JEE Physics Thermodynamics MCQs Set 20

Answer: (b) \(T_2 = 0\ \text{K}\)

Explanation:
1. \(\eta = 1 - \frac{T_2}{T_1} = 1\) requires \(T_2 = 0\ \text{K}\).
2. \(T_1 = 0\) would give an undefined or negative efficiency.

Teacher's Note:
1. Absolute zero cannot be reached, so 100% is never achieved.
2. Use Kelvin always.

Answer: (a) isothermal expansion

Explanation:
1. The Carnot cycle consists of isothermal expansion at \(T_1\), adiabatic expansion, isothermal compression at \(T_2\) and adiabatic compression, in that order.

Teacher's Note:
1. Two isothermal and two adiabatic steps.
2. Isothermal expansion at the hot source starts the cycle.

Answer: (c) reach absolute zero temperature

Explanation:
1. \(\eta = 1 - \frac{T_2}{T_1}\) equals 1 only if \(T_2 = 0\ \text{K}\).
2. Absolute zero is unattainable.

Teacher's Note:
1. The limit is set by the sink temperature.
2. Friction is not part of the ideal Carnot cycle.

Answer: (d) Adiabatic

Explanation:
1. Work done on the gas is the area under the P-V curve.
2. Starting from the same state and compressing to the same volume, the adiabatic curve is steepest and encloses the largest area.
3. Isochoric gives no change in volume.

Teacher's Note:
1. Adiabatic compression also raises temperature, so pressure is higher.
2. Higher pressure over the same volume change means more work.

Answer: (c) entropy

Explanation:
1. Entropy measures disorder of a system.

Teacher's Note:
1. Entropy is a state function.
2. In natural processes the entropy of the universe increases.

Answer: (a) I only

Explanation:
1. Specific heat is heat per unit mass per degree, so it does not depend on the mass: I is false.
2. It varies slightly with temperature (II is true) and depends on the material (III is true).

Teacher's Note:
1. Heat capacity depends on mass; specific heat does not.
2. Specific heat is a material property.

Answer: (a) I only

Explanation:
1. For an ideal gas \(U\) depends on temperature only.
2. Isothermal: \(\Delta U = 0\), so I is true and III is false.
3. Adiabatic: \(\Delta U = -W \neq 0\), so II is false.

Teacher's Note:
1. Internal energy of an ideal gas follows temperature.
2. Adiabatic changes alter temperature.

Answer: (c) III only

Explanation:
1. Two isothermal curves belong to different temperatures and cannot meet: I is true.
2. Rising air expands adiabatically and cools: II is true.
3. Compression of a gas (adiabatic) heats it: III is false.

Teacher's Note:
1. Compression heats, expansion cools.
2. That is the reason for cooling of rising air.

Answer: (a) I and II

Explanation:
1. An isothermal change needs thermal contact with the surroundings, so I is true.
2. An adiabatic change has no heat exchange, so II is true.
3. For an ideal gas internal energy is constant in an isothermal change, so III is false.

Teacher's Note:
1. Isothermal: good conductor and slow change.
2. Adiabatic: insulated or very fast change.

Answer: (c) I and II

Explanation:
1. I is the Kelvin-Planck statement and II is the Clausius statement.
2. Both are forms of the second law.

Teacher's Note:
1. Two classic statements of the second law.
2. They are equivalent to each other.

Answer: (b) II only

Explanation:
1. No engine can beat a Carnot engine working between the same temperatures: I false, II true.
2. The efficiency of a Carnot engine does not depend on the working substance, so III is false.

Teacher's Note:
1. Carnot efficiency depends only on the two temperatures.
2. The substance is irrelevant.

Answer: (d) I, II and III

Explanation:
1. I follows from the second law.
2. II is the zeroth law.
3. During melting the temperature is constant, so the change in internal energy is that of potential energy.
4. All three are correct.

Teacher's Note:
1. Zeroth law defines temperature.
2. Latent heat changes potential energy of molecules.

Answer: (d) I and III

Explanation:
1. A Carnot cycle has two isothermal and two adiabatic steps, so I is incorrect.
2. It is a reversible engine, so II is correct.
3. Its efficiency is less than 100%, so III is incorrect.

Teacher's Note:
1. Count the steps: two isothermal and two adiabatic.
2. Efficiency is always below 1.

Answer: (d) I, II, III and IV

Explanation:
1. All four are natural, dissipative processes that cannot spontaneously reverse.
2. So all four are irreversible.

Teacher's Note:
1. Free expansion, combustion, leakage and heat conduction all increase entropy.
2. Natural processes are irreversible.

Answer: (a) (A)-(4), (B)-(3), (C)-(2), (D)-(1)

Explanation:
1. Isothermal means constant temperature (4).
2. Isobaric means constant pressure (3).
3. Isochoric means constant volume (2).
4. Adiabatic means \(\Delta Q = 0\) (1).

Teacher's Note:
1. Prefixes iso-thermal, iso-baric and iso-choric refer to T, P and V.
2. Adiabatic means no heat flow.

Answer: (a) (A)-(3), (B)-(1), (C)-(2), (D)-(4)

Explanation:
1. The coefficient of volume expansion of an ideal gas is \(\frac{1}{273}\) per degree for all gases (A-3).
2. Boyle's law gives a decrease in pressure on expansion (B-1).
3. An ideal gas obeys the laws at all temperatures (C-2).
4. A real gas behaves ideally at high temperature and low pressure (D-4).

Teacher's Note:
1. Ideal gas laws are limiting laws of real gases.
2. Real gases approach ideal behaviour at low pressure and high temperature.

Answer: (b) 6 cal

Explanation:
1. Along iaf: \(\Delta U = Q - W = 50 - 20 = 30\ \text{cal}\).
2. Internal energy change is path independent, so along ibf \(W = Q - \Delta U = 36 - 30 = 6\ \text{cal}\).

Teacher's Note:
1. Internal energy change is the same for every path between two states.
2. Work and heat change with the path.

Answer: (b) 460 J

Explanation:
1. Work along BC: \(W = 6\times10^4 \times 2\times10^{-3} = 120\ \text{J}\); AB does no work.
2. \(\Delta U_{AC} = Q - W = 500 - 120 = 380\ \text{J}\).
3. Work along the straight line AC is the trapezoid area: \(\frac{2+6}{2}\times10^4 \times 2\times10^{-3} = 80\ \text{J}\).
4. \(Q_{AC} = 380 + 80 = 460\ \text{J}\).

Teacher's Note:
1. Internal energy change is the same on both paths.
2. Work is the area under each path.

Answer: (d) The change in internal energy in the process BC is \(-500R\)

Explanation:
1. For a diatomic gas \(C_V = \frac{5}{2}R\).
2. Whole cycle: \(\Delta U = 0\). CA: \(\frac{5}{2}R(400-600) = -500R\). AB: \(\frac{5}{2}R(800-400) = 1000R\).
3. BC: \(\frac{5}{2}R(600-800) = -500R\), so (d) is correct.

Teacher's Note:
1. Use \(\Delta U = nC_V\Delta T\) for any process of an ideal gas.
2. Check each option separately.

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

Explanation:
1. Heat can be added during a phase change or isothermal expansion without a temperature rise, so the Assertion is false.
2. The Reason is also false for the same cases.

Teacher's Note:
1. Latent heat changes the state at constant temperature.
2. Isothermal expansion absorbs heat at constant temperature.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. In an isothermal process of an ideal gas \(\Delta U = 0\), so the heat is converted into work, not internal energy: the Assertion is false.
2. The first law as written is correct, so the Reason is true.

Teacher's Note:
1. Isothermal means no change in internal energy of an ideal gas.
2. Heat becomes work.

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

Explanation:
1. The first law is the conservation of energy including heat, so the Assertion is true.
2. Energy is a fundamental quantity: true, but this does not explain why the law is a restatement of energy conservation.

Teacher's Note:
1. Both statements are true.
2. The second does not explain the first.

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

Explanation:
1. \(C_P > C_V\), so the Assertion is false.
2. At constant pressure extra heat is needed for expansion work, so the Reason is also false as stated.

Teacher's Note:
1. Constant pressure allows the gas to expand and do work.
2. So \(C_P\) is the larger one.

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

Explanation:
1. \(c = \frac{Q}{m\Delta T}\).
2. Adiabatic: \(Q = 0\), so \(c = 0\). Isothermal: \(\Delta T = 0\), so \(c \to \infty\).
3. The Reason gives the formula that explains the Assertion.

Teacher's Note:
1. The definition of specific heat explains both extreme cases.
2. Use \(c = Q/(m\Delta T)\).

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

Explanation:
1. The gas rushes out and expands quickly, which is adiabatic.
2. The temperature falls and water vapour condenses into fog, so the Reason explains the Assertion.

Teacher's Note:
1. Rapid expansion cools a gas.
2. Cooling makes vapour condense.

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

Explanation:
1. Isothermal curves belong to different temperatures and never intersect: the Assertion is false.
2. The slope of an isothermal curve is \(-P/V\), not necessarily small: the Reason is false.

Teacher's Note:
1. A point on the P-V diagram has only one temperature.
2. So two isotherms cannot cross.

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

Explanation:
1. At a common point \(P\) and \(V\) are equal, so \(T = \frac{PV}{nR}\) would be the same for both curves, which contradicts different temperatures.
2. The Reason explains the Assertion.

Teacher's Note:
1. Temperature is fixed by \(P\) and \(V\).
2. Two isotherms at different temperatures cannot share a point.