JEE Physics Thermodynamics MCQs Set 21

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. A reversible engine has no dissipative losses, so it gives the greatest possible efficiency between two given temperatures.
2. The Reason explains the Assertion.

Teacher's Note:
1. Dissipation (friction, heat leaks) lowers efficiency.
2. A reversible engine has none.

Answer: (d) Both Assertion and Reason are false.

Explanation:
1. In a cyclic process \(\Delta U = 0\), so the Reason is true.
2. But the net work equals the net heat absorbed, which is not zero in general (it is the area of the loop): the Assertion is false.

Teacher's Note:
1. Net work in a cycle is the loop area.
2. Only the change in internal energy vanishes.

Answer: (b) Both Assertion and Reason are true but the Reason is not a correct explanation of the Assertion.

Explanation:
1. \(\eta = 1 - \frac{T_2}{T_1}\) increases when \(T_2\) falls: the Assertion is true.
2. The Reason is the definition of efficiency and is true, but it does not by itself explain the temperature dependence.

Teacher's Note:
1. The definition does not mention temperature.
2. The temperature dependence comes from \(\eta = 1 - T_2/T_1\).

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. The Carnot efficiency is an upper limit that any engine can be compared with.
2. So the Reason explains the Assertion.

Teacher's Note:
1. Carnot efficiency is the benchmark.
2. Real engines are always below it.

Answer: (b) 15.4 J

Explanation:
1. Adiabatic A to B: \(Q = 0\), work done on the system is 35 J, so \(\Delta U = +35\ \text{J}\).
2. Other path: \(Q = 12\times4.2 = 50.4\ \text{J}\).
3. \(W = Q - \Delta U = 50.4 - 35 = 15.4\ \text{J}\).

Teacher's Note:
1. Internal energy change is path independent.
2. Convert calories to joules before using the first law.

Answer: (c) 7900 J

Explanation:
1. \(Q = 2000\times4.2 = 8400\ \text{J}\).
2. \(\Delta U = Q - W = 8400 - 500 = 7900\ \text{J}\).

Teacher's Note:
1. Work done by the system is subtracted from the heat absorbed.
2. Convert kcal to joules first.

Answer: (b) 675.2 K and \(3.6\times10^{6}\ \text{N/m}^2\)

Explanation:
1. \(n = \frac{PV}{RT} = \frac{1.6\times10^6\times0.0083}{8.3\times300} = 5.33\ \text{mol}\).
2. Closed vessel means constant volume, \(C_V = C_P - R = \frac{3R}{2}\).
3. \(Q = nC_V\Delta T\) gives \(\Delta T = \frac{2.49\times10^4}{5.33\times1.5\times8.3} \approx 375\ \text{K}\), so \(T = 675\ \text{K}\).
4. \(P = 1.6\times10^6\times\frac{675}{300} = 3.6\times10^6\ \text{N/m}^2\).

Teacher's Note:
1. In a closed rigid vessel use \(C_V\), not \(C_P\).
2. Pressure scales with absolute temperature at constant volume.

Answer: (a) 2.08 kJ

Explanation:
1. \(n = \frac{40}{28} = 1.43\ \text{mol}\).
2. Nitrogen is diatomic: \(C_P = \frac{7}{2}R\).
3. \(Q = nC_P\Delta T = 1.43\times3.5\times8.3\times50 \approx 2075\ \text{J} \approx 2.08\ \text{kJ}\).

Teacher's Note:
1. Use \(C_P\) for constant pressure heating.
2. Nitrogen is diatomic, so \(C_P = 7R/2\).

Answer: (d) 293 cal/K

Explanation:
1. \(Q = mL = 1000\times80 = 8\times10^4\ \text{cal}\).
2. \(\Delta S = \frac{Q}{T} = \frac{8\times10^4}{273} \approx 293\ \text{cal/K}\).

Teacher's Note:
1. Latent heat per gram must be multiplied by the mass in grams.
2. Use absolute temperature.

Answer: (c) 256 atm

Explanation:
1. \(T_1 = 300\ \text{K}\), \(T_2 = 1200\ \text{K}\).
2. Adiabatic: \(P \propto T^{\gamma/(\gamma-1)} = T^{3.5}\).
3. \(P_2 = 2\times4^{3.5} = 2\times128 = 256\ \text{atm}\).

Teacher's Note:
1. Use the form of the adiabatic law that links \(P\) and \(T\).
2. \(\frac{\gamma}{\gamma-1} = 3.5\) for a diatomic gas.

Answer: (b) 668.4 K

Explanation:
1. \(T_1 = 291\ \text{K}\), \(\gamma = 1.4\).
2. \(T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} = 291\times8^{0.4} = 291\times2.297 \approx 668.4\ \text{K}\).

Teacher's Note:
1. Convert Celsius to kelvin before using the adiabatic law.
2. \(8^{0.4} = 2^{1.2}\).

Answer: (c) Change in pressure will be more than three times the initial pressure

Explanation:
1. \(P_2 = P_1\left(\frac{V_1}{V_2}\right)^{\gamma} = 3^{\gamma}P_1\).
2. For \(\gamma \geq 1.4\), \(3^{\gamma} \geq 4.66\), so the change \(P_2 - P_1 = (3^\gamma - 1)P_1 > 3P_1\).

Teacher's Note:
1. In an adiabatic compression the pressure rises more than in an isothermal one.
2. Isothermal compression would give exactly three times.

Answer: (a) \(6.25\times10^{5}\ \text{J}\)

Explanation:
1. Assume a Carnot engine: \(\eta = 1 - \frac{375}{500} = 0.25\).
2. \(W = \eta Q = 0.25\times25\times10^5 = 6.25\times10^5\ \text{J}\).

Teacher's Note:
1. Heat consumed is the heat absorbed from the source.
2. Work = efficiency times heat absorbed.

Answer: (d) 150 J

Explanation:
1. \(\beta = \frac{Q_2}{W} = \frac{1}{3}\) gives \(Q_2 = \frac{W}{3}\).
2. Heat released is \(Q_1 = Q_2 + W = \frac{4W}{3} = 200\ \text{J}\).
3. So \(W = 150\ \text{J}\).

Teacher's Note:
1. Heat released to the hot reservoir includes the work supplied.
2. \(Q_1 = Q_2 + W\).

Answer: (b) 250 K

Explanation:
1. \(\beta = \frac{T_2}{T_1 - T_2} = 5\) with \(T_1 = 300\ \text{K}\).
2. \(T_2 = 5(300 - T_2)\), so \(6T_2 = 1500\) and \(T_2 = 250\ \text{K}\).

Teacher's Note:
1. Take room temperature as the hot reservoir.
2. Solve the linear equation in \(T_2\).

Answer: (c) the room will get a little hotter

Explanation:
1. The refrigerator rejects heat \(Q_1 = Q_2 + W\) to the room, which is more than the heat \(Q_2\) it removes from the room.
2. So the room gets a little hotter.

Teacher's Note:
1. The work supplied by the compressor ends up as heat in the room.
2. A refrigerator cannot cool a closed room.

Answer: (c) \(\frac{P}{64}\)

Explanation:
1. Isothermal: \(P_1 = \frac{P}{2}\) at volume \(2V\).
2. Adiabatic from \(2V\) to \(16V\): ratio 8, so \(P_2 = \frac{P}{2}\times8^{-5/3} = \frac{P}{2}\times\frac{1}{32} = \frac{P}{64}\).

Teacher's Note:
1. Apply the isothermal law first and then the adiabatic law.
2. \(8^{5/3} = 2^5 = 32\).

Answer: (c) 3 J

Explanation:
1. Stored energy is \(\frac{1}{2}kx^2 \propto x^2\).
2. At 1 mm: 1 J. At 2 mm: 4 J.
3. Additional work \(= 4 - 1 = 3\ \text{J}\).

Teacher's Note:
1. Energy grows with the square of the compression.
2. Subtract the energy already stored.

Answer: (b) 8 kJ

Explanation:
1. \(Q = 2000\times4.2 = 8400\ \text{J}\).
2. \(\Delta U = Q - W = 8400 - 400 = 8000\ \text{J} = 8\ \text{kJ}\).

Teacher's Note:
1. Convert kcal to J before subtracting.
2. Work done by the system is subtracted.

Answer: (a) Kinetic energy of the box

Explanation:
1. Internal energy is the energy of the molecular motions and interactions in the frame of the gas.
2. The bulk kinetic energy of the moving box is mechanical energy and is not included.

Teacher's Note:
1. Internal energy is measured relative to the centre of mass.
2. Bulk motion is excluded.

Answer: (d) increases

Explanation:
1. \(\eta = \frac{T_1 - T_2}{T_1}\).
2. After the change \(\eta' = \frac{T_1 - T_2}{T_1 - 100}\) has the same numerator and a smaller denominator.
3. So the efficiency increases.

Teacher's Note:
1. The difference stays the same while the source temperature falls.
2. Hence the ratio increases.

Answer: (b) 0.75

Explanation:
1. \(TV^{\gamma-1}\) is constant with \(\gamma = 1.4\): \(\frac{T_2}{T_1} = \left(\frac{1}{32}\right)^{0.4} = 2^{-2} = \frac{1}{4}\).
2. Efficiency of the cycle \(\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{1}{4} = 0.75\).

Teacher's Note:
1. \(32^{0.4} = 2^{2} = 4\).
2. Efficiency of an Otto-type cycle depends on the expansion ratio.

Answer: (d) 372 K and 310 K

Explanation:
1. \(\frac{T_2}{T_1} = \frac{5}{6}\) and \(\frac{T_2 - 62}{T_1} = \frac{2}{3}\).
2. Subtracting: \(\frac{62}{T_1} = \frac{1}{6}\), so \(T_1 = 372\ \text{K}\).
3. \(T_2 = \frac{5}{6}\times372 = 310\ \text{K}\).

Teacher's Note:
1. Write both efficiencies as temperature ratios.
2. Subtract to eliminate \(T_2\).

Answer: (a) \(16.8\times10^{6}\ \text{J}\)

Explanation:
1. \(\eta = 1 - \frac{300}{1200} = 0.75\).
2. \(Q_1 = \frac{W}{\eta} = \frac{12.6\times10^6}{0.75} = 16.8\times10^6\ \text{J}\).

Teacher's Note:
1. Efficiency links work output with heat absorbed.
2. Convert Celsius to kelvin.

Answer: (b) 250 K, 200 K

Explanation:
1. \(T_2 = 0.8T_1\) and \(T_2 - 50 = 0.6T_1\).
2. Subtracting: \(50 = 0.2T_1\), so \(T_1 = 250\ \text{K}\).
3. \(T_2 = 200\ \text{K}\).

Teacher's Note:
1. Write efficiency as \(1 - T_2/T_1\) twice.
2. Subtract the two equations.

Answer: (d) \(\frac{2}{3}\)

Explanation:
1. \(Q_1 = 3W\), so \(Q_2 = Q_1 - W = 2W\).
2. Fraction rejected \(= \frac{Q_2}{Q_1} = \frac{2}{3}\).

Teacher's Note:
1. Energy in equals work plus energy rejected.
2. The fraction is taken with respect to input.