Thermodynamics Questions and Answers – Work in a Reversible Process-2

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This set of Thermodynamics Multiple Choice Questions & Answers (MCQs) focuses on “Work in a Reversible Process-2”.

1. Water at 250°C, 1000 kPa is brought to saturated vapour in a piston/cylinder with an isobaric process. Find the specific work.
a) -18.28 kJ/kg
b) -48.28 kJ/kg
c) -28.28 kJ/kg
d) -38.28 kJ/kg
View Answer

Answer: d
Explanation: Process: P = C => w = ∫ P dv = P(v2 − v1)
1: v1 = 0.23268 m3/kg, s1= 6.9246 kJ/kgK, u1 = 2709.91 kJ/kg
2: v2 = 0.19444 m3/kg, s2 = 6.5864 kJ/kg K,
u2 = 2583.64 kJ/kg, T2 = 179.91°C
1w2 = P (v2 − v1) = 1000 (0.1944 – 0.23268) = -38.28 kJ/kg.
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2. A heavily insulated cylinder/piston contains ammonia at 60°C, 1200 kPa. The piston is moved, expanding the ammonia in a reversible process until the temperature is −20°C during which 600 kJ of work is given out by ammonia. What was the initial volume of the cylinder?
a) 0.285 m3
b) 0.385 m3
c) 0.485 m3
d) 0.585 m3
View Answer

Answer: b
Explanation: State 1: v1 = 0.1238 m3/kg, s1 = 5.2357 kJ/kg K,
u1 = h – Pv = 1553.3 – 1200×0.1238 = 1404.9 kJ/kg
Process: reversible (1S2(gen) = 0) and adiabatic (dQ = 0) => s2 = s1
State 2: T2, s2 ⇒ x2 = (5.2357 – 0.3657)/5.2498 = 0.928
u2 = 88.76 + 0.928×1210.7 = 1211.95 kJ/kg
1Q2 = 0 = m(u2 – u1) + 1W2 = m(1211.95 – 1404.9) + 600 ⇒ m = 3.110 kg
V1 = mv1 = 3.11 × 0.1238 = 0.385 m3.

3. Water at 250°C, 1000 kPa is brought to saturated vapor in a piston/cylinder with an adiabatic process. Find the specific work.
a) 139.35 kJ/kg
b) 149.35 kJ/kg
c) 159.35 kJ/kg
d) 169.35 kJ/kg
View Answer

Answer: c
Explanation: State 1: v1 = 0.23268 m3/kg, u1 = 2709.91 kJ/kg, s1 = 6.9246 kJ/kg K
State 2: x = 1 and s2 = s1 = 6.9246 kJ/kg K
T2 = 140.56°C, P2 = 367.34 kPa, v2 = 0.50187 m3/kg, u2= 2550.56 kJ/kg
1w2 = u1 – u2 = 2709.91 – 2550.56 = 159.35 kJ/kg.
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4. A piston/cylinder contains 2kg water at 200°C, 10 MPa. The water expands in an isothermal process to a pressure of 200 kPa. Any heat transfer takes place with an ambient at 200°C and whole process is be assumed reversible. Calculate the total work.
a) 1290.3 kJ
b) 1390.3 kJ
c) 1490.3 kJ
d) 1590.3 kJ
View Answer

Answer: a
Explanation: State 1: v1 = 0.001148 m3/kg, u1 = 844.49 kJ/kg,
s1 = 2.3178 kJ/kg K, V1 = mv1 = 0.0023 m3
State 2: v2 = 1.08034 m3/kg, u2 = 2654.4 kJ/kg,
s2 = 7.5066 kJ/kg K, V2 = mv2 = 2.1607 m3
1Q2 = mT(s2 − s1) = 2 × 473.15 (7.5066 – 2.3178) = 4910 kJ
1W2 = 1Q2 – m(u2 – u1) = 1290.3 kJ.

5. A piston/cylinder of total 1kg steel contains 0.5 kg ammonia at 1600 kPa both masses at 120°C with minimum volume being 0.02 m3. The whole system is cooled down to 30°C by heat transfer to the ambient at 20°C, and during the process the steel keeps same temperature as the ammonia. Find the work.
a) – 28.14 kJ
b) – 38.14 kJ
c) – 48.14 kJ
d) – 58.14 kJ
View Answer

Answer: d
Explanation: 1 : v1 = 0.11265 m3/kg, u1 = 1516.6 kJ/kg,
s1 = 5.5018 kJ/kg K, V1 = mv1 = 0.05634 m3
2 : 30°C < T(stop) so v2 = v(stop) = 0.04 m3/kg
1W2= ∫ P dV = (P1)m(v2-v1) = 1600 × 0.5 (0.04 – 0.11265)
= – 58.14 kJ.
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6. A mass of 1 kg of air contained in a cylinder at 1000 K, 1.5 MPa, expands in a reversible isothermal process to a volume 10 times larger. Calculate the heat transfer during the process.
a) 460.84 kJ
b) 560.84 kJ
c) 660.84 kJ
d) 760.84 kJ
View Answer

Answer: c
Explanation: Process: T = constant so with ideal gas => u2 = u1
1Q2 = 1W2 = ⌠PdV = P1V1 ln (V2/V1) = mRT1 ln (V2/V1)
= 1 × 0.287 × 1000 ln (10) = 660.84 kJ.

7. A piston/cylinder contains air at 400 K, 100 kPa which is compressed to a final pressure of 1000 kPa. Consider the process to be a reversible adiabatic process. Find the specific work.
a) -166.7 kJ/kg
b) -266.7 kJ/kg
c) -366.7 kJ/kg
d) -466.7 kJ/kg
View Answer

Answer: b
Explanation: We have constant s, an isentropic process
T2 = T1( P2 / P1)^[(k-1)/k] = 400(1000/100)^(0.4/1.4)
= 400 × 10^(0.28575) = 772 K
1w2 = u1 – u2 = Cv(T1 – T2) = 0.717(400 – 772) = -266.7 kJ/kg.
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8. A piston/cylinder contains air at 400 K, 100 kPa which is compressed to a final pressure of 1000 kPa. Consider the process to be a reversible isothermal process. Find the specific work.
a) −264 kJ/kg
b) −364 kJ/kg
c) −464 kJ/kg
d) −564 kJ/kg
View Answer

Answer: a
Explanation: For this process T2 = T1 so since ideal gas we get u2 = u1 and also 1w2 = 1q2
1w2 = 1q2 = T(s2 – s1) = −RT ln(P2/P1)
= − 0.287 × 400 ln 10 = −264 kJ/kg.

9. Consider a small air pistol with a cylinder volume of 1 cm3 at 27°C, 250 kPa. The bullet acts as a piston and is released so the air expands in an adiabatic process. If the pressure should be 100 kPa as the bullet leaves the cylinder find the work done by the air.
a) 0.115 J
b) 0.125 J
c) 0.135 J
d) 0.145 J
View Answer

Answer: d
Explanation: Process: Adiabatic 1q2 = 0 Reversible 1s2(gen) = 0
this is an isentropic expansion process giving s2 = s1
T2 = T1( P2 / P1)^[(k-1)/k] = 300(100/250)^(0.4/1.4) = 300 × 0.4^(0.28575) = 230.9 K
V2 = V1 P1 T2/P2 T1 = 1 × 250 × 230.9/100 × 300 = 1.92 cm3
Work = [1/(k-1)](P2V2 – P1V1) = [1/(1-1.4)](100 × 1.92 – 250 × 1) ×10^(-6)
= 0.145 J.
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10. A spring loaded piston cylinder contains 1.5 kg air at 160 kPa and 27°C. It is heated in a process where pressure is linear in volume, P = A + BV, to twice the initial volume where it reaches 900 K. Find the work assuming a source at 900 K.
a) 61.4 kJ
b) 161.4 kJ
c) 261.4 kJ
d) 361.4 kJ
View Answer

Answer: b
Explanation: State 1: u1 = 214.36 kJ/kg, V1 = mRT1/ P1 = (1.5 × 0.287 ×300) / 160 = 0.8072 m3
State 2: u2 = 674.824 kJ/kg,
P2 = RT2/v2 = RT2/(2v1) = T2 P1/(2T1)= P1(T2/2)T1
= 160 × 900 / 2 × 300 = 240 kPa
1W2 = ∫ PdV = 0.5 × (P1 + P2) (V2 – V1) = 0.5 × (P1 + P2) V1
= 0.5 × (160 + 240) 0.8072 = 161.4 kJ.

11. Helium contained in a cylinder at ambient conditions, 100 kPa, 20°C, is compressed in a reversible isothermal process to 600 kPa, after which the gas is expanded back to 100 kPa in a reversible adiabatic process. Calculate the net work per kilogram of helium.
a) -623.6 kJ/kg
b) +467.4 kJ/kg
c) -1091.0 kJ/kg
d) none of the mentioned
View Answer

Answer: a
Explanation: The adiabatic reversible expansion gives constant s
T3 = T2(P3/P2)^[(k-1)/k] = 293.15 (100/600)^0.4 = 143.15 K
The isothermal process: 1w2 = -RT1 ln(P2/P1)
= -2.0771 × 293.15 × ln(600/100) = -1091.0 kJ/kg
The adiabatic process: 2w3 = CVo(T2-T3) = 3.116 (293.15 – 143.15) = +467.4 kJ/kg
The net work is the sum: w(NET) = -1091.0 + 467.4 = -623.6 kJ/kg.

12. A cylinder/piston contains 1kg methane gas at 100 kPa, 20°C. The gas is compressed reversibly to a pressure of 800 kPa. Calculate the work required if the process is adiabatic.
a) -112.0 kJ
b) -212.0 kJ
c) -312.0 kJ
d) -412.0 kJ
View Answer

Answer: c
Explanation: Process: 1Q2 = 0 => s2 = s1 thus isentropic process
T2 = T1(P2/P1)^[(k-1)/k] = 293.2(800/100)^0.230 = 473.0 K
1W2 = -mCv(T2 – T1) = -1 × 1.7354 (473.0 – 293.2)
= -312.0 kJ.

13. A piston/cylinder contains air at 100 kPa, 300 K. It is now compressed in a reversible adiabatic process to a volume 7 times as small. Use constant heat capacity and find the specific work.
a) -233.6 kJ/kg
b) -243.6 kJ/kg
c) -253.6 kJ/kg
d) -263.6 kJ/kg
View Answer

Answer: c
Explanation: v2/ v1 = 1/7; P2 /P1 = (v2/v1)^(-k) = 7^(1.4) = 15.245
P2 = P1[7^(1.4)] = 100 × 15.245 = 1524.5 kPa
T2 = T1 (v1/v2)^(k-1) = 300 × 7^(0.4) = 653.4 K
1q2 = 0 kJ/kg; work = R(T2-T1)/(1-k) = 0.287*(653.4 – 300)/(-0.4)
= -253.6 kJ/kg.

14. A gas confined in a piston-cylinder is compressed in a quasi-static process from 80 kPa and 0.1 m3 to 400 kPa and 0.03m3. If the pressure and volume are related by PV^n= constant, calculate the work involved in the process.
a) – 12.87 kJ
b) 12.87 kJ
c) – 11.87 kJ
d) 11.87 kJ
View Answer

Answer: c
Explanation: n = ln(P2/P1)/ln(V1/V2) = ln(400/80)/ln(0.1/0.03) = 1.337
Work involved in the process (1W2) = (P2V2 – P1V1)(1 – n) = – 11.87 kJ.

15. One kg of steam at 200 kPa with 20% quality is heated at constant pressure to 400°C. Calculate the work done by the system .
a) 274.261 kJ
b) 374.261 kJ
c) 474.261 kJ
d) 574.261 kJ
View Answer

Answer: a
Explanation: v1 = 0.001061 + 0.2*0.88467 = 0.177995 m3/kg; v2 = 1.5493 m3/kg
work done by the system during this process (1W2) = mP(v2 – v1)
= 1*200*(1.5493 – 0.177995) = 274.261 kJ.

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