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
View Answer
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
View Answer
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
View Answer
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.
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
View Answer
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
View Answer
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
View Answer
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
View Answer
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
View Answer
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
View Answer
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
View Answer
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
View Answer
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
View Answer
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
View Answer
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
View Answer
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
View Answer
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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