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 m^{3}. 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 m^{3} 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

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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

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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

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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

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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

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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

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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

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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

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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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