# Gas Dynamics Questions and Answers – Second Law of Thermodynamics (Entropy Equation)

This set of Gas Dynamics Multiple Choice Questions & Answers (MCQs) focuses on “Second Law of Thermodynamics (Entropy Equation)”.

1. What does the second law of thermodynamics provide?
a) Heat transfer during the process
b) Total work during the process
c) Direction of process
d) Nature of the process

Explanation: According to the Second law of thermodynamics the process always proceeds in the direction so that the entropy of the system and surroundings always increases. The first law of thermodynamics lacks such information. Thus the second law of thermodynamics provides information about the direction of the process.

2. The change in entropy for a reversible process is given by which of the following expression?
a) ΔS = ΔQ / T
b) ΔS = ΔW / T
c) ΔS = ΔQ / T + ΔSirrev
d) ΔS = ΔW / T + ΔSirrev

Explanation: For any thermodynamic process, the change in entropy is defined as the ratio of heat transferred during a process to the absolute temperature of the system in addition to the change in entropy due to the irreversible phenomenon.”
I.e. ΔS = (ΔQ / T) + ΔSirreversible
However since the process is reversible, ΔSirreversible = 0.
Hence the change in entropy for a reversible process is expressed as ΔS = ΔQ / T.

3. It is possible to build a perpetual motion machine of a second kind.
a) True
b) False

Explanation: A perpetual motion machine of a second kind refers to a heat engine with only a heat source i.e. without any cooling source. And as per Planck’s statement, “It is impossible to construct an engine that while working in a complete cycle, will produce no effect other than increasing weight and cooling of a heat source.” Hence it is impossible to build a perpetual motion machine of a second kind.

4. For 1 kg of air moving at 70 m / s, and having a static enthalpy of 260 J, calculate specific stagnation enthalpy.
a) 2.53 KJ
b) 2.64 KJ
c) 2.71 KJ
d) 2.94 KJ

Explanation: The specific enthalpy is given by:
h = H / m
And for 1 kg of mass, h = H
Now specific stagnation enthalpy,
h0 = h + $$\frac {V^2}{2}$$
Thus
h0 = 260 + $$\frac {70^2}{2}$$
h0 = 2710 J = 2.71 KJ

5. A reversible heat engine transfers 700 J of heat to constant temperature source at 390 K. If the surrounding is having a temperature of 288 K. Calculate the change in entropy of the system.
a) 0.78 J / K
b) -1.03 J / K
c) -1.05 J / K
d) -0.87 J / K

Explanation: For a reversible process change in entropy is given by ΔS = ΔQ / T
Therefore, ΔS = -700 / (390 + 288)
ΔS = -1.03 J / K

6. The following refers to which kind of thermodynamic cycle?

a) Carnot cycle
b) Rankine cycle
c) Otto cycle
d) Brayton cycle

Explanation: Here the indicated cycle is the Carnot cycle, as during process 1 – 2 heat is adiabatic compression occurs. Then during process 2 – 3 fluid is isothermally expanded as heat is added to the system. During process 3 – 1 adiabatic expansion occurs and in process 4 – 1 fluid is isothermally compressed as heat is removed from the system.

7. An engine operating at 475 K receives the heat of 860 J and produces work of 412 J. The temperature of surrounding is 288 K.
a) True
b) False

Explanation: The Carnot efficiency of an engine is given by:
η = [(TH – Tc) / TH] * 100
η = [(475 – 288) / 475] * 100
η = 39.36 %
Thus, a given engine having efficiency of 39.36 %, produces a work of 338.5 J.

8. What will be the coefficient of performance for a perfect refrigerator?
a) 0
b) 1
c) Infinite
d) 0.5

Explanation: As indicated by the second law of thermodynamics given by the Kelvin – Plank, a perfect engine cannot be created that utilizes all of the heat energy to produce the work, without any loss. Now since the perfect refrigerator should not do any work to cool the object, as per the Kelvin – Plank statement, then W = 0.
And since the coefficient of performance for a refrigerator is:
COP = Qc / W
Then, COP = Infinite     (because W = 0)

9. In an adiabatic process, the entropy of the system _________
a) Decreases in the direction of the process
b) Increases in the direction of the process
c) Remains constant
d) Becomes zero

Explanation: For an adiabatic process there is no heat transfer in the system i.e. ΔQ = 0. Now since the change in entropy is given by;
ΔS = (ΔQ / T) + ΔSirreversible
ΔS = ΔSirreversible,
Thus ΔS > 0
Hence the entropy of the system always increases in the direction of the process.

10. Calculate the efficiency of a perfect heat engine.
a) 0.5
b) Infinite
c) 0
d) 1

Explanation: As per the Kelvin – Plank statement, a perfect heat engine does not have a heat loss while converting the heat into useful work. Hence for a perfect heat engine the temperature of a cooling source (Qc) = 0.
Thus,the efficiency of a perfect heat engine; η = (QH – Qc) / QH
η = 1

Sanfoundry Global Education & Learning Series – Gas Dynamics.

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