JEE Physics Thermodynamics MCQs Set 11

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

Explanation:
1. For one mole, \( PV=RT\Rightarrow T=\frac{aV^3}{R} \), so \( dT=\frac{3aV^2}{R}dV \).
2. \( dU=C_VdT=\frac52R\,dT=\frac{15}{2}aV^2dV \) and \( dW=PdV=aV^2dV \).
3. \( dQ=\frac{17}{2}aV^2dV \), so \( C=\frac{dQ}{dT}=\frac{17R}{6} \).

Teacher's Note:
1. Write dQ = dU + dW and express everything through dV.
2. General result for PV^x = constant: C = Cv + R/(1 - x); here x = -2.

Answer: (c) \( f=\frac{2}{\gamma-1} \)

Explanation:
1. \( C_V=\frac f2R \) and \( C_P=\left(\frac f2+1\right)R \).
2. \( \gamma=\frac{C_P}{C_V}=1+\frac2f\Rightarrow f=\frac{2}{\gamma-1} \).

Teacher's Note:
1. Check with a monatomic gas: gamma = 5/3 gives f = 3.
2. gamma decreases as the degrees of freedom increase.

Answer: (b) \( \frac54R \)

Explanation:
1. For \( PV^x=\text{const} \): \( C=C_V+\frac{R}{1-x} \).
2. \( C=\frac32R+\frac{R}{1-5}=\frac32R-\frac R4=\frac54R \).

Teacher's Note:
1. The formula includes the isothermal (x = 1) and adiabatic (x = gamma) cases as limits.
2. A negative term here shows the gas gives out heat while it expands in this process.

Answer: (b) \( \frac13R(T_2-T_1) \)

Explanation:
1. \( W=\int P\,dV=\int aV^2dV=\frac a3(V_2^3-V_1^3) \).
2. With \( P=aV^2 \): \( aV^3=PV=RT \), so \( W=\frac13R(T_2-T_1) \).

Teacher's Note:
1. Replace aV cubed by PV, which equals RT for one mole.
2. Check with P constant: W = R (T2 - T1), three times larger.

Answer: (b) \( \frac38N_Ak_B(T_2-T_1) \)

Explanation:
1. 1 g of helium (molar mass 4 g/mol) is \( \frac14 \) mole.
2. Helium is monatomic: \( C_V=\frac32R=\frac32N_Ak_B \).
3. \( Q=nC_V\Delta T=\frac14\cdot\frac32N_Ak_B(T_2-T_1)=\frac38N_Ak_B(T_2-T_1) \).

Teacher's Note:
1. Find the number of moles first, then use the molar heat capacity at constant volume.
2. Heat at constant volume is nCv times the temperature rise.

Answer: (d) 13 kJ

Explanation:
1. First process: \( \Delta U=Q-W=16-20=-4 \) kJ.
2. \( \Delta U \) is the same for the second process: \( W=Q-\Delta U=9-(-4)=13 \) kJ.

Teacher's Note:
1. Internal energy is a state function, so it is the same between the same two states.
2. Apply the first law Q = delta U + W in each process.

Answer: (d) -60 cal

Explanation:
1. Along iaf: \( \Delta U_{if}=80-50=30 \) cal.
2. So \( \Delta U_{fi}=-30 \) cal.
3. Along the return path: \( Q=\Delta U+W=-30+(-30)=-60 \) cal (the gas gives out 60 cal).

Teacher's Note:
1. Going backwards changes the sign of the change in internal energy.
2. Negative work means the surroundings do work on the gas.

Answer: (d) 24 J

Explanation:
1. Along acb: \( \Delta U=200-80=120 \) J.
2. Along adb the change in internal energy is the same: \( W=Q-\Delta U=144-120=24 \) J.

Teacher's Note:
1. Different paths between the same states have the same dU but different Q and W.
2. Apply the first law along each path.

Answer: (b) 25%

Explanation:
1. At constant pressure: \( \frac WQ=\frac{R}{C_P}=\frac{\gamma-1}{\gamma} \).
2. \( \frac{4/3-1}{4/3}=\frac14=25\% \).

Teacher's Note:
1. The same formula gives 40% for a monatomic gas and about 28.6% for a diatomic gas.
2. The more complex the molecule, the smaller the fraction of heat converted to work.

Answer: (d) \( \frac{10}{2} \)

Explanation:
1. For an adiabatic process, \( P^{1-\gamma}T^\gamma=\text{const}\Rightarrow P\propto T^{\gamma/(\gamma-1)} \).
2. For \( \gamma=\frac53 \): \( \frac{\gamma}{\gamma-1}=\frac{5/3}{2/3}=\frac52 \), so \( P\propto T^{5/2} \) and \( P^2\propto T^5 \).
3. \( C=5=\frac{10}{2} \).

Teacher's Note:
1. Square the relation, because the question asks about P squared.
2. Combine PV^gamma = constant with PV = nRT to eliminate V.

Answer: (b) \( 81P_0 \)

Explanation:
1. \( PV^\gamma=\text{const}\Rightarrow P_2=P_0\left(\frac{V_1}{V_2}\right)^{\gamma}=P_0(27)^{4/3} \).
2. \( 27^{4/3}=(3^3)^{4/3}=3^4=81 \), so \( P_2=81P_0 \).

Teacher's Note:
1. 27 is a perfect cube, which makes the fractional power easy.
2. The pressure rises much more than in an isothermal compression, where it would be 27 P0.

Answer: (a) \( 2^{1.4}P \)

Explanation:
1. For a diatomic gas \( \gamma=\frac75=1.4 \).
2. \( P_2=P\left(\frac{V}{V/2}\right)^{1.4}=2^{1.4}P\approx2.64P \).

Teacher's Note:
1. Isothermal compression to half the volume would only double the pressure.
2. In adiabatic compression the temperature also rises, so the pressure rises more.

Answer: (a) \( \gamma:\gamma-1 \)

Explanation:
1. V proportional to T means the pressure is constant (isobaric).
2. \( \Delta Q=nC_P\Delta T \) and \( \Delta W=nR\Delta T \).
3. \( \frac{\Delta Q}{\Delta W}=\frac{C_P}{R}=\frac{\gamma}{\gamma-1} \).

Teacher's Note:
1. A straight line through the origin on a V-T graph is an isobaric process (Charles' law).
2. Cp = gamma R/(gamma - 1).

Answer: (c) in the first vessel, both pressure and work done are more

Explanation:
1. In an adiabatic expansion the gas cools and the pressure falls faster (\( PV^\gamma=\text{const} \)) than in the isothermal expansion (\( PV=\text{const} \)).
2. On a P-V diagram the adiabat lies below the isotherm, so the area under it (the work done) is also smaller.
3. Hence in the first (isothermal) vessel both the final pressure and the work done are more.

Teacher's Note:
1. The adiabatic curve is steeper than the isothermal curve through the same point.
2. In an isothermal expansion heat flows in to make up for the work done, so the gas does more work.

Answer: (a) 26.8%

Explanation:
1. \( T_2=273 \) K and \( T_1=373 \) K.
2. \( \eta=1-\frac{T_2}{T_1}=1-\frac{273}{373}=\frac{100}{373}\approx0.268=26.8\% \).

Teacher's Note:
1. Always use absolute temperatures in the Carnot efficiency formula.
2. The result is low because the temperature difference is only 100 K.

Answer: (b) 250 K

Explanation:
1. Initially \( 0.4=1-\frac{300}{T_1}\Rightarrow T_1=500 \) K.
2. New efficiency \( =0.4\times1.5=0.6=1-\frac{300}{T_1'}\Rightarrow T_1'=750 \) K.
3. Increase \( =750-500=250 \) K.

Teacher's Note:
1. An increase of 50 percent of 40 percent is 60 percent, not 90 percent.
2. Find the source temperature in each case from eta = 1 - T2/T1.

Answer: (c) 40 W

Explanation:
1. \( T_2=270 \) K and \( T_1=300 \) K, so \( \text{COP}=\frac{T_2}{T_1-T_2}=\frac{270}{30}=9 \).
2. \( W=\frac{Q_2}{\text{COP}}=\frac{360}{9}=40 \) J per second, that is 40 W.

Teacher's Note:
1. Coefficient of performance is heat extracted divided by work input.
2. The smaller the temperature gap, the less power is needed per joule extracted.

Answer: (b) \( 31^\circ\)C

Explanation:
1. \( \text{COP}=\frac{T_2}{T_1-T_2}\Rightarrow5=\frac{253}{T_1-253} \).
2. \( T_1-253=50.6\Rightarrow T_1\approx303.6 \) K \( \approx30.6^\circ\)C \( \approx31^\circ\)C.

Teacher's Note:
1. Use kelvin for T2 = 253 K in the coefficient of performance.
2. The surroundings are warmer than the freezer, as heat must be pumped uphill.

Answer: (c) 118.25 W

Explanation:
1. \( T_2=277 \) K, \( T_1=303 \) K, so \( \text{COP}=\frac{277}{26} \).
2. Heat removed per second \( =300\times4.2=1260 \) J.
3. \( P=\frac{1260}{277/26}=\frac{1260\times26}{277}\approx118.25 \) W.

Teacher's Note:
1. Convert calories to joules before computing the power.
2. Power is the work done per second.

Answer: (a) 380 J

Explanation:
1. \( \eta=1-\frac{T_2}{T_1}=\frac1{20}\Rightarrow\frac{T_2}{T_1}=\frac{19}{20} \).
2. \( \text{COP}=\frac{T_2}{T_1-T_2}=\frac{19/20}{1/20}=19 \).
3. \( Q_2=19\times W=19\times20=380 \) J.

Teacher's Note:
1. The coefficient of performance of a reversible refrigerator is (1 - eta)/eta.
2. A small efficiency as an engine means a large COP as a refrigerator.

Answer: (b) 15.4%

Explanation:
1. Work done per cycle = area of the rectangle \( =(2V_0-V_0)(2P_0-P_0)=P_0V_0 \).
2. Heat absorbed (in AB and BC) \( =\frac32P_0V_0+5P_0V_0=\frac{13}{2}P_0V_0 \).
3. \( \eta=\frac{W}{Q_{in}}=\frac{P_0V_0}{\frac{13}{2}P_0V_0}=\frac2{13}\approx15.4\% \).

Teacher's Note:
1. The work done in a cycle equals the area enclosed on the P-V diagram.
2. Efficiency is the work done divided by the heat absorbed.