Category: Thermodynamics Objective Questions

Thermodynamics Multiple Choice Questions on “Entropy Generation in a Closed and Open System-1”.

1. The entropy of any closed system can increase in which if the following way?
a) by heat interaction in which there is entropy transfer
b) dissipative effects or internal irreversibilities
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: These two processes increase the entropy of a closed system.

2. Entropy increase dS of the system can be expressed as
a) dS=dS(due to external heat interaction)-dS(due to internal irreversibility)
b) dS=dS(due to external heat interaction)+dS(due to internal irreversibility)
c) dS=-dS(due to external heat interaction)-dS(due to internal irreversibility)
d) dS=-dS(due to external heat interaction)+dS(due to internal irreversibility)
Answer: b
Clarification: Total entropy increase of the system is the sum of these two entropies.

3. The entropy increase due to internal irreversibility is also called entropy production or entropy generation.
a) true
b) false
Answer: a
Clarification: This entropy is generated during the process within the system.

4. Which of the following statement is true?
a) if the isentropic process is reversible, it must be adiabatic
b) if the isentropic process is adiabatic, it cannot but be reversible
c) if the process is adiabatic and reversible, it must be isentropic
d) all of the mentioned
Answer: d
Clarification: An adiabatic process need not be isentropic, since entropy can also increase due to friction.

5. Lost work is given by
a) pdV-dW
b) pdV+dW
c) -pdV-dW
d) pdV*dW
Answer: a
Clarification: The lost work d(LW) indicates the work that is lost due to irreversibility.

6. The amount of entropy generation is given by
a) S2+S1+∫(dQ/T)
b) S2-S1+∫(dQ/T)
c) S2-S1-∫(dQ/T)
d) none of the mentioned
Answer: c
Clarification: Here (S2-S1) is the entropy change of the system and ∫(dQ/T) is the entropy transfer.

7. Any thermodynamic process is accompanied by entropy generation.
a) true
b) false
Answer: a
Clarification: This comes from the second law.

8. Which of the following statement is false?
a) for a reversible process, entropy generation is zero
b) the entropy generation does not depend on the path the system follows
c) for an irreversible process, entropy generation is greater than zero
d) none of the mentioned
Answer: b
Clarification: Entropy generation is not a thermodynamic property and depends on the path that system follows.

9. If the path A causes more entropy generation than path B, then
a) path A is more irreversible than path B
b) path A involves more lost work
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: The amount of entropy generation quantifies the intrinsic irreversibility of the process.

10. In an open system, there is a transfer of which of the following quantity?
a) mass
b) energy
c) entropy
d) all of the mentioned
Answer: d
Clarification: In an open system, there is a transfer of all these three quantities.

11. The rate of entropy increase of the control volume ____ or ____ the net rate of entropy transfer to it.
a) exceeds or is less than
b) exceeds, is equal to
c) is less than, or equal to
d) none of the mentioned
Answer: b
Clarification: The difference is the entropy generated within the control volume due to irreversibility.

12. Mass and energy are conserved quantities, but entropy is generally not conserved.
a) true
b) false
Answer: a
Clarification: This is a basic fact about entropy.

13. The rate at which entropy is transferred out must ____ the rate at which entropy enters the control volume.
a) be less than
b) equal to
c) exceed
d) none of the mentioned
Answer: c
Clarification: The difference is the rate of entropy generated within the control volume owing to irreversibilities.

14. A chip dissipates 2 kJ of electric work and rejects it as heat transfer from its surface which is at 50°C to 25°C air. How much entropy is generated in the chip?thermodynamics-questions-answers-entropy-generation-closed-open-system-1-q14″ alt=”thermodynamics-questions-answers-entropy-generation-closed-open-system-1-q14″ width=”385″ height=”161″ class=”alignnone size-full wp-image-166425″ srcset=”/2017/06/thermodynamics-questions-answers-entropy-generation-closed-open-system-1-q14 385w, /2017/06/thermodynamics-questions-answers-entropy-generation-closed-open-system-1-q14-300×125 300w” sizes=”(max-width: 385px) 100vw, 385px”>
a) 4.19 J/K
b) 5.19 J/K
c) 6.19 J/K
d) 7.19 J/K
Answer: c
Clarification: C.V.1 Chip with surface at 50°C, we assume chip state is constant.
U2 – U1 = 0 = 1Q2 – 1W2 = W(electrical,in) – Q(out,1)
S2 – S1 = 0 = – [Q(out,1)]/[T(surf)] + 1S2(gen1)
1S2(gen1) = [Q(out,1)]/[T(surf)] = W(electrical,in) / T(surf)
= 2/323.15 = 6.19 J/K.

15. A chip dissipates 2 kJ of electric work and rejects it as heat transfer from its surface which is at 50°C to 25°C air. How much entropy is generated outside the chip?thermodynamics-questions-answers-entropy-generation-closed-open-system-1-q14″ alt=”thermodynamics-questions-answers-entropy-generation-closed-open-system-1-q14″ width=”385″ height=”161″ class=”alignnone size-full wp-image-166425″ srcset=”/2017/06/thermodynamics-questions-answers-entropy-generation-closed-open-system-1-q14 385w, /2017/06/thermodynamics-questions-answers-entropy-generation-closed-open-system-1-q14-300×125 300w” sizes=”(max-width: 385px) 100vw, 385px”>
a) 0.419 J/K
b) 0.519 J/K
c) 0.619 J/K
d) 0.719 J/K
Answer: b
Clarification: C.V.2 From chip surface at 50°C to air at 25°C, assume constant state.
U2 – U1 = 0 = 1Q2 – 1W2 = Q(out,1) – Q(out,2)
S2 – S1 = 0 = [Q(out,1) / T(surf)] – [Q(out,2) / T(air)] + 1S2(gen2)
1S2(gen2) = [Q(out,2) / T(air)] – [Q(out,1) / T(surf)]
= (2/298.15) – (2/323.15) = 0.519 J/K.

Thermodynamics Multiple Choice Questions on “Second Law Efficiency-1”.

1. The first law efficiency is defined as the ratio of the output energy to the input energy.
a) true
b) false
Answer: a
Clarification: First law efficiency = output energy / input energy.

2. Which of the following statement is true about the first law?
a) it is concerned only with the quantities of energy
b) it disregards the form in which the energy exists
c) it does not discriminate between the energies available at different temperatures
d) all of the mentioned
Answer: d
Clarification: It is the second law which provides a means of assigning a quality index to energy.

3. With the concept of exergy available, which of the following is possible?
a) to analyse means of minimizing the consumption of available energy to perform a given process
b) to ensure most efficient possible conversion of energy
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: These statements tell us why the concept of exergy is so important.

4. Second law efficiency is defined as
a) actual exergy intake / minimum exergy intake
b) minimum exergy intake / actual exergy intake
c) actual exergy intake / maximum exergy intake
d) maximum exergy intake / minimum exergy intake
Answer: b
Clarification: It is the ratio of minimum exergy which must be consumed to do a task divided by the actual amount of exergy consumed in performing the task.

5. For a power plant, second law efficiency can be given by desired output(W) / available energy(Wmax).
a) true
b) false
Answer: a
Clarification: Here, A=Wmax and Amin=W, hence second law efficiency = Amin/A = W/Wmax.

6. Second law efficiency can also be given as
a) 1 / ( first law efficiency * Carnot efficiency)
b) Carnot efficiency * first law efficiency
c) Carnot efficiency / first law efficiency
d) first law efficiency / Carnot efficiency
Answer: d
Clarification: First law efficiency = W/Q1 = (W/Wmax)*(Wmax/Q1) = second law efficiency * Carnot efficiency.

7. If work is involved, Amin= ____ and if heat is involved, Amin= ____
a) w, Q(1+To/T)
b) W, Q(1-To/T)
c) Q(1+To/T), W
d) Q(1-To/T), W
Answer: b
Clarification: This is because, Wmax=Q1(1-To/T).

8. If solar energy Qr is available at a reservoir storage temperature of Tr and if quantity of heat Qa is transferred by the solar collector at temperature Ta, then which of the following is true?
a) first law efficiency = Qa/Qr
b) second law efficiency = exergy output / exergy input
c) second law efficiency = (first law efficiency)*(1-To/Ta)/(1-To/Tr)
d) all of the mentioned
Answer: d
Clarification: Second law efficiency = first law efficiency / Carnot efficiency.

9. In case of a heat pump, second law efficiency is given as
a) (first law efficiency)*(1-Ta/To)
b) (first law efficiency)*(1+To/Ta)
c) (first law efficiency)*(1-To/Ta)
d) none of the mentioned
Answer: c
Clarification: First law efficiency = Qa/Wi and second law efficiency = Qa*(1-To/Ta)/Wi.

10. Both first law efficiency and second law efficiency indicate how effectively the input has been converted into the product.
a) true
b) false
Answer: a
Clarification: First law of efficiency does this on energy basis and second law efficiency does it on exergy basis.

11. For proper utilization of exergy, it is desirable to make first law efficiency ____ and the source and use temperatures should ____
a) as close to unity, be different
b) as close to unity, match
c) as close to zero, match
d) as close to zero, be different
Answer: b
Clarification: If first law efficiency is close to unity, the all the energy carried in by heat transfer is used and no heat is lost to the surroundings.

Thermodynamics Multiple Choice Questions & Answers (MCQs) on “Dalton’s Law of Partial Pressures and Gibbs Function”.

1. The expression which represents the pressure exerted by a gas is
a) nVRT
b) nRT/V
c) V/nRT
d) 1/nVRT
Answer: b
Clarification: This expression comes from the gas equation where V is the volume occupied by the gas at temperature T.

2. The expression nRT/V is called the partial pressure of a gas.
a) true
b) false
Answer: a
Clarification: This is the partial pressure that a gas exerts.

3. According to the Dalton’s law of partial pressures, the total pressure of a mixture of ideal gases is equal to the
a) difference of the highest and lowest pressure
b) product of the partial pressures
c) sum of the partial pressures
d) none of the mentioned
Answer: c
Clarification: According to the Dalton’s law of partial pressures, p=p1+p2+p3+…..+pc.

4. Which of the following relation is correct?
a) mole fraction of the Kth gas = moles of the Kth gas / total number of moles of gas
b) partial pressure of Kth gas = (mole fraction of the Kth gas)*(sum of the partial pressures)
c) sum of mole fractions of all the gases is unity
d) all of the mentioned
Answer: d
Clarification: All these statements come from the Dalton’s law of partial pressures.

5. The gas constant of the mixture is the ____ of the gas constants of the components.
a) average
b) weighted mean
c) sum
d) difference of the highest and the lowest
Answer: b
Clarification: It can be found from the Dalton’s law and gas equation.

6. A quantity called partial volume of a component of mixture is used.
a) true
b) false
Answer: a
Clarification: It is the volume which the component alone would occupy at the pressure and temperature of the mixture.

7. Which of the following statement is true?
a) V=V1+V2+….+Vc , where V is the partial volume of the component
b) 1/v = 1/(v1) + 1/(v2) + …….. + 1/(vc) , where v is the specific volume of the component
c) total density is equal to the sum of the densities of the components
d) all of the mentioned
Answer: d
Clarification: these relations come from the Dalton’s law and the gas equation.

8. The total entropy of a mixture of gases is the ____ of the partial entropies.
a) average
b) weighted mean
c) sum
d) difference of the highest and the lowest
Answer: c
Clarification: This is given by the Gibbs theorem.

9. When gases which are at equal pressure and temperature are mixed adiabatically without work, then
a) internal energy of the gaseous system remains constant
b) heat transfer of the gaseous system remains constant
c) entropy of the gaseous system remains constant
d) all of the mentioned
Answer: a
Clarification: This is because of the first law.

10. The fact that internal energy of a mixture is equal to the sum of the partial internal energies of the gases can also be applied to properties like H,Cv,Cp,S,F, and G.
a) true
b) false
Answer: a
Clarification: This statement comes from the Gibbs theorem.

Thermodynamics assessment questions on “Characteristics of an Ideal Working Fluid in Vapor Power Cycles”.

1. Which of the following statement is true about steam?
a) the maximum temperature that can be used in steam cycles is 600 degree Celsius
b) the critical temperature of steam is around 375 degree Celsius
c) large superheating is required
d) all of the mentioned
Answer: d
Clarification: These are certain drawbacks with steam as working fluid.

2. With steam as working fluid, as pressure increases
a) metal stresses increases
b) thickness of walls of tubes, boiler drums, etc increases
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: These changes take place very rapidly because of prevalence of high temperature.

3. At the heat rejection temperature of 40 degree Celsius, the saturation pressure of steam is 0.075 bar.
a) true
b) false
Answer: a
Clarification: This pressure is considerably lower than the atmospheric pressure.

4. The working fluid should have ____ critical temperature.
a) low
b) high
c) it does not matter
d) none of the mentioned
Answer: b
Clarification: This is to have a low saturation pressure at the maximum allowed temperature.

5. The saturation pressure at heat rejection temperature should be ____ atmospheric pressure.
a) equal to
b) below
c) above
d) none of the mentioned
Answer: c
Clarification: This is done to avoid the necessity of vacuum in the condenser.

6. The specific heat of the working fluid should be ____
a) small
b) large
c) very large
d) none of the mentioned
Answer: a
Clarification: This is done so that little heat transfer is required to raise the liquid to its boiling point.

7. The freezing point of fluid should be ____ room temperature.
a) equal to
b) below
c) above
d) none of the mentioned
Answer: b
Clarification: So that the fluid does not get solidified when it flows through pipelines.

8. The fluid should be chemically ____ and ____ contaminate the material of construction.
a) unstable, should
b) unstable, should not
c) stable, should
d) stable, should not
Answer: d
Clarification: The working fluid used should be stable and should not contaminate at any temperature.

9. The fluid should not be toxic, corrosive or excessively viscous.
a) true
b) false
Answer: a
Clarification: Also the working fluid should be low in cost.

10. Why is superheat desired?
a) to reduce piping losses
b) to improve efficiency
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: The thermal efficiency of the cycle is close to Carnot efficiency.

To practice assessment questions on all areas of Thermodynamics,

Thermodynamics Multiple Choice Questions on “Properties of Atmospheric Air”.

1. Dry air consists of
a) oxygen, nitrogen
b) carbon dioxide
c) hydrogen, argon
d) all of the mentioned
Answer: d
Clarification: Dry air is a mixture of all these gases.

2. Complete dry air exists in nature.
a) true
b) false
Answer: b
Clarification: Complete dry air does not exist in nature.

3. Which of the following is true?(here pa=partial pressure of dry air, pw=partial pressure of water vapour, p=atmospheric pressure)
a) p=pw
b) p=pa
c) p=pw + pa
d) all of the mentioned
Answer: c
Clarification: This comes by Dalton’s law of partial pressures.

4. In a mixture of dry air and water vapour,
a) mole fraction of dry air = pa/p
b) mole fraction of water vapour = pw/p
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: This comes from the concepts of Dalton’s law of partial pressures.

5. When pw is very small,
a) saturation temperature of water vapour at pw is less than atmospheric temperature
b) water vapour in air exists in superheated state
c) air is said to be in unsaturated state
d) all of the mentioned
Answer: d
Clarification: Mostly partial pressure of water vapour is very small.

6. Relative humidity is defined as
a) (saturation pressure of pure water) / pw
b) pw / (saturation pressure of pure water)
c) (saturation pressure of pure water) / p
d) p / (saturation pressure of pure water)
Answer: b
Clarification: Here pw is the partial pressure of water vapour and ps is the saturation pressure of pure water at same temperature of mixture.

7. For saturated air, relative humidity is 0%.
a) true
b) false
Answer: b
Clarification: For saturated air, relative humidity is 100%.

8. If water is injected into a container with has unsaturated air,
a) water will evaporate
b) moisture content of air will decrease
c) pw will decrease
d) all of the mentioned
Answer: a
Clarification: The moisture content of air will increase and pw will increase.

9. Humidity ratio is given by the ratio of
a) (mass of dry air per unit mass of water vapour)^2
b) 1/(mass of dry air * mass of water vapour)
c) water vapour mass per unit mass of dry air
d) mass of dry air per unit mass of water vapour
Answer: c
Clarification: Humidity ratio is also called specific humidity.

10. The degree of saturation is the ratio of
a) (saturated specific humidity / actual specific humidity)^2
b) 1/(saturated specific humidity * actual specific humidity)
c) saturated specific humidity / actual specific humidity
d) actual specific humidity / saturated specific humidity
Answer: d
Clarification: Here both saturated specific humidity and specific humidity are at same temperature.

11. The degree of saturation varies between -1 and 0.
a) true
b) false
Answer: b
Clarification: The degree of saturation varies between 0 and 1.

12. Which of the following statement is true?
a) dew point temperature is the temperature at which water vapour starts condensing
b) dry bulb temperature is recorded by thermometer with dry bulb
c) wet bulb temperature is recorded by thermometer when bulb is covered with a cotton wick which is saturated with water
d) all of the mentioned
Answer: d
Clarification: These are the definitions of dew point temperature, dry bulb temperature and wet bulb temperature.

13. The wet bulb temperature is the ____ temperature recorded by moistened bulb.
a) lowest
b) highest
c) atmospheric
d) none of the mentioned
Answer: a
Clarification: This is a property of wet bulb temperature.

14. At any dbt, the ____ the difference of wbt reading below below dbt, ____ is the amount of water vapour held in mixture.
a) smaller, smaller
b) greater, greater
c) greater, smaller
d) smaller, greater
Answer: c
Clarification: This is an important property of dbt and wbt.

15. When unsaturated air flows over a sheet of water in an insulated chamber
a) specific humidity of air decreases
b) the water evaporates
c) both air and water are cooled during evaporation
d) all of the mentioned
Answer: a
Clarification: The specific humidity of air increases during this process.

Thermodynamics Multiple Choice Questions 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
Answer: c
Clarification: 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
Answer: a
Clarification: 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
Answer: d
Clarification: 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
Answer: b
Clarification: 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
Answer: b
Clarification: 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
Answer: a
Clarification: 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
Answer: d
Clarification: 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
Answer: c
Clarification: 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
Answer: a
Clarification: 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
Answer: b
Clarification: 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
Answer: d
Clarification: 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
Answer: c
Clarification: 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
Answer: b
Clarification: 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.