JEE Physics Thermodynamics MCQs Set 19

Answer: (c) Internal energy

Explanation:
1. A state variable depends only on the present state of the system, not on the path followed.
2. Heat and work are path dependent, whereas internal energy is fixed by the state.
3. So internal energy is the state variable.

Teacher's Note:
1. Heat and work are energy in transit, not properties stored in the system.
2. Internal energy, pressure, volume and temperature are state variables.

Answer: (b) pressure of the gas

Explanation:
1. For an ideal gas \(U = \frac{f}{2}nRT\).
2. It depends on the number of moles, on the temperature and on the degrees of freedom (atomicity).
3. It does not depend on pressure separately.

Teacher's Note:
1. Ideal gas molecules have no intermolecular forces, so only kinetic energy counts.
2. Pressure matters only through the temperature it implies.

Answer: (b) in liquid state is less than that in gaseous state

Explanation:
1. Converting a liquid to vapour at constant temperature needs latent heat.
2. This heat raises the potential energy of the molecules, so the vapour has more internal energy.
3. Hence the liquid has less internal energy than the gas.

Teacher's Note:
1. Latent heat changes internal energy without changing temperature.
2. Compare states at the same temperature using the latent heat absorbed.

Answer: (c) It introduces the concept of entropy

Explanation:
1. The first law is energy conservation and defines internal energy.
2. It applies to every process including cyclic ones.
3. Entropy is introduced by the second law, so (c) is incorrect.

Teacher's Note:
1. First law: energy; zeroth law: temperature; second law: entropy.
2. Remember which law introduces which quantity.

Answer: (b) Internal energy and entropy are state functions

Explanation:
1. Entropy change is zero in a reversible adiabatic process, so (a) is false.
2. Internal energy does not change in an isothermal process of an ideal gas, so (c) is false.
3. Adiabatic work equals the fall in internal energy, so (d) is false.
4. Both internal energy and entropy depend only on the state, so (b) is correct.

Teacher's Note:
1. A state function depends only on the initial and final states.
2. Test each option with one counter-example.

Answer: (b) \(C = 3R\)

Explanation:
1. Each atom in a solid vibrates in three dimensions, giving \(6\) quadratic terms.
2. Energy per mole is \(U = 3RT\).
3. So \(C = \frac{dU}{dT} = 3R\) (law of Dulong and Petit).

Teacher's Note:
1. This law holds at ordinary and high temperatures.
2. It is independent of the temperature.

Answer: (c) \(PV^{\gamma} = \text{constant}\)

Explanation:
1. For a reversible adiabatic change of an ideal gas \(dQ = 0\).
2. Combining this with \(PV = nRT\) and \(C_P/C_V = \gamma\) gives \(PV^{\gamma} = \text{constant}\).

Teacher's Note:
1. \(PV = \text{constant}\) is for the isothermal case.
2. The adiabatic curve is steeper because of the factor \(\gamma\).

Answer: (d) \(C_P - C_V = 2R\)

Explanation:
1. Mayer's relation is \(C_P - C_V = R\) for one mole.
2. Using \(\gamma = C_P/C_V\) gives \(C_V = \frac{R}{\gamma-1}\) and \(C_P = \frac{\gamma R}{\gamma-1}\).
3. So (d) is wrong.

Teacher's Note:
1. Mayer's relation has no factor of 2.
2. Derive \(C_V\) and \(C_P\) from Mayer's relation and \(\gamma\) to check any variant.

Answer: (d) more than one

Explanation:
1. \(C_P = C_V + R\) with \(R > 0\), so \(C_P > C_V\).
2. Therefore \(\gamma = C_P/C_V > 1\).

Teacher's Note:
1. At constant pressure the gas also does expansion work, so it needs more heat.
2. That is why \(C_P\) exceeds \(C_V\).

Answer: (a) Adiabatic compression

Explanation:
1. With no heat exchange, \(\Delta U = -W_{by}\).
2. In adiabatic compression work is done on the gas, so \(\Delta U > 0\) and the temperature rises.

Teacher's Note:
1. Compressing a gas quickly heats it, as in a bicycle pump.
2. Isothermal changes keep the temperature fixed.

Answer: (a) infinite

Explanation:
1. Specific heat is \(c = \frac{Q}{m\,\Delta T}\).
2. In an isothermal process heat is absorbed while \(\Delta T = 0\), so \(c \to \infty\).

Teacher's Note:
1. Adiabatic process has \(Q = 0\), so specific heat is zero.
2. Isothermal has \(\Delta T = 0\), so specific heat is infinite.

Answer: (b) change in temperature

Explanation:
1. In an adiabatic change \(W_{by} = -\Delta U = -nC_V\Delta T\).
2. So the work depends only on the change in temperature (for a given amount of gas).

Teacher's Note:
1. Use the first law with \(Q = 0\).
2. Work depends on the temperature change only.

Answer: (b) Bursting of a tyre

Explanation:
1. A tyre bursts so rapidly that the escaping air has no time to exchange heat.
2. So it is an adiabatic process.
3. Melting of ice needs heat from outside.

Teacher's Note:
1. Fast processes are nearly adiabatic.
2. Very slow processes in conducting walls are nearly isothermal.

Answer: (c) \(TV^{\gamma-1} = \text{constant}\)

Explanation:
1. Put \(P = \frac{nRT}{V}\) in \(PV^{\gamma} = \text{constant}\).
2. This gives \(TV^{\gamma-1} = \text{constant}\).

Teacher's Note:
1. There are three equivalent forms: in \(P,V\), in \(T,V\) and in \(P,T\).
2. Remember \(\gamma - 1\) goes with \(TV\).

Answer: (a) \(\Delta U = -W\) in an adiabatic process

Explanation:
1. First law: \(Q = \Delta U + W\).
2. In an adiabatic process \(Q = 0\), so \(\Delta U = -W\).
3. For an ideal gas isothermal process \(\Delta U = 0\), so (b) and (c) fail.

Teacher's Note:
1. Write the first law with the sign convention given in the question.
2. Then set \(Q = 0\) for adiabatic changes.

Answer: (c) adiabatic slope = \(\gamma \times\) isothermal slope

Explanation:
1. Isothermal: \(PV = c\), so \(\frac{dP}{dV} = -\frac{P}{V}\).
2. Adiabatic: \(PV^{\gamma} = c\), so \(\frac{dP}{dV} = -\gamma\frac{P}{V}\).
3. So the adiabatic slope is \(\gamma\) times the isothermal slope.

Teacher's Note:
1. Since \(\gamma > 1\), the adiabatic curve is steeper.
2. Differentiate both laws at the same point.

Answer: (b) isobaric

Explanation:
1. Work is the area under the P-V curve.
2. Starting at the same point, the isobaric line stays highest, the isothermal curve falls, and the adiabatic curve falls fastest.
3. So the isobaric expansion does the most work.

Teacher's Note:
1. Compare areas under the three curves from the same starting point.
2. The adiabatic expansion does least work.

Answer: (c) Heat given to a system is used up in raising the temperature only

Explanation:
1. At constant pressure \(V/T\) is constant and \(W = P\Delta V\).
2. Heat supplied is shared between increasing internal energy and doing work, so \(\Delta Q > W\).
3. So (c) is incorrect.

Teacher's Note:
1. In an isobaric expansion some heat becomes work.
2. That is why \(C_P > C_V\).

Answer: (a) In an isochoric process pressure remains constant

Explanation:
1. In an isochoric process the volume stays constant, not the pressure.
2. The other three statements are correct.

Teacher's Note:
1. Iso-choric means same volume; iso-baric means same pressure.
2. Learn the prefixes.

Answer: (d) For free expansion of a gas \(\Delta E_{int} > 0\)

Explanation:
1. In free expansion \(Q = 0\) and \(W = 0\), so \(\Delta E_{int} = 0\).
2. Hence (d) is wrong.
3. Statements (a), (b) and (c) follow from the first law.

Teacher's Note:
1. Free expansion of an ideal gas keeps the temperature constant.
2. Use the first law for each option.

Answer: (c) adiabatic

Explanation:
1. The definition of an adiabatic process is \(Q = 0\).

Teacher's Note:
1. Adiabatic means no heat exchange.
2. It does not mean the temperature is constant.

Answer: (a) \(\frac{\theta_2}{\theta_1 - \theta_2}\)

Explanation:
1. Coefficient of performance is \(\beta = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2}\).
2. For a reversible refrigerator \(\frac{Q_2}{Q_1} = \frac{\theta_2}{\theta_1}\), giving \(\beta = \frac{\theta_2}{\theta_1 - \theta_2}\).

Teacher's Note:
1. The numerator is the heat removed from the cold space.
2. The denominator is the work supplied.

Answer: (b) No heat engine can be 100% efficient

Explanation:
1. Kelvin-Planck statement: no cyclic engine can convert all absorbed heat into work.
2. So no heat engine can be 100% efficient.

Teacher's Note:
1. Some heat must always be rejected to a colder sink.
2. Absolute zero cannot be reached by finite steps.

Answer: (d) \(\Delta U = 0\)

Explanation:
1. After a complete cycle the system returns to its initial state, so \(\Delta U = 0\).
2. Then \(\Delta Q = \Delta W\), which is not zero in general.

Teacher's Note:
1. Internal energy is a state function.
2. Net work in a cycle equals net heat absorbed.

Answer: (d) \(\eta = 1 - \frac{\theta_2}{\theta_1}\)

Explanation:
1. Efficiency is \(\eta = \frac{W}{Q_{in}} = \frac{Q_{in} - Q_{out}}{Q_{in}} = 1 - \frac{\theta_2}{\theta_1}\).

Teacher's Note:
1. Efficiency is always less than one.
2. Options with a plus sign would exceed one.

Answer: (b) \(\frac{\theta_1}{\theta_2} = \frac{T_1}{T_2}\)

Explanation:
1. For a Carnot cycle \(\frac{Q_1}{Q_2} = \frac{T_1}{T_2}\), where \(Q\) is heat and \(T\) the absolute temperature of the reservoir.
2. The other relations do not hold.

Teacher's Note:
1. Heats exchanged are proportional to the temperatures of the reservoirs.
2. Use absolute temperature.

Answer: (d) \(T_1\) is high and \(T_2\) is low

Explanation:
1. \(\eta = 1 - \frac{T_2}{T_1}\) is largest when \(\frac{T_2}{T_1}\) is smallest.
2. That needs a high \(T_1\) and a low \(T_2\).

Teacher's Note:
1. A large temperature difference helps.
2. The ratio matters, not the difference alone.