Q: The SI unit of entropy is:

\text{J}/\text{K}. From \Delta S = \dfrac{Q}{T}, entropy has the units of energy divided by temperature, i.e. \text{J}/\text{K}.

Q: One kilogram of ice melts at 0^\circ\text{C}. Taking the latent heat of fusion as 3.34 \times 10^5\text{ J/kg}, the entropy change of the ice is approximately:

+1223\text{ J/K}. Melting occurs at the constant temperature T = 273\text{ K}, so \Delta S = \dfrac{Q}{T} = \dfrac{3.34 \times 10^5}{273} \approx 1223\text{ J/K}, positive because the liquid is more disordered.

Q: A system absorbs 100\text{ J} of heat reversibly at a constant temperature of 400\text{ K}. Its entropy change is:

0.25\text{ J/K}. \Delta S = \dfrac{Q}{T} = \dfrac{100}{400} = 0.25\text{ J/K}.

Question 1

The change in entropy of a system that absorbs heat Q reversibly at constant absolute temperature T is:

Accepted Answer

\dfrac{Q}{T}. By definition \Delta S = \dfrac{Q_{rev}}{T} for an isothermal reversible exchange of heat.

Question 2

The SI unit of entropy is:

Accepted Answer

ext{J}/	ext{K}. From \Delta S = \dfrac{Q}{T}, entropy has the units of energy divided by temperature, i.e. 	ext{J}/	ext{K}.

Question 3

For any irreversible process occurring in an isolated system, the total entropy:

Accepted Answer

Always increases. This is the entropy statement of the second law: natural, spontaneous processes always increase the entropy of the universe.

Question 4

For a reversible process, the total entropy change of the system together with its surroundings is:

Accepted Answer

Zero. A reversible process is the limiting ideal case in which the entropy gained by one part exactly equals that lost by the other, so \Delta S_{universe} = 0.

Question 5

Entropy is best described physically as a measure of the:

Accepted Answer

Disorder or randomness of a system. Statistically, entropy counts the number of microscopic arrangements consistent with the macroscopic state; more disorder means more arrangements and higher entropy.

Question 6

Entropy is:

Accepted Answer

A state function, independent of the path followed. The entropy change between two states is the same whatever path connects them, which is why \Delta S can be computed along a convenient reversible path.

Question 7

One kilogram of ice melts at 0^\circ	ext{C}. Taking the latent heat of fusion as 3.34 	imes 10^5	ext{ J/kg}, the entropy change of the ice is approximately:

Accepted Answer

+1223	ext{ J/K}. Melting occurs at the constant temperature T = 273	ext{ K}, so \Delta S = \dfrac{Q}{T} = \dfrac{3.34 	imes 10^5}{273} \approx 1223	ext{ J/K}, positive because the liquid is more disordered.

Question 8

A system absorbs 100	ext{ J} of heat reversibly at a constant temperature of 400	ext{ K}. Its entropy change is:

Accepted Answer

0.25	ext{ J/K}. \Delta S = \dfrac{Q}{T} = \dfrac{100}{400} = 0.25	ext{ J/K}.

Question 9

In a reversible adiabatic process, the entropy of the system:

Accepted Answer

Remains constant, so the process is called isentropic. With Q = 0 and the process reversible, \Delta S = \dfrac{Q_{rev}}{T} = 0, so the entropy is unchanged throughout.

Question 10

The entropy change of n moles of an ideal gas during a reversible isothermal expansion from V_1 to V_2 is:

Accepted Answer

nR\ln\dfrac{V_2}{V_1}. For an isothermal process Q = W = nRT\ln\dfrac{V_2}{V_1}, so \Delta S = \dfrac{Q}{T} = nR\ln\dfrac{V_2}{V_1}.

Question 11

The free expansion of a gas into a vacuum is accompanied by:

Accepted Answer

An increase in entropy, since the process is irreversible. Although Q = 0, the process is irreversible. The gas occupies a larger volume with more accessible microstates, so its entropy rises.

Question 12

Heat flows spontaneously from a hot body to a cold body because this direction of flow:

Accepted Answer

Increases the total entropy of the universe. The cold body gains \dfrac{Q}{T_C} while the hot body loses \dfrac{Q}{T_H}. Since T_C < T_H, the gain exceeds the loss and the net entropy change is positive.

Question 13

The second law of thermodynamics is needed, over and above the first law, because it:

Accepted Answer

Determines the direction in which a natural process proceeds. Many energy-conserving processes never happen spontaneously. The second law, through entropy, distinguishes the possible direction from the impossible one.

Question 14

The third law of thermodynamics states that as the absolute temperature approaches zero, the entropy of a perfect crystal:

Accepted Answer

Approaches zero. At 0	ext{ K} a perfect crystal has only one possible microscopic arrangement, so its disorder — and hence its entropy — tends to zero.

Question 15

Of the following, the process with the largest increase in entropy is:

Accepted Answer

Ice melting into water at 0^\circ	ext{C}. Melting converts an ordered crystal into a disordered liquid, so entropy increases. Freezing and isothermal compression both decrease it, and reversible adiabatic change leaves it unaltered.

Question 16

For a complete cycle of a reversible heat engine, the total entropy change of the working substance is:

Accepted Answer

Zero. Entropy is a state function, so returning the working substance to its initial state makes its net entropy change zero over the cycle.

Question 17

A system absorbs 500	ext{ J} of heat reversibly at a constant temperature of 250	ext{ K}. Its entropy change is:

Accepted Answer

2	ext{ J/K}. \Delta S = \dfrac{Q}{T} = \dfrac{500}{250} = 2	ext{ J/K}.

Question 18

Two kilograms of ice melt at 0^\circ	ext{C}. Taking the latent heat of fusion as 3.34 	imes 10^5	ext{ J/kg}, the entropy change is approximately:

Accepted Answer

2447	ext{ J/K}. Q = mL = 2 	imes 3.34 	imes 10^5 = 6.68 	imes 10^5	ext{ J} at T = 273	ext{ K}, so \Delta S = \dfrac{6.68 	imes 10^5}{273} \approx 2447	ext{ J/K}.

Question 19

Two moles of an ideal gas expand reversibly and isothermally until the volume doubles. The entropy change is (R = 8.314, \ln 2 = 0.693):

Accepted Answer

11.5	ext{ J/K}. \Delta S = nR\ln\dfrac{V_2}{V_1} = 2 	imes 8.314 	imes 0.693 \approx 11.5	ext{ J/K}.

Question 20

An amount of heat 1000	ext{ J} flows from a body at 400	ext{ K} to one at 300	ext{ K}. The net entropy change of the system is approximately:

Accepted Answer

+0.83	ext{ J/K}. The cold body gains \dfrac{1000}{300} = 3.33	ext{ J/K} and the hot body loses \dfrac{1000}{400} = 2.50	ext{ J/K}. The net change is +0.83	ext{ J/K}, positive as required for an irreversible process.

Entropy & Second Law — MH-CET Physics MCQs with Solutions

Questions with solutions