JEE Physics Thermodynamics MCQs Set 05

Answer: (a) 40 %

Explanation:
1. At constant pressure \( \dfrac{W}{Q} = \dfrac{R}{C_P} = 1 - \dfrac{1}{\gamma} \).
2. \( = 1 - \dfrac{3}{5} = \dfrac{2}{5} = 40\ \% \).

Teacher's Note:
1. The rest, 60 %, increases the internal energy.

Answer: (c) \( \dfrac{2}{5}Q \)

Explanation:
1. \( Q = nC_P\Delta T = \dfrac{5}{2}nR\Delta T \) and \( W = nR\Delta T \).
2. \( W = \dfrac{2}{5}Q \).

Teacher's Note:
1. Monatomic: C_P = 5R/2.

Answer: (b) \( \dfrac{1}{3}R(T_2-T_1) \)

Explanation:
1. \( P = aV^2 \), so \( W = \int PdV = \dfrac{a}{3}(V_2^3 - V_1^3) \).
2. With \( PV = RT \) we get \( aV^3 = RT \).
3. \( W = \dfrac{R}{3}(T_2 - T_1) \).

Teacher's Note:
1. Write P in terms of V and integrate.

Answer: (d) \( \dfrac{5}{7} \)

Explanation:
1. \( \dfrac{\Delta U}{Q} = \dfrac{nC_V\Delta T}{nC_P\Delta T} = \dfrac{1}{\gamma} = \dfrac{5}{7} \).

Teacher's Note:
1. The remaining 2/7 of the heat is the work done.

Answer: (d) \( P^{1-\gamma}T^{\gamma} = \text{constant} \)

Explanation:
1. Combine \( PV^\gamma = c \) with \( V = \dfrac{nRT}{P} \).
2. \( P\left(\dfrac{T}{P}\right)^\gamma = c \), so \( P^{1-\gamma}T^\gamma = \text{constant} \).

Teacher's Note:
1. Check with TV^(γ−1) = constant for consistency.

Answer: (d) may diminish and may also increase

Explanation:
1. In an adiabatic process \( \Delta U = -W \).
2. During compression W is negative and U increases; during expansion U decreases.
3. So it may either increase or diminish.

Teacher's Note:
1. The direction depends on whether the gas is compressed or expanded.

Answer: (b) \( 12.48\times10^2 \) J

Explanation:
1. \( \Delta U = nC_V\Delta T = 1\times\dfrac{3}{2}\times8.31\times100 = 1246.5 \) J.
2. \( \approx 12.48\times10^2 \) J.

Teacher's Note:
1. ΔU depends only on ΔT for an ideal gas, whether at constant pressure or not.

Answer: (a) work is done in the expansion of the gas at constant pressure

Explanation:
1. At constant pressure the gas expands, so part of the heat supplied goes into external work.
2. The extra heat per degree is the work, \( C_P - C_V = R \).

Teacher's Note:
1. At constant volume no work is done.

Answer: (b) internal energy is increased

Explanation:
1. In adiabatic compression \( Q = 0 \), so \( \Delta U = -W > 0 \) (W is negative).
2. The work done on the gas raises its internal energy and hence its temperature.

Teacher's Note:
1. No heat is supplied, only work is done on the gas.

Answer: (a) −33°C

Explanation:
1. The sudden burst is adiabatic and the pressure falls to 1 atm.
2. \( T_2 = T_1\left(\dfrac{P_2}{P_1}\right)^{(\gamma-1)/\gamma} = 300\times\left(\dfrac{1}{2}\right)^{1/3} = \dfrac{300}{1.251} = 240 \) K.
3. \( 240 - 273 = -33 \)°C.

Teacher's Note:
1. Sudden expansion cools the gas.

Answer: (a) zero

Explanation:
1. Specific heat \( = \dfrac{Q}{m\Delta T} \).
2. In an adiabatic change \( Q = 0 \) while ΔT is not zero, so the specific heat is zero.

Teacher's Note:
1. In an isothermal change it is infinite.

Answer: (b) rise

Explanation:
1. A sudden compression is adiabatic and work is done on the gas.
2. Its internal energy and temperature rise.

Teacher's Note:
1. Fast compression leaves no time for heat to escape.

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

Explanation:
1. For the isothermal expansion \( P_2 = \dfrac{P_1}{2} \). For the adiabatic expansion \( P_2 = \dfrac{P_1}{2^\gamma} \), which is smaller.
2. The adiabatic curve falls more steeply, so the area under it (work) is also smaller.
3. So the first vessel has both more pressure and more work.

Teacher's Note:
1. The adiabatic curve is steeper than the isothermal one.

Answer: (b) heat is rejected by the gas

Explanation:
1. In an isothermal process \( \Delta U = 0 \).
2. Work is done on the gas, so \( Q = W < 0 \): the gas gives out heat.

Teacher's Note:
1. Pressure rises on isothermal compression.

Answer: (a) slope of adiabatic curve = γ × (slope of isothermal curve)

Explanation:
1. Isothermal: \( \dfrac{dP}{dV} = -\dfrac{P}{V} \). Adiabatic: \( \dfrac{dP}{dV} = -\gamma\dfrac{P}{V} \).
2. So the adiabatic slope is γ times the isothermal slope.

Teacher's Note:
1. γ is greater than 1, so the adiabatic curve is steeper.

Answer: (c) \( 3P_1V_1 \)

Explanation:
1. Work equals the area of the triangle.
2. Base \( = 3V_1 - V_1 = 2V_1 \), height \( = 4P_1 - P_1 = 3P_1 \).
3. Area \( = \dfrac{1}{2}\times2V_1\times3P_1 = 3P_1V_1 \).
4. The expansion A to B is at higher pressure than the compression B to C, so W is positive.

Teacher's Note:
1. Area of a triangle is half base times height.

Answer: (d) −1245 J

Explanation:
1. \( W = \dfrac{nR(T_1 - T_2)}{\gamma - 1} = \dfrac{8.3\times(300 - 375)}{0.5} = -1245 \) J.

Teacher's Note:
1. The negative sign means work is done on the gas.

Answer: (a) \( \delta Q = dU \)

Explanation:
1. At constant volume \( \delta W = P\,dV = 0 \).
2. The first law \( \delta Q = dU + \delta W \) becomes \( \delta Q = dU \).

Teacher's Note:
1. All heat goes into internal energy.

Answer: (b) 250 J

Explanation:
1. \( W = P\Delta V = 10^3\times0.25 = 250 \) J.

Teacher's Note:
1. Work is measured in joules.

Answer: (c) isothermal

Explanation:
1. In an isothermal change the internal energy of an ideal gas is constant, so \( Q = W \).
2. All the heat supplied becomes work.

Teacher's Note:
1. This does not violate the second law because the gas ends up in a different state.