) J.
2. At constant volume \( W=0 \), so \( Q=\Delta U=72 \) J.

Teacher's Note:
1. No volume change means no work.
2. The heat absorbed equals the rise in internal energy.

Answer: (a) \( C_p \) and \( C_v \) both increase by the same amount

Explanation:
1. At 30 K the gas is effectively monatomic: \( C_v=\tfrac32R \), \( C_p=\tfrac52R \).
2. At 300 K it behaves as diatomic: \( C_v=\tfrac52R \), \( C_p=\tfrac72R \).
3. Both rise by R, since \( C_p-C_v=R \) always.

Teacher's Note:
1. The difference \( C_p-C_v=R \) stays fixed for an ideal gas.
2. Rotational modes switch on and raise both capacities.

Answer: (d) infinite

Explanation:
1. Heat is absorbed at constant temperature, so \( \Delta T=0 \).
2. \( c=\dfrac{Q}{m\Delta T}\to\infty \).

Teacher's Note:
1. Specific heat is heat per unit mass per unit temperature change.
2. The same holds for any substance undergoing a change of state.

Answer: (c) positive work is done by the gas in expanding against the external pressure

Explanation:
1. At constant pressure the gas expands and does work \( P\Delta V \).
2. This work is supplied from the extra heat, so \( C_p=C_v+R \).

Teacher's Note:
1. An ideal gas has no intermolecular forces.
2. The extra heat goes into work, not into temperature.

Answer: (b) infinite

Explanation:
1. The heat capacity depends on the process: \( C=C_v+\dfrac{R}{1-n} \) for \( PV^n=\text{const} \).
2. Every value of n gives a different heat capacity, so there are infinitely many.

Teacher's Note:
1. Only \( C_p \) and \( C_v \) are conventionally tabulated.
2. Isothermal heat capacity is infinite and adiabatic is zero.

Answer: (a) \( C_p \) is slightly greater than \( C_v \)

Explanation:
1. A solid expands very little when heated at constant pressure, so the work \( P\Delta V \) is tiny.
2. The extra heat is therefore very small and \( C_p\) exceeds \( C_v \) only slightly.

Teacher's Note:
1. The relation \( C_p-C_v=R \) holds only for ideal gases.
2. Never expect \( C_p\lt C_v \).

Answer: (d) decreases

Explanation:
1. The sudden compression is adiabatic and heats the gas.
2. The metal walls then conduct heat away until the gas returns to room temperature.
3. At fixed volume, lower temperature means lower pressure.

Teacher's Note:
1. At constant volume \( P\propto T \).
2. The final temperature equals that of the surroundings.

Answer: (c) \( T_1>T_2 \)

Explanation:
1. Sample 1 stays at the initial temperature.
2. In sample 2 the gas does work at the cost of its internal energy, so it cools: \( T_2\lt T_1 \).

Teacher's Note:
1. Adiabatic expansion lowers the temperature.
2. The isothermal gas receives heat from the surroundings.

Answer: (b) \( P_1>P_2 \)

Explanation:
1. At the same final volume \( P\propto T \).
2. Since \( T_1>T_2 \), \( P_1>P_2 \).

Teacher's Note:
1. Both gases have the same n and V at the end.
2. The adiabatic curve falls faster than the isotherm.

Answer: (a) \( W_1>W_2 \)

Explanation:
1. The isothermal curve lies above the adiabatic curve on the P–V diagram.
2. The area under it is larger, so \( W_1>W_2 \).

Teacher's Note:
1. Work is the area under the P–V curve.
2. The adiabatic pressure drops faster during expansion.

Answer: (d) (i) and (iii) only

Explanation:
1. (i) is the definition of an adiabatic change.
2. (iii) \( \Delta U=nC_v\Delta T \) holds for any process of an ideal gas.
3. (ii) is wrong: \( W=-\Delta U \), so the work done by the gas equals the decrease, not the change itself.

Teacher's Note:
1. Sign matters: the gas does work at the expense of its internal energy.
2. \( \Delta U=nC_v\Delta T \) is valid for every ideal gas process.

Answer: (c) P

Explanation:
1. Isothermal: \( PV=\text{const} \), so \( dP=-P\dfrac{dV}{V} \).
2. \( K=-V\dfrac{dP}{dV}=P \).

Teacher's Note:
1. The adiabatic bulk modulus is γP.
2. The isothermal value does not depend on the gas type.

Answer: (b) 3/2

Explanation:
1. Adiabatic: \( P^{1-\gamma}T^{\gamma}=\text{const} \), so \( P\propto T^{\gamma/(\gamma-1)} \).
2. \( \dfrac{\gamma}{\gamma-1}=3 \), giving \( \gamma=\dfrac32 \).

Teacher's Note:
1. Eliminate V between \( PV^\gamma=\text{const} \) and \( PV=nRT \).
2. Cross-check: \( 1.5/0.5=3 \).

Answer: (a) P/n

Explanation:
1. \( V=cP^{-n} \), so \( \dfrac{dV}{dP}=-n\dfrac VP \).
2. \( K=-V\dfrac{dP}{dV}=\dfrac Pn \).

Teacher's Note:
1. Here the exponent is on P, not V.
2. With n = 1 this reduces to the isothermal modulus P.

Answer: (d) \( \left(\dfrac{T_0}{2a}\right)^{1/3} \)

Explanation:
1. \( P=\dfrac{RT}{V}=R\left(\dfrac{T_0}{V}+aV^2\right) \).
2. \( \dfrac{dP}{dV}=R\left(-\dfrac{T_0}{V^2}+2aV\right)=0 \).
3. \( V^3=\dfrac{T_0}{2a} \).

Teacher's Note:
1. Set the derivative of pressure with respect to volume to zero.
2. The second derivative is positive there, so it is a minimum.

Answer: (c) \( \dfrac{3}{2^{2/3}}\,a^{1/3}R\,T_0^{2/3} \)

Explanation:
1. At the minimum \( aV^2=\dfrac{T_0}{2V} \), so \( P=R\left(\dfrac{T_0}{V}+\dfrac{T_0}{2V}\right)=\dfrac{3RT_0}{2V} \).
2. With \( V=(T_0/2a)^{1/3} \): \( P=\dfrac{3RT_0}{2}\left(\dfrac{2a}{T_0}\right)^{1/3}=\dfrac{3\cdot2^{1/3}}{2}a^{1/3}RT_0^{2/3} \).
3. Since \( 2^{1/3}/2=2^{-2/3} \), \( P_{min}=\dfrac{3}{2^{2/3}}a^{1/3}RT_0^{2/3} \).

Teacher's Note:
1. Substitute the volume found in the previous step.
2. Check the powers: \( T_0\cdot T_0^{-1/3}=T_0^{2/3} \).

Answer: (b) \( \gamma>n>1 \)

Explanation:
1. \( C=C_v+\dfrac{R}{1-n}=\dfrac{R(\gamma-n)}{(\gamma-1)(1-n)} \).
2. For \( C\lt 0 \) the numerator and denominator must have opposite signs: \( \gamma-n>0 \) and \( 1-n\lt 0 \), i.e. \( 1\lt n\lt \gamma \).

Teacher's Note:
1. Negative heat capacity means heat is absorbed while the temperature falls.
2. Use the factored form of C.

Answer: (a) 1600 J

Explanation:
1. \( \eta=1-\dfrac{300}{600}=0.5 \).
2. \( Q_1=\dfrac{W}{\eta}=\dfrac{800}{0.5}=1600 \) J.

Teacher's Note:
1. Efficiency is work output divided by heat input.
2. The heat rejected to the sink is the other 800 J.

Answer: (d) 9

Explanation:
1. \( \text{COP}=\dfrac{T_L}{T_H-T_L}=\dfrac{273}{30} \).
2. \( =9.1\approx9 \).

Teacher's Note:
1. The temperature difference is 30 K.
2. A refrigerator can have COP greater than 1.

Answer: (c) is heated

Explanation:
1. The refrigerator takes in electrical work W and moves heat from inside to outside.
2. Total heat given to the room = heat removed + W, which exceeds the heat removed, so the room warms up.

Teacher's Note:
1. Energy supplied by the compressor ends up as heat in the room.
2. The net effect is like an electric heater.