# Thermodynamics Questions and Answers – Work in a Reversible Process-1

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This set of Thermodynamics Multiple Choice Questions & Answers (MCQs) focuses on “Work in a Reversible Process-1”.

1. Hot air at 1500 K expands in a polytropic process to a volume 6 times as large with n = 1.5. Find the specific boundary work.
a) 309.5 kJ/kg
b) 409.5 kJ/kg
c) 509.5 kJ/kg
d) 609.5 kJ/kg

Explanation: u1 = 444.6 kJ/kg, u2 = 1205.25 kJ/kg
T2 = T1(v1/v2)^(n-1) = 1500(1/6)^0.5 = 612.4 K
1w2 = R(T2-T1)/(1-n) = 0.287(612.4 – 1500)/(1 – 1.5) = 509.5 kJ/kg.

2. In a Carnot-cycle heat pump, heat is rejected from R-22 at 40°C, during which the R-22 changes from saturated vapor to saturated liquid. The heat is transferred to the R-22 at 0°C. Determine the COP for the cycle.
a) 6.83
b) 7.83
c) 8.83
d) 9.83

Explanation: s4 = s3 = 0.3417 kJ/kg K = 0.1751 + x4(0.7518) => x4 = 0.2216
s1 = s2 = 0.8746 kJ/kg K = 0.1751 + x1(0.7518) => x1 = 0.9304
β′ = q/w = Th/(Th – Tl) = 313.2/40 = 7.83.

3. 1kg of ammonia in a piston/cylinder at 50°C, 1000 kPa is expanded in a reversible isothermal process to 100 kPa. Find the work for this process.
a) 333.75 kJ
b) 343.75 kJ
c) 353.75 kJ
d) 363.75 kJ

Explanation: 1W2 = ⌠ PdV
State 1: u1 = 1391.3 kJ/kg; s1 = 5.265 kJ/kg K
State 2: u2 = 1424.7 kJ/kg; s2 = 6.494 kJ/kg K;
v2 = 1.5658 m^3/kg; h2 = 1581.2 kJ/kg
1Q2 = 1 kg (273 + 50) K (6.494 – 5.265) kJ/kg K = 396.967 kJ
1W2 = 1Q2 – m(u2 – u1) = 363.75 kJ.
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4. 1kg of ammonia in a piston/cylinder at 50°C, 1000 kPa is expanded in a reversible isobaric process to 140°C. Find the work in the process.
a) 50.5 kJ
b) 60.5 kJ
c) 70.5 kJ
d) 80.5 kJ

Explanation: 1W2 = mP(v2 – v1)
v1 = 0.145 m^3/kg, u1 = 1391.3 kJ/kg
v2 = 0.1955 m^3/kg, u2 = 1566.7 kJ/kg
1W2 = 1 × 1000(0.1955 – 0.145) = 50.5 kJ.

5. 1kg of ammonia in a piston/cylinder at 50°C, 1000 kPa is expanded in a reversible adiabatic process to 100 kPa. Find the work for this process.
a) 222.4 kJ
b) 232.4 kJ
c) 242.4 kJ
d) 252.4 kJ

Explanation: 1Q2 = 0 ⇒ s2 = s1 and u1 = 1391.3 kJ/kg, s1 = 5.2654 kJ/kg K
sg2 = 5.8404 kJ/kg K, sf = 0.1192 kJ/kg K; x2 = (s – sf)/sfg
x2 = (5.2654 − 0.1192)/5.7212 = 0.90;
u2 = uf + x2 ufg = 27.66 + 0.9×1257.0 = 1158.9 kJ/kg
1W2 = 1 × (1391.3 – 1158.9) = 232.4 kJ.

6. A cylinder-piston contains ammonia at 50°C, 20% quality, volume being 1 L. The ammonia expands slowly, and heat is transferred to maintain a constant temperature. The process continues until all liquid is gone. Determine the work for this process.
a) 7.11 kJ
b) 9.11 kJ
c) 5.11 kJ
d) 8.11 kJ

Explanation: T1 = 50°C, x1 = 0.20, V1 = 1 L, v1 = 0.001777 + 0.2 ×0.06159 = 0.014095 m^3/kg
s1 = 1.5121 + 0.2 × 3.2493 = 2.1620 kJ/kg K,
m = V1/v1 = 0.001/0.014095 = 0.071 kg
v2 = vg = 0.06336 m^3/kg, s2 = sg = 4.7613 kJ/kg K
Process: T = constant to x2 = 1.0, P = constant = 2.033 MPa
1W2 = ⌠PdV = Pm(v2 – v1) = 2033 × 0.071 × (0.06336 – 0.014095)
= 7.11 kJ.

7. An insulated cylinder fitted with a piston contains 0.1 kg of water at 100°C and 90% quality. The piston is moved, compressing the water till it reaches a pressure of 1.2 MPa. How much work is required in the process?
a) -27.5 kJ
b) -47.5 kJ
c) -17.5 kJ
d) -37.5 kJ

Explanation: 1Q2 = 0 = m(u2 – u1) + 1W2
State 1: 100°C, x1 = 0.90: s1 = 1.3068 + 0.90×6.048 = 6.7500 kJ/kg K
u1 = 418.91 + 0.9 × 2087.58 = 2297.7 kJ/kg
State 2: s2 = s1 = 6.7500 and P2 = 1.2 MPa which gives
T2 = 232.3°C and u2 = 2672.9 kJ/kg
1W2 = -m(u2 – u1) = -0.1(2672.9 – 2297.7) = -37.5 kJ.

8. Compression and heat transfer brings R-134a from 50°C, 500 kPa to saturated vapour in an isothermal process. Find the specific work.
a) -24.25 kJ/kg
b) -25.25 kJ/kg
c) -26.25 kJ/kg
d) -27.25 kJ/kg

Explanation: Process: T = C and assume reversible ⇒ 1q2 = T (s2 – s1)
u1 = 415.91 kJ/kg, s1 = 1.827 kJ/kg K
u2 = 403.98 kJ/kg, s2 = 1.7088 kJ/kg K
1q2 = (273 + 50) × (1.7088 – 1.827) = -38.18 kJ/kg
w2 = 1q2 + u1 – u2 = -38.18 + 415.91 – 403.98 = -26.25 kJ/kg.

9. 1kg of water at 300°C expands against a piston in a cylinder until it reaches 100 kPa, at which point the water has a quality of 90.2%. The expansion is reversible and adiabatic. How much work is done by the water?
a) 371.2 kJ
b) 471.2 kJ
c) 571.2 kJ
d) 671.2 kJ

Explanation: Process: Adiabatic Q = 0 and reversible => s2 = s1
P2 = 100 kPa, x2 = 0.902, thus s2 = 1.3026 + 0.902 × 6.0568 = 6.7658 kJ/kg K
s2 = 1.3026 + 0.902 × 6.0568 = 6.7658 kJ/kg K
State 1 At T1 = 300°C, s1 = 6.7658 and ⇒ P1 = 2000 kPa, u1 = 2772.6 kJ/kg
1W2 = m(u1 – u2) = 1(2772.6 – 2301.4) = 471.2 kJ.

10. A piston/cylinder has 2kg ammonia at 100 kPa, 50°C which is compressed to 1000 kPa. The temperature is assumed to be constant. Find the work for the process assuming it to be reversible.
a) -727.6 kJ
b) -794.2 kJ
c) -723.6 kJ
d) -743.2 kJ

Explanation: Process: T = constant and assume reversible process
v1 = 1.5658 m^3/kg, u1 = 1424.7 kJ/kg, s1 = 6.4943 kJ/kg K
v2 = 0.1450 m^3/kg, u2 = 1391.3 kJ/kg, s2 = 5.2654 kJ/kg K
1Q2 = mT(s2 − s1) = 2 × 323.15 (5.2654 – 6.4943) = -794.2 kJ
1W2 = 1Q2 – m(u2 – u1) = -794.24 – 2(1391.3 – 1424.62)
= -727.6 kJ.

11. A piston cylinder has R-134a at 100 kPa, –20°C which is compressed to 500 kPa in a reversible adiabatic process. Find the specific work.
a) -41.63 kJ/kg
b) -11.63 kJ/kg
c) -21.63 kJ/kg
d) -31.63 kJ/kg

Explanation: Process: Adiabatic and reversible => s2 = s1
u1 = 367.36 kJ/kg, s1 = 1.7665 kJ/kg K
P2 = 500 kPa, s2 = s1 = 1.7665 kJ/kg K
very close at 30°C, u2 = 398.99 kJ/kg
1w2 = u2 – u1 = 367.36 – 398.99 = -31.63 kJ/kg.

12. A cylinder containing R-134a at 150 kPa, 10°C has an initial volume of 20 L. A piston compresses the R-134a in a isothermal, reversible process until it reaches the saturated vapour state. Calculate the required work in the process.
a) -1.197 kJ
b) -2.197 kJ
c) -3.197 kJ
d) -4.197 kJ

Explanation: Process: T = constant, reversible
u1 = 388.36 kJ/kg, s1 = 1.822 kJ/kg K, m = V/v1 = 0.02/0.148283 = 0.1349 kg
u2 = 383.67 kJ/kg, s2 = 1.7218 kJ/kg K
1Q2 = ⌠Tds = mT(s2 – s1) = 0.1349 × 283.15 × (1.7218 – 1.822) = -3.83 kJ
1W2 = m(u1 – u2) + 1Q2 = 0.1349 × (388.36 – 383.67) – 3.83 = -3.197 kJ.

13. A piston/cylinder has 2kg water at 250°C, 1000 kPa which is now cooled with a constant load on the piston. This isobaric process ends when the water has reached a state of saturated liquid. Find the work.
a) -363.1 kJ
b) -463.1 kJ
c) -563.1 kJ
d) -663.1 kJ

Explanation: Process: P = C => W = ∫ P dV = P(V2 − V1)
State 1: v1 = 0.23268 m^3/kg, s1 = 6.9246 kJ/kg K, u1 = 2709.91 kJ/kg
State 2: v2 = 0.001127 m^3/kg, s2 = 2.1386 kJ/kg K, u2 = 761.67 kJ/kg
1W2 = m P (v2 − v1) = 2 × 1000 (0.001127 – 0.23268) = -463.1 kJ.

14. Water at 250°C, 1000 kPa is brought to saturated vapour in a piston/cylinder with an isothermal process. Find the specific work.
a) -38 kJ/kg
b) -138 kJ/kg
c) -238 kJ/kg
d) -338 kJ/kg

Explanation: Process: T = constant, reversible
State 1: v1 = 0.23268 m^3/kg; u1 = 2709.91 kJ/kg; s1 = 6.9246 kJ/kg K
State 2: v2 = 0.05013 m^3/kg, u2 = 2602.37 kJ/kg, s2 = 6.0729 kJ/kg K
1q2 = ∫ T ds = T(s2 − s1) = (250 + 273) (6.0729 – 6.9246) = -445.6 kJ/kg
1w2 = 1q2 + u1 − u2 = -445.6 + 2709.91 – 2602.37 = -338 kJ/kg.

15. Water at 250°C, 1000 kPa is brought to saturated vapour in a rigid container. Find the specific heat transfer in this isometric process.
a) −132 kJ/kg
b) −232 kJ/kg
c) −332 kJ/kg
d) −432 kJ/kg

Explanation: Process: v = constant => 1w2 = 0
State 1: u1 = 2709.91 kJ/kg, v1 = 0.23268 m^3/kg
State 2: x = 1 and v2 = v1, thus P2=800 kPa
T2 = 170 + 5 × (0.23268 – 0.24283)/(0.2168 – 0.24283)
= 170 + 5 × 0.38993 = 171.95°C
u2 = 2576.46 + 0.38993 × (2580.19 – 2576.46) = 2577.9 kJ/kg
1q2 = u2 − u1 = 2577.9 – 2709.91 = −132 kJ/kg.

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