JEE Physics Thermodynamics MCQs Set 08

Answer: (d) −123 °C

Explanation:
1. Average kinetic energy \( =\dfrac{3}{2}kT\propto T \).
2. \( \dfrac{T'}{300}=\dfrac{1}{2} \), so \( T'=150 \) K.
3. \( 150-273=-123 \) °C.

Teacher's Note:
1. Average translational kinetic energy depends only on the absolute temperature.
2. Halving the energy means halving the kelvin temperature, not the Celsius value.

Answer: (b) more than V

Explanation:
1. After the increase, \( P_1=1.1P \). After decreasing this by 10%, \( P_2=0.9\times1.1P=0.99P \).
2. At constant temperature \( PV=P_2V_2 \), so \( V_2=\dfrac{PV}{0.99P}=\dfrac{V}{0.99}\approx1.01V \).
3. The final volume is slightly more than V.

Teacher's Note:
1. The second 10% change is taken on the new pressure, not on the original pressure.
2. The two changes do not cancel: a 10% rise followed by a 10% fall leaves the pressure 1% lower than the start.

Answer: (c) 18

Explanation:
1. \( P=\dfrac{\rho RT}{M} \), so at constant pressure \( \rho T=\text{constant} \).
2. \( \rho'=24\times\dfrac{300}{400}=18 \).

Teacher's Note:
1. At constant pressure the density of a gas varies inversely with the absolute temperature.
2. Heating a gas makes it expand, so its density must fall.

Answer: (d) \( P_aV_a=P_bV_b \)

Explanation:
1. Thermal equilibrium means the two gases are at the same temperature.
2. For N molecules, \( PV=NkT \).
3. With the same N and the same T, \( P_aV_a=P_bV_b \). The pressures or the volumes need not be equal separately.

Teacher's Note:
1. Thermal equilibrium means equal temperatures, nothing more.
2. Use \( PV=NkT \) and compare the right-hand sides.

Answer: (a) Q = W = 0 and ΔU = 0

Explanation:
1. In a free expansion the gas expands into a vacuum, so it does no work: W = 0.
2. The process is adiabatic, so Q = 0.
3. By the first law \( \Delta U=Q-W=0 \).

Teacher's Note:
1. Free expansion into vacuum means there is no opposing pressure, so no work is done.
2. For an ideal gas \( \Delta U=0 \) also means the temperature does not change.

Answer: (d) 11RT

Explanation:
1. Oxygen is diatomic with 5 degrees of freedom: \( U_1=2\times\dfrac{5}{2}RT=5RT \).
2. Argon is monatomic with 3 degrees of freedom: \( U_2=4\times\dfrac{3}{2}RT=6RT \).
3. \( U=5RT+6RT=11RT \).

Teacher's Note:
1. Each degree of freedom contributes \( \dfrac{1}{2}RT \) per mole to the internal energy.
2. Add the internal energies of the components of a mixture.

Answer: (d) 3PV

Explanation:
1. The cycle is a triangle on the P-V diagram. Its base (along the volume axis) is \( 3V-V=2V \) and its height (along the pressure axis) is \( 4P-P=3P \).
2. Work done = area of the triangle \( =\dfrac{1}{2}\times2V\times3P=3PV \).
3. The path A to B to C to A runs clockwise on the P-V diagram, so the work is positive: 3PV.

Teacher's Note:
1. Work done in a cyclic process equals the area enclosed on the P-V diagram.
2. Clockwise cycles give positive work and anticlockwise cycles give negative work.

Answer: (c) 450 kcal

Explanation:
1. Efficiency \( \eta=1-\dfrac{T_2}{T_1}=1-\dfrac{375}{500}=0.25 \).
2. Work done \( W=\eta Q_1=0.25\times600=150 \) kcal.
3. Heat rejected \( Q_2=Q_1-W=600-150=450 \) kcal.

Teacher's Note:
1. For a Carnot engine \( \dfrac{Q_2}{Q_1}=\dfrac{T_2}{T_1} \), so \( Q_2=600\times\dfrac{375}{500}=450 \) kcal directly.
2. Temperatures must be in kelvin.

Answer: (b) \( \dfrac{3}{2}\rho_0 \)

Explanation:
1. \( P=\dfrac{\rho RT}{M} \), so \( \rho\propto\dfrac{P}{T} \).
2. \( \dfrac{\rho_B}{\rho_A}=\dfrac{3P/2T_0}{P/T_0}=\dfrac{3}{2} \).
3. \( \rho_B=\dfrac{3}{2}\rho_0 \).

Teacher's Note:
1. Density of a gas is proportional to pressure and inversely proportional to absolute temperature.
2. Use ratios so that R and the molar mass cancel.

Answer: (b) −4.6 \( T_0 \)

Explanation:
1. Along the adiabat \( TV^{2/3}=\text{constant} \): \( T_Q(2V_0)^{2/3}=T_0V_0^{2/3} \), so \( T_Q=\dfrac{T_0}{1.587}=0.63\,T_0 \).
2. \( \Delta U=C_V(T_Q-T_0)=\dfrac{3}{2}R(0.63-1)T_0 \).
3. \( \Delta U=1.5\times8.3\times(-0.37)\,T_0\approx-4.6\,T_0 \) (in joules).

Teacher's Note:
1. Internal energy depends only on temperature, so any path between the two states gives the same \( \Delta U \).
2. The temperature at Q is lower than \( T_0 \), so \( \Delta U \) is negative.

Answer: (d) \( \dfrac{1}{2}nRT \)

Explanation:
1. Initial energy \( U_1=\dfrac{5}{2}NRT \).
2. Converting n moles of diatomic gas into monatomic gas gives 2n moles of monatomic gas (each molecule splits into 2 atoms). Final energy \( U_2=\dfrac{3}{2}(2n)RT+\dfrac{5}{2}(N-n)RT=\dfrac{5}{2}NRT+\dfrac{1}{2}nRT \).
3. \( \Delta U=\dfrac{1}{2}nRT \).

Teacher's Note:
1. Each diatomic molecule that dissociates gives two monatomic atoms, so the number of moles doubles.
2. Compare the energy before and after at the same temperature.

Answer: (c) 3R

Explanation:
1. \( Q=\Delta U+W \) with \( W=\dfrac{Q}{2} \), so \( \Delta U=\dfrac{Q}{2} \).
2. For one mole \( \Delta U=\dfrac{3}{2}R\Delta T \), hence \( \dfrac{Q}{2}=\dfrac{3}{2}R\Delta T \) and \( Q=3R\Delta T \).
3. \( C=\dfrac{Q}{\Delta T}=3R \).

Teacher's Note:
1. Molar specific heat is the heat per mole per kelvin for the particular process.
2. Use \( \Delta U=C_V\Delta T \) for any process of an ideal gas.