This set of Thermodynamics Multiple Choice Questions & Answers (MCQs) focuses on “Polytropic Process-1”.
1. A polytropic process(n = − 1) starts with P = 0, V = 0 and ends with P= 600 kPa, V = 0.01 m3. Find the boundary work done.
a) 1 kJ
b) 2 kJ
c) 3 kJ
d) 4 kJ
View Answer
Explanation: W = ⌠ PdV
= (1/2)(P1 + P2)(V2 – V1)
= (1/2)(P2 + 0)( V2 – 0)
= (1/2)(600*0.1)
= 3 kJ.
2. The piston/cylinder contains carbon dioxide at 300 kPa, with volume of 0.2 m3 and at 100°C. Mass is added at such that the gas compresses with PV^(1.2) = constant to a final temperature of 200°C. Determine the work done during the process.
a) -80.4 kJ
b) -40.4 kJ
c) -60.4 kJ
d) -50.4 kJ
View Answer
Explanation: Work done = (P2V2 – P1V1)/(1-n) and mR = (P1V1)/T1 = 0.1608 kJ/K
Work done = 0.1608(473.2 – 373.2)/(1 – 1.2) = -80.4 kJ.
3. Neon at 400 kPa, 20°C is brought to 100°C in a polytropic process with n = 1.4. Find the work done.
a) -52.39 kJ/kg
b) -62.39 kJ/kg
c) -72.39 kJ/kg
d) -82.39 kJ/kg
View Answer
Explanation: For Neon, k = γ = 1.667 so n < k, Cv = 0.618, R = 0.412
1w2 = [R/(1-n)](T2 – T1) = -82.39 kJ/kg.
4. A mass of 1kg of air contained in a cylinder at 1000 K, 1.5 MPa, expands in a reversible adiabatic process to 100 kPa. Calculate the work done during the process using Constant specific heat.
a) 286.5 kJ
b) 386.5 kJ
c) 486.5 kJ
d) 586.5 kJ
View Answer
Explanation: Process: 1Q2 = 0, 1S2 gen = 0 => s2 = s1
T2 = T1(P2/P1)^[(k-1)/k] = 1000(0.1/1.5)0.286 = 460.9 K
1W2 = -(U2 – U1) = mCv(T1 – T2)
= 1 × 0.717(1000 – 460.9) = 386.5 kJ.
5. 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 isothermal.
a) -216.0 kJ
b) -316.0 kJ
c) -416.0 kJ
d) -516.0 kJ
View Answer
Explanation: Process: T = constant. For ideal gas then u2 = u1 1W2 = 1Q2 and ∫ dQ/T = 1Q2/T
1W2 = 1Q2 = mT(s2 – s1) = -mRT ln(P2/P1)
= -0.51835× 293.2 ln(800/100) = -316.0 kJ.
6. 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 polytropic, with exponent n = 1.15.
a) -314.5 kJ
b) -414.5 kJ
c) -514.5 kJ
d) -614.5 kJ
View Answer
Explanation: Process: Pv^(n) = constant with n = 1.15 ;
T2 = T1(P2/P1)^[(n-1)/n] = 293.2(800/100)^0.130 = 384.2 K
1W2 = ∫ mP dv = m(P2v2 – P1v1)/(1 – n) = mR (T2 – T1)/(1 – n)
= 1*0.51835(384.2 – 293.2)/(1 – 1.15) = -314.5 kJ.
7. Helium in a piston/cylinder at 20°C, 100 kPa is brought to 400 K in a reversible polytropic process with exponent n = 1.25. Helium can be assumed to be an ideal gas with constant specific heat. Find the specific work.
a) -587.7 kJ/kg
b) -687.7 kJ/kg
c) -787.7 kJ/kg
d) -887.7 kJ/kg
View Answer
Explanation: Process: Pv^(n) = C & Pv = RT => Tv^(n-1) = C
Cv = 3.116 kJ/kg K, R = 2.0771 kJ/kg K
v2 / v1 = (T1 / T2 )^[1/(n-1)] = 0.2885
P2 / P1 = (v1 / v2)^(n) = 4.73 => P2 = 473 kPa
W = (P2 v2 – P1 v1)/(1-n) = R(T2-T1)/(1-n) = -887.7 kJ/kg.
8. Consider air in a cylinder volume of 0.2 L at 7 MPa, 1800K. It now expands in a reversible polytropic process with exponent, n = 1.5, through a volume ratio of 8:1. Calculate the work for the process.
a) 1.61 kJ
b) 1.71 kJ
c) 1.81 kJ
d) 1.91 kJ
View Answer
Explanation: Process: PV^(1.50) = constant, V2/V1 = 8
State 1: P1 = 7 MPa, T1 = 1800 K, V1 = 0.2 L, m1=P1V1/RT1 = 2.71×10-3 kg
State 2: T2 = T1 (V1/V2)^(n-1) = 1800(1/8)^(0.5) = 636.4 K
1W2 = ⌠ PdV = mR(T2 – T1)/(1 – n)
= 2.71×10^(-3) × 0.287(636.4 – 1800)/(1-1.5) = 1.81 kJ.
9. A cylinder/piston contains carbon dioxide at 300°C, 1 MPa with a volume of 200L. The total external force acting on the piston is proportional to V3. This system is allowed to cool to room temperature, 20°C. Find the work.
a) -24.4 kJ
b) -34.4 kJ
c) -44.4 kJ
d) -54.4 kJ
View Answer
Explanation: PV^(-3) = constant
State 1: m = P1V1/RT1 = (1000 × 0.2)/(0.18892 × 573.2) = 1.847 kg
P2 = P1(T2/T1)^[n/(n-1)] = 1000(293.2/573.2)^(3/4) = 604.8 kPa
V2 = V1(T1/T2)^[1/(n-1)] = 0.16914 m^3
Work = ⌠ PdV = (P2V2 – P1V1)/(1-n) = [604.8 × 0.16914 – 1000 × 0.2] / [1-(-3)] = -24.4 kJ.
10. A cylinder/piston contains 100L of air at 25°C, 110 kPa. The air is compressed in a reversible polytropic process to a final state of 200°C, 800 kPa. Assume the heat transfer is with the ambient at 25°C. Find the work done by the air.
a) -11.28 kJ
b) -21.28 kJ
c) -31.28 kJ
d) -41.28 kJ
View Answer
Explanation: m = P1V1 /(RT1) = 110 × 0.1/(0.287 × 298.15) = 0.1286 kg
T2/T1 = (P2/P1)^[(n-1)/n] => 473.15/298.15 = (800/110)^[(n-1)/n] ⇒ (n-1)/n = 0.2328 hence n = 1.3034
V2 = V1(P1/P2)^(1/n) = 0.1(110/800)^(0.7672) = 0.02182 m^3
Work = ⌠PdV = (P2V2 – P1V1)/(1-n) = (800 × 0.02182 – 110 × 0.1)/(1 – 1.3034)
= -21.28 kJ.
11. A mass of 2 kg ethane gas at 100°C, 500 kPa, undergoes a reversible polytropic expansion with n = 1.3, to a final temperature of 20°C. Find the work done.
a) 43.7 kJ/kg
b) 53.7 kJ/kg
c) 63.7 kJ/kg
d) 73.7 kJ/kg
View Answer
Explanation: P2 = P1(T2/T1)^[n/(n-1)] = 500(293.2/373.2)^(4.333) = 175.8 kPa
Work = ⌠PdV = (P2V2 – P1V1)/(1-n) = R(T2-T1)/(1-n)
= 0.2765(293.2-373.2)/(1-1.30) = 73.7 kJ/kg.
12. A piston/cylinder contains air at 100 kPa, 300 K. A reversible polytropic process with n = 1.3 brings the air to 500 K. Any heat transfer if it comes in is from a 325°C reservoir and if it goes out it is to the ambient at 300 K. Find the specific work.
a) -171.3 kJ/kg
b) -181.3 kJ/kg
c) -191.3 kJ/kg
d) -201.3 kJ/kg
View Answer
Explanation: Process : Pv^(n) = C
Work = ⌠PdV = (P2V2 – P1V1)/(1-n) = R(T2-T1)/(1-n)
= 0.287 (500 – 300)/(1 – 1.3) = -191.3 kJ/kg.
13. A cylinder/piston contains saturated vapour R-22 at 10°C; the volume is 10 L. The R-22 is compressed to 60°C, 2 MPa in a reversible polytropic process. If all the heat transfer during the process is with the ambient at 10°C, calculate the work done.
a) −6.26 kJ
b) −7.26 kJ
c) −8.26 kJ
d) −9.26 kJ
View Answer
Explanation: State 1: P1 = 0.681 MPa, v1 = 0.03471; m = V1/v1 = 0.01/0.03471 = 0.288 kg
State 2: v2 = 0.01214 m^3/kg; P2/P1 = 2.0/0.681 = (0.03471/0.01214)^(n)
=> n = 1.0255
Work = ⌠PdV = m(P2v2 – P1v1)/(1-n)
= 0.288(2000 × 0.01214 – 681 × 0.03471)/(1 – 1.0255) = −7.26 kJ.
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