Category: Thermodynamics Objective Questions

Thermodynamics Multiple Choice Questions on “Work in a Reversible Process-2”.

1. Water at 250°C, 1000 kPa is brought to saturated vapour in a piston/cylinder with an isobaric process. Find the specific work.
a) -18.28 kJ/kg
b) -48.28 kJ/kg
c) -28.28 kJ/kg
d) -38.28 kJ/kg
Answer: d
Clarification: Process: P = C => w = ∫ P dv = P(v2 − v1)
1: v1 = 0.23268 m³/kg, s1= 6.9246 kJ/kgK, u1 = 2709.91 kJ/kg
2: v2 = 0.19444 m³/kg, s2 = 6.5864 kJ/kg K,
u2 = 2583.64 kJ/kg, T2 = 179.91°C
1w2 = P (v2 − v1) = 1000 (0.1944 – 0.23268) = -38.28 kJ/kg.

2. A heavily insulated cylinder/piston contains ammonia at 60°C, 1200 kPa. The piston is moved, expanding the ammonia in a reversible process until the temperature is −20°C during which 600 kJ of work is given out by ammonia. What was the initial volume of the cylinder?
a) 0.285 m³
b) 0.385 m³
c) 0.485 m³
d) 0.585 m³
Answer: b
Clarification: State 1: v1 = 0.1238 m³/kg, s1 = 5.2357 kJ/kg K,
u1 = h – Pv = 1553.3 – 1200×0.1238 = 1404.9 kJ/kg
Process: reversible (1S2(gen) = 0) and adiabatic (dQ = 0) => s2 = s1
State 2: T2, s2 ⇒ x2 = (5.2357 – 0.3657)/5.2498 = 0.928
u2 = 88.76 + 0.928×1210.7 = 1211.95 kJ/kg
1Q2 = 0 = m(u2 – u1) + 1W2 = m(1211.95 – 1404.9) + 600 ⇒ m = 3.110 kg
V1 = mv1 = 3.11 × 0.1238 = 0.385 m³.

3. Water at 250°C, 1000 kPa is brought to saturated vapor in a piston/cylinder with an adiabatic process. Find the specific work.
a) 139.35 kJ/kg
b) 149.35 kJ/kg
c) 159.35 kJ/kg
d) 169.35 kJ/kg
Answer: c
Clarification: State 1: v1 = 0.23268 m³/kg, u1 = 2709.91 kJ/kg, s1 = 6.9246 kJ/kg K
State 2: x = 1 and s2 = s1 = 6.9246 kJ/kg K
T2 = 140.56°C, P2 = 367.34 kPa, v2 = 0.50187 m³/kg, u2= 2550.56 kJ/kg
1w2 = u1 – u2 = 2709.91 – 2550.56 = 159.35 kJ/kg.

4. A piston/cylinder contains 2kg water at 200°C, 10 MPa. The water expands in an isothermal process to a pressure of 200 kPa. Any heat transfer takes place with an ambient at 200°C and whole process is be assumed reversible. Calculate the total work.
a) 1290.3 kJ
b) 1390.3 kJ
c) 1490.3 kJ
d) 1590.3 kJ
Answer: a
Clarification: State 1: v1 = 0.001148 m³/kg, u1 = 844.49 kJ/kg,
s1 = 2.3178 kJ/kg K, V1 = mv1 = 0.0023 m³
State 2: v2 = 1.08034 m³/kg, u2 = 2654.4 kJ/kg,
s2 = 7.5066 kJ/kg K, V2 = mv2 = 2.1607 m³
1Q2 = mT(s2 − s1) = 2 × 473.15 (7.5066 – 2.3178) = 4910 kJ
1W2 = 1Q2 – m(u2 – u1) = 1290.3 kJ.

5. A piston/cylinder of total 1kg steel contains 0.5 kg ammonia at 1600 kPa both masses at 120°C with minimum volume being 0.02 m³. The whole system is cooled down to 30°C by heat transfer to the ambient at 20°C, and during the process the steel keeps same temperature as the ammonia. Find the work.
a) – 28.14 kJ
b) – 38.14 kJ
c) – 48.14 kJ
d) – 58.14 kJ
Answer: d
Clarification: 1 : v1 = 0.11265 m³/kg, u1 = 1516.6 kJ/kg,
s1 = 5.5018 kJ/kg K, V1 = mv1 = 0.05634 m³
2 : 30°C < T(stop) so v2 = v(stop) = 0.04 m³/kg
1W2= ∫ P dV = (P1)m(v2-v1) = 1600 × 0.5 (0.04 – 0.11265)
= – 58.14 kJ.

6. A mass of 1 kg of air contained in a cylinder at 1000 K, 1.5 MPa, expands in a reversible isothermal process to a volume 10 times larger. Calculate the heat transfer during the process.
a) 460.84 kJ
b) 560.84 kJ
c) 660.84 kJ
d) 760.84 kJ
Answer: c
Clarification: Process: T = constant so with ideal gas => u2 = u1
1Q2 = 1W2 = ⌠PdV = P1V1 ln (V2/V1) = mRT1 ln (V2/V1)
= 1 × 0.287 × 1000 ln (10) = 660.84 kJ.

7. A piston/cylinder contains air at 400 K, 100 kPa which is compressed to a final pressure of 1000 kPa. Consider the process to be a reversible adiabatic process. Find the specific work.
a) -166.7 kJ/kg
b) -266.7 kJ/kg
c) -366.7 kJ/kg
d) -466.7 kJ/kg
Answer: b
Clarification: We have constant s, an isentropic process
T2 = T1( P2 / P1)^[(k-1)/k] = 400(1000/100)^(0.4/1.4)
= 400 × 10^(0.28575) = 772 K
1w2 = u1 – u2 = Cv(T1 – T2) = 0.717(400 – 772) = -266.7 kJ/kg.

8. A piston/cylinder contains air at 400 K, 100 kPa which is compressed to a final pressure of 1000 kPa. Consider the process to be a reversible isothermal process. Find the specific work.
a) −264 kJ/kg
b) −364 kJ/kg
c) −464 kJ/kg
d) −564 kJ/kg
Answer: a
Clarification: For this process T2 = T1 so since ideal gas we get u2 = u1 and also 1w2 = 1q2
1w2 = 1q2 = T(s2 – s1) = −RT ln(P2/P1)
= − 0.287 × 400 ln 10 = −264 kJ/kg.

9. Consider a small air pistol with a cylinder volume of 1 cm3 at 27°C, 250 kPa. The bullet acts as a piston and is released so the air expands in an adiabatic process. If the pressure should be 100 kPa as the bullet leaves the cylinder find the work done by the air.
a) 0.115 J
b) 0.125 J
c) 0.135 J
d) 0.145 J
Answer: d
Clarification: Process: Adiabatic 1q2 = 0 Reversible 1s2(gen) = 0
this is an isentropic expansion process giving s2 = s1
T2 = T1( P2 / P1)^[(k-1)/k] = 300(100/250)^(0.4/1.4) = 300 × 0.4^(0.28575) = 230.9 K
V2 = V1 P1 T2/P2 T1 = 1 × 250 × 230.9/100 × 300 = 1.92 cm³
Work = [1/(k-1)](P2V2 – P1V1) = [1/(1-1.4)](100 × 1.92 – 250 × 1) ×10^(-6)
= 0.145 J.

10. A spring loaded piston cylinder contains 1.5 kg air at 160 kPa and 27°C. It is heated in a process where pressure is linear in volume, P = A + BV, to twice the initial volume where it reaches 900 K. Find the work assuming a source at 900 K.
a) 61.4 kJ
b) 161.4 kJ
c) 261.4 kJ
d) 361.4 kJ
Answer: b
Clarification: State 1: u1 = 214.36 kJ/kg, V1 = mRT1/ P1 = (1.5 × 0.287 ×300) / 160 = 0.8072 m³
State 2: u2 = 674.824 kJ/kg,
P2 = RT2/v2 = RT2/(2v1) = T2 P1/(2T1)= P1(T2/2)T1
= 160 × 900 / 2 × 300 = 240 kPa
1W2 = ∫ PdV = 0.5 × (P1 + P2) (V2 – V1) = 0.5 × (P1 + P2) V1
= 0.5 × (160 + 240) 0.8072 = 161.4 kJ.

11. Helium contained in a cylinder at ambient conditions, 100 kPa, 20°C, is compressed in a reversible isothermal process to 600 kPa, after which the gas is expanded back to 100 kPa in a reversible adiabatic process. Calculate the net work per kilogram of helium.
a) -623.6 kJ/kg
b) +467.4 kJ/kg
c) -1091.0 kJ/kg
d) none of the mentioned
Answer: a
Clarification: The adiabatic reversible expansion gives constant s
T3 = T2(P3/P2)^[(k-1)/k] = 293.15 (100/600)^0.4 = 143.15 K
The isothermal process: 1w2 = -RT1 ln(P2/P1)
= -2.0771 × 293.15 × ln(600/100) = -1091.0 kJ/kg
The adiabatic process: 2w3 = CVo(T2-T3) = 3.116 (293.15 – 143.15) = +467.4 kJ/kg
The net work is the sum: w(NET) = -1091.0 + 467.4 = -623.6 kJ/kg.

12. 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 adiabatic.
a) -112.0 kJ
b) -212.0 kJ
c) -312.0 kJ
d) -412.0 kJ
Answer: c
Clarification: Process: 1Q2 = 0 => s2 = s1 thus isentropic process
T2 = T1(P2/P1)^[(k-1)/k] = 293.2(800/100)^0.230 = 473.0 K
1W2 = -mCv(T2 – T1) = -1 × 1.7354 (473.0 – 293.2)
= -312.0 kJ.

13. A piston/cylinder contains air at 100 kPa, 300 K. It is now compressed in a reversible adiabatic process to a volume 7 times as small. Use constant heat capacity and find the specific work.
a) -233.6 kJ/kg
b) -243.6 kJ/kg
c) -253.6 kJ/kg
d) -263.6 kJ/kg
Answer: c
Clarification: v2/ v1 = 1/7; P2 /P1 = (v2/v1)^(-k) = 7^(1.4) = 15.245
P2 = P1[7^(1.4)] = 100 × 15.245 = 1524.5 kPa
T2 = T1 (v1/v2)^(k-1) = 300 × 7^(0.4) = 653.4 K
1q2 = 0 kJ/kg; work = R(T2-T1)/(1-k) = 0.287*(653.4 – 300)/(-0.4)
= -253.6 kJ/kg.

14. A gas confined in a piston-cylinder is compressed in a quasi-static process from 80 kPa and 0.1 m3 to 400 kPa and 0.03m³. If the pressure and volume are related by PV^n= constant, calculate the work involved in the process.
a) – 12.87 kJ
b) 12.87 kJ
c) – 11.87 kJ
d) 11.87 kJ
Answer: c
Clarification: n = ln(P2/P1)/ln(V1/V2) = ln(400/80)/ln(0.1/0.03) = 1.337
Work involved in the process (1W2) = (P2V2 – P1V1)(1 – n) = – 11.87 kJ.

15. One kg of steam at 200 kPa with 20% quality is heated at constant pressure to 400°C. Calculate the work done by the system .
a) 274.261 kJ
b) 374.261 kJ
c) 474.261 kJ
d) 574.261 kJ
Answer: a
Clarification: v1 = 0.001061 + 0.2*0.88467 = 0.177995 m3/kg; v2 = 1.5493 m3/kg
work done by the system during this process (1W2) = mP(v2 – v1)
= 1*200*(1.5493 – 0.177995) = 274.261 kJ.

Thermodynamics Multiple Choice Questions on “Equations of State of a Gas”.

1. A mole of a substance has a mass equal to the molecular weight of the substance.
a) true
b) false
Answer: a
Clarification: 1gm of oxygen has mass of 32gm and 1gm of nitrogen has a mass of 28gm and so on.

2. According to Avogadro’s law, volume of a g mol of all gases at the pressure of ____ and temperature of ____ is same.
a) 760 mm Hg, 100 degree Celsius
b) 760 mm Hg, 0 degree Celsius
c) 750 mm Hg, 100 degree Celsius
d) 750 mm Hg, 0 degree Celsius
Answer: b
Clarification: This is the Avogadro law and these temperature and pressure condition is know as normal temperature and pressure(NTP).

3. At NTP, the volume of a g mol of all gases is(in litres)
a) 22.1
b) 22.2
c) 22.3
d) 22.4
Answer: d
Clarification: This comes from the Avogadro’s law.

4. Which of the following statement is true?
a) number of kg moles of a gas = mass / molecular weight
b) molar volume = total volume of the gas / number of kg moles
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: Number of moles is in kgmoles and molar volume is in (m^3)/kg mol.

5. The equation of state is a functional relationship between
a) pressure
b) molar or specific volume
c) temperature
d) all of the mentioned
Answer: d
Clarification: It is expressed in the form f(p,v,T)=0.

6. If two properties (out of p,v,T) of a gas are known, the third can be evaluated.
a) true
b) false
Answer: a
Clarification: The third property can be calculated from the equation of state.

7. Which of the following statement is true about a gas?
a) lim(pv) with p tending to 0 is independent of the nature of gas
b) lim(pv) with p tending to 0 depends only on the temperature
c) this holds true for all the gases
d) all of the mentioned
Answer: d
Clarification: This is one of the fundamental property for all the gases.

8. Universal gas constant is given by
a) lim(pv) / 273.16
b) R
c) 0.083 litre-atm/gmol K
d) all of the mentioned
Answer: d
Clarification: Putting the value of lim(pv) = 22.4 litre-atm/gmol we get the value of R.

9. The equation of state of a gas is lim(pv)=RT .
a) true
b) false
Answer: a
Clarification: The limit is calculated with p tending to 0 and v is the molar volume here.

10. For which of the following gases, does the product (pv) when plotted against p gives depends only on temperature?
a) nitrogen
b) hydrogen
c) air and oxygen
d) all of the mentioned
Answer: d
Clarification: It is a fundamental property of gas that lim(pv) with p tending to 0 is independent of the nature of gas and depends only on the temperature.

Thermodynamics Multiple Choice Questions on “Mixtures of Variable Composition”.

1. For a system of variable composition, the internal energy depends on
a) entropy
b) volume
c) moles
d) all of the mentioned
Answer: d
Clarification: If some substance is added to the system, then energy of the system increases.

2. If the composition of system does not change, then dU=TdS-pdV .
a) true
b) false
Answer: a
Clarification: If the composition changes, then the relation includes many other terms.

3. The molal chemical potential is given by
a) ∂U/∂S
b) ∂U/∂n
c) ∂U/∂V
d) all of the mentioned
Answer: b
Clarification: The molal chemical potential signifies the change in internal energy per unit mole of a component when S,V and number of moles of all other components are constant.

4. The Gibbs entropy equation is given by
a) TdS = dU – pdV – Σ(molal chemical potential)*dn
b) TdS = dU + pdV + Σ(molal chemical potential)*dn
c) TdS = dU + pdV – Σ(molal chemical potential)*dn
d) TdS = dU – pdV + Σ(molal chemical potential)*dn
Answer: c
Clarification: Here summation is taken for all the components present in the system.

5. An equation in Gibbs energy is be given by
a) dG = Vdp + SdT + Σ(molal chemical potential)*dn
b) dG = Vdp – SdT – Σ(molal chemical potential)*dn
c) dG = Vdp + SdT – Σ(molal chemical potential)*dn
d) dG = Vdp – SdT + Σ(molal chemical potential)*dn
Answer: d
Clarification: For this we use G=U+pV-TS and here the summation is taken for all the components present in the system.

6. The equation written for Gibbs energy can also be written for
a) H
b) F
c) Both of the mentioned
d) None of the mentioned
Answer: c
Clarification: From the equation dG = Vdp – SdT + Σ[(molal chemical potential)*dn], we can write similar equations for F and H..

7. Chemical potential is an extensive property.
a) true
b) false
Answer: b
Clarification: Chemical potential is an intensive property.

8. If the phase of a multi-component system is enlarged, which of the following will happen?
a) U,S and V will increase and T,p and chemical potential will remain same
b) U,S and V will decrease and T,p and chemical potential will remain same
c) U,S and V will increase and T,p and chemical potential will decrease
d) U,S and V will decrease and T,p and chemical potential will increase
Answer: a
Clarification: This depends on the type of property.

9. The Gibbs-Duhem equation is given by
a) SdT + Vdp – Σ(n)*d(molal chemical potential)
b) -SdT + Vdp – Σ(n)*d(molal chemical potential)
c) SdT + Vdp – Σ(n)*d(molal chemical potential)
d) -SdT – Vdp – Σ(n)*d(molal chemical potential)
Answer: b
Clarification: This equation shows the relationship for simultaneous changes in p,T and chemical potential.

10. For a phase which has only one constituent,
a) chemical potential = n/G
b) chemical potential = 1/(G*n)
c) chemical potential = G*n
d) chemical potential = G/n
Answer: d
Clarification: This means that chemical potential is the molar Gibbs function and is a function of p and T only.

11. If a closed system is in equilibrium, which of the following remains constant?
a) entropy
b) volume
c) internal energy
d) all of the mentioned
Answer: d
Clarification: In a closed system, there is no interaction with surroundings hence these quantities along with mass remains constant.

12. At chemical equilibrium, G will be minimum subjected to the equations of constraint.
a) true
b) false
Answer: a
Clarification: This minimum occurs at a constant p and T.

13. The Gibbs phase rule for a non-reactive system is given by
a) f = C + (number of phases) + 2
b) f = C – (number of phases) – 2
c) f = C – (number of phases) + 2
d) f = C – (number of phases) – 2
Answer: c
Clarification: Here f is the variance or the degree of freedom and C is the number of constituents.

14. For a pure substance existing in a single phase,
a) C=1
b) number of phases = 1
c) f=2
d) all of the mentioned
Answer: d
Clarification: Hence we need to know two properties to fix up the state of the system at equilibrium.

15. Which of the following statement is true?
a) if C=1 and number of phases=2, then f=1
b) if C=1 and number of phases=3, then f=0
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: This comes from the Gibbs phase rule.

Thermodynamics Multiple Choice Questions on “Brayton Cycle-2”.

1. How can regeneration be used to improve the efficiency of Brayton cycle?
a) the energy of exhaust gas can be used to heat up the air which leaves the compressor
b) heat supplied from external source thus decreases
c) the amount of heat rejected also decreases
d) all of the mentioned
Answer: d
Clarification: This is how a regenerator can be used in Brayton cycle.

2. The temperature of air leaving turbine is less than that of air leaving compressor.
a) true
b) false
Answer: b
Clarification: The temperature of air leaving turbine is more than that of air leaving compressor.

3. In the regenerator,
a) temperature of air leaving the compressor is raised
b) temperature of air leaving the turbine is raised
c) both of the mentioned
d) none of the mentioned
Answer: a
Clarification: This is done by using the heat from the turbine exhaust.

4. Which of the following statement is true for a regenerator?
a) mean temperature of heat addition decreases
b) mean temperature of heat rejection decreases
c) both of the mentioned
d) none of the mentioned
Answer: b
Clarification: The mean temperature of heat addition increases by using a regenerator.

5. By using a regenerator,
a) efficiency increases, work output decreases
b) both efficiency and work output increases
c) efficiency increases, work output remains unchanged
d) efficiency remains same, work output increases
Answer: c
Clarification: This is because mean temperature of heat addition increases and mean temperature of heat rejection decreases by using regenerator.

6. Which of the following is true about a regenerator?
a) it is costly
b) it is heavy and bulky
c) it causes pressure losses
d) all of the mentioned
Answer: d
Clarification: This loss in pressure decreases the cycle efficiency.

7. When we add a regenerator, cycle efficiency always increases.
a) true
b) false
Answer: b
Clarification: The addition of regenerator after a certain pressure ratio decreases the cycle efficiency as compared to Brayton cycle.

8. When the turbine efficiency and compressor efficiency decreases, the cycle efficiency ____
a) decreases
b) increases
c) remains same
d) none of the mentioned
Answer: a
efficiency: The Brayton cycle is very sensitive to the efficiency of the turbine and compressor.

9. The ____ the pressure ratio, the ____ will be efficiency.
a) less, more
b) less, less
c) more, more
d) more, less
Answer: c
Clarification: This comes from the Brayton cycle efficiency in terms of pressure ratio.

10. As the pressure ratio increases, the efficiency steadily ____
a) decreases
b) increases
c) remains constant
d) none of the mentioned
Answer: b
Clarification: As pressure ratio increases, the mean temperature of heat addition increases and the mean temperature of heat rejection decreases.

11. The maximum pressure ratio is given by
a) (Tmin/Tmax)^((ɣ-1)/ɣ)
b) (Tmin/Tmax)^(ɣ/(ɣ-1))
c) (Tmax/Tmin)^((ɣ-1)/ɣ)
d) (Tmax/Tmin)^(ɣ/(ɣ-1))
Answer: d
Clarification: Here Tmax is the maximum temperature and Tmin is the minimum temperature which is the temperature of surroundings.

12. The optimum value of pressure ratio at which work capacity becomes maximum is given by
a) (Tmin/Tmax)^((ɣ-1)/2ɣ)
b) (Tmax/Tmin)^(ɣ/2(ɣ-1))
c) (Tmax/Tmin)^(2(ɣ-1)/ɣ)
d) (Tmax/Tmin)^(2ɣ/(ɣ-1))
Answer: b
Clarification: This is the optimum value of pressure ratio.

13. The relation between maximum pressure ratio and optimum pressure ratio is given by
a) optimum pressure ratio = (maximum pressure ratio)/2
b) optimum pressure ratio = maximum pressure ratio
c) optimum pressure ratio = sqrt(maximum pressure ratio)
d) optimum pressure ratio = (maximum pressure ratio)^2
Answer: c
Clarification: This relation comes from the expressions of optimum pressure ratio and maximum pressure ratio.

14. Which of the following statement is true?
a) maximum work done = (specific heat at constant preferred)*(sqrt(Tmax) – sqrt(Tmin))^2
b) efficiency of cycle = 1- sqrt(Tmin/Tmax)
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: These expressions come from the maximum pressure ratio.

15. The efficiency of Brayton cycle can be increased by using staged heat supply or by use of staged compression with intercooling.
a) true
b) false
Answer: a
Clarification: The staged heat supply is also called reheat.

Thermodynamics Multiple Choice Questions on “First Law for a Closed System”.

1. Energy has different forms which include
a) heat
b) work
c) all of the mentioned
d) none of the mentioned
Answer: c
Clarification: Basic fact about energy.

2. Work input is directly proportional to heat and the constant of proportionality is called
a) joule’s equivalent
b) mechanical equivalent of heat
c) all of the mentioned
d) none of the mentioned
Answer: c
Clarification: True for a closed system undergoing a cycle.

3. The value of constant of proportionality, J, has the value
a) 1
b) 0
c) -1
d) infinity
Answer: a
Clarification: In the S.I. system, both heat and work are measured in the derived unit of energy, the Joule.

4. It was Joule who first established that heat is a form of energy, and thus laid the foundation of the first law of thermodynamics.
a) true
b) false
Answer: a
Clarification: Prior to Joule, heat was considered to be an invisible fluid flowing from a body of higher calorie to a body of lower calorie.

5. Which of the following represents the energy in storage?
a) heat
b) work
c) internal energy
d) none of the mentioned
Answer: c
Clarification: Energy in storage is internal energy or the energy of the system.

6. By first law of thermodynamics,
a) Q=ΔE-W
b) Q=ΔE+W
c) Q=-ΔE-W
d) Q=-ΔE+W
Answer: b
Clarification: Q-W is the net energy stored in system and is called internal energy of system.

7. The expression (ΣW)cycle=(ΣQ)cycle applies only to systems undergoing cycles.
a) true
b) false
Answer: a
Clarification: The above expression holds for a closed cycle.

8. Which of the following is the first law for a closed system undergoing a cycle?
a) ∫dW=∫dQ
b) J∫dW=∫dQ
c) ∫dW=J∫dQ
d) none of the mentioned
Answer: c
Clarification: This is the expression for first law of thermodynamics where ∫ denotes the cyclic integral for the closed path.

9. Which of the following an be considered as the definition of energy?
a) Q=ΔE+W
b) Q-W=ΔE
c) first law of thermodynamics
d) all of the mentioned
Answer: d
Clarification: The first law is a particular formulation of the principle of the conservation of energy.

10. The first law of thermodynamics gives only the change on energy ΔE for the process.
a) true
b) false
Answer: a
Clarification: An absolute value of energy E, is not given by the first law.