250+ TOP MCQs on Polytropic Process-2 and Answers

Thermodynamics Inteview Questions and Answers for freshers on “Polytropic Process-2”.

1. A cylinder/piston contains air at 100 kPa and 20°C with a V=0.3 m^3. The air is compressed to 800 kPa in a reversible polytropic process with n = 1.2, after which it is expanded back to 100 kPa in a reversible adiabatic process. Find the net work.
a) -174.6 kJ/kg
b) -154.6 kJ/kg
c) -124.6 kJ/kg
d) -194.6 kJ/kg
Answer: a
Clarification: m = P1V1/RT1 = (100 × 0.3)/(0.287 × 293.2) = 0.3565 kg
T2/T1 = (P2/P1)^[(n-1)/n] = 293.2(800/100)^(0.167) = 414.9 K
W = (P2 v2 – P1 v1)/(1-n) = R(T2-T1)/(1-n)
= 0.287(414.9-293.2)/(1-1.20) = -174.6 kJ/kg.

2. A piston-cylinder contains carbon dioxide at 2MPa with V=50 L. The device has a mass of 4 kg. Everything is initially at 200°C. By heat transfer the whole system cools to 25°C, at which point the gas pressure is 1.5 MPa. Find the work done.
a) -10.0 kJ
b) -12.0 kJ
c) -14.0 kJ
d) -16.0 kJ
Answer: c
Clarification: CO2: m = P1V1/RT1 = 2000 × 0.05/(0.18892 × 473.2) = 1.1186 kg
V2 = V1(P1/P2)(T2/T1) = 0.05(2/1.5)(298.2/473.2) = 0.042 m^3
Work = ⌠PdV = (P1 + P2)(V2 – V1)/2 = (2000 + 1500)(0.042 – 0.050)/2
= -14.0 kJ.

3. A gas initially at 500°C, 1 MPa is contained in a piston-cylinder arrangement with an initial volume of 0.1 m^3. It is then slowly expanded according to the relation PV = constant until a final pressure of 100 kPa is attained. Determine the work for this process.
a) 200.3 kJ
b) 210.3 kJ
c) 220.3 kJ
d) 230.3 kJ
Answer: d
Clarification: Process: PV = C ⇒ V2 = P1V1/P2 = 1000 × 0.1/100 = 1 m^3
1W2 = ∫ P dV = ⌠ CV^(-1)dV = C ln(V2/V1)
1W2 = P1V1 ln(V2/V1) = 1000 × 0.1 ln (1/0.1) = 230.3 kJ.

4. Helium gas expands from 350 K, 125 kPa and 0.25 m^3 to 100 kPa in a polytropic process with n = 1.667. How much work does it give out?
a) 3.09 kJ
b) 4.09 kJ
c) 5.09 kJ
d) 6.09 kJ
Answer: b
Clarification: Process: PV^n = constant = P1(V1)^n = P2(V2)^n
V2 = V1 (P1/P2)^(1/n) = 0.25 × (125/100)^(0.6) = 0.2852 m^3
Work = (P2V2 – P1V1)/(1-n) = (100× 0.2852 – 125× 0.25)/(1 – 1.667)
= 4.09 kJ.

5. Air goes through a polytropic process from 325 K, 125 kPa to 500 K, 300 kPa. Find the specific work in the process.
a) -51.8 kJ/kg
b) -61.8 kJ/kg
c) -71.8 kJ/kg
d) -81.8 kJ/kg
Answer: a
Clarification: Process: Pv^(n) = Const = P1(v1)^n = P2(v2)^n
Ideal gas Pv = RT hence v1 = RT/P = 0.287 × 325/125 = 0.7462 m^3/kg
v1 = RT/P = 0.287 × 500/300 = 0.47833 m^3/kg
n = ln(P2/P1) / ln(v1/v2) = ln 2.4 / ln 1.56 = 1.969
Work = (P2v2 – P1v1)/(1-n) = R(T2-T1)/(1-n) = 0.287(500 – 325)/(1-1.969)
= -51.8 kJ/kg.

6. A piston-cylinder contains 0.1 kg air at 400 K, 100 kPa which goes through a polytropic compression process (n = 1.3) to a pressure of 300 kPa. How much work has been done by air in the process?
a) -277 kJ
b) -377 kJ
c) -477 kJ
d) -577 kJ
Answer: c
Clarification: Process: Pv^(n) = Const;
T2 = T1 ( P2 V2 / P1V1) = T1 ( P2 / P1)(P1 / P2 )^(1/n)
= 400 × (300/100)^(1 – 1/1.3) = 515.4 K
Work = (P2V2 – P1V1)/(1-n) = mR(T2-T1)/(1-n)
= (0.2 × 0.287)(515.4-400)/(1 − 1.3) = -477 kJ.

7. A balloon behaves according to the equation P = (C2)V^(1/3), C2 = 100 kPa/m. The balloon is blown up with air from a volume of 1 m^3 to a volume of 3 m^3. Find the work done by the air assuming it is at 25°C.
a) 219.5 kJ
b) 229.5 kJ
c) 239.5 kJ
d) 249.5 kJ
Answer: d
Clarification: The process is polytropic with exponent n = -1/3.
P1 = (C2)V^(1/3) = 100 × 1^(1/3) = 100 kPa
P1 = (C2)V^(1/3) = 100 × 3^(1/3) = 144.22 kPa
Work = ⌠PdV = (P2V2 – P1V1)/(1-n) = (144.22 × 3 – 100 × 1)/(1 – (-1/3))
= 249.5 kJ.

8. A balloon behaves such that pressure inside it is proportional to the diameter squared. It contains 2kg of ammonia at 0°C, 60% quality. They are now heated so that the final pressure is 600 kPa. Find the work done in the process.
a) 117.5 kJ
b) 127.5 kJ
c) 137.5 kJ
d) 147.5 kJ
Answer: a
Clarification: Process : P∝D^2, with V ∝ D^3 this implies P∝D^2 ∝ V^(2/3)
so PV^(-2/3) = constant, hence n = −2/3
V1 = mv1 = 2(0.001566 + 0.6 × 0.28783) = 0.3485 m^3
V2 = V1*(P2/P1)^(3/2) = 0.3485(600/429.3)^(3/2) = 0.5758 m^3
Work = ⌠PdV = (P2V2 – P1V1)/(1-n) = (600 × 0.5758 – 429.3 × 0.3485)/[1 – (-2/3)]
= 117.5 kJ.

9. Consider a piston-cylinder with 0.5 kg of R-134a as saturated vapour at -10°C. It is compressed to a pressure of 500 kPa in a polytropic process with n = 1.5. Determine the work done during the process.
a) -6.07 kJ
b) -7.07 kJ
c) -8.07 kJ
d) -9.07 kJ
Answer: b
Clarification: Pv^(1.5) = constant until P = 500 kPa
1: v1 = 0.09921 m3/kg, P = Psat = 201.7 kPa
2: v2 = v1(P1/P2)^(1/1.5) = 0.09921×(201.7/500)^(2/3) = 0.05416
hence it is superheated vapour at T2 = 79°C
Work = ⌠PdV = m(P2 v2 – P1 v1)/(1-1.5) = 2*(500 × 0.05416 – 201.7 × 0.09921)/(-0.5)
= -7.07 kJ.

10. R-12 in a piston-cylinder arrangement is initially at 50°C, x = 1. It is then expanded in a process so that P = Cv^(−1) to a pressure of 100 kPa. Find the work.
a) 23.2 kJ/kg
b) 33.2 kJ/kg
c) 43.2 kJ/kg
d) 53.2 kJ/kg
Answer: c
Clarification: State 1: 50°C, x=1, P1 = 1219.3 kPa, v1 = 0.01417 m^3/kg
Process: P = Cv^(-1) ⇒ Work = ∫ P dv = C ln(v2/v1)
State 2: 100 kPa thus v2 = (v1)(P1)/P2 = 0.1728 m^3/kg hence T = – 13.2°C
Work = P1v1[ln(v2/v1)] = 1219.3 × 0.01417 × ln(0.1728/0.01417)
= 43.2 kJ/kg.

11. A piston-cylinder contains water at 3 MPa, 500°C. It is cooled in a polytropic process to 1 MPa, 200°C. Find the specific work in the process.
a) 155.2 kJ
b) 165.2 kJ
c) 175.2 kJ
d) 185.2 kJ
Answer: a
Clarification: Pv^(n) = C thus (P1/P2) = (v2/v1)^n
n= ln(P1/P2) / ln(v2/v1) = 1.0986/0.57246 = 1.919
Work = ⌠PdV = (P2 v2 – P1 v1)/(1-n) = (1000 × 0.20596 – 3000 × 0.11619)/(1 – 1.919)
= 155.2 kJ.

12. A piston/cylinder contains carbon dioxide at 300 kPa, 100°C with a volume of 0.2 m^3. Weights are added at such a rate that the gas compresses according to the relation PV^1.2 = constant to a final temperature of 200°C. Determine the work done during the process.
a) 70.4 kJ
b) -70.4 kJ
c) 80.4 kJ
d) -80.4 kJ
Answer: d
Clarification: For the Polytropic process PV^n = constant
1W2 = ∫PdV = (P2V2 – P1V1)/(1 – n )
Assuming ideal gas, PV = mRT
But mR = P1V1/T1 = 300 × 0.2/373.15 = 0.1608 kJ/K
1W2 = 0.1608(473.15 – 373.15)/(1 – 1.2) = -80.4 kJ.

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