Category: Thermodynamics Objective Questions

250+ TOP MCQs on Thermodynamics Basic Concepts and Answers

Thermodynamics Multiple Choice Questions on “Basic Concepts”.

1. One kg of diatomic Oxygen is present in a 500 L tank. Find the specific volume on both mass and mole basis.
a) 0.6 m3/kg , 0.260 m3/mole
b) 0.5 m3/kg , 0.0160 m3/mole
c) 0.56 m3/kg , 0.0215 m3/mole
d) 0.7 m3/kg , 0.0325 m3/mole
Answer: b
Clarification: The specific volume on mass basis= 0.5/1 = 0.5 m3/kg
specific volume on mole basis= 0.5/(1/0.032) = 0.0160 m3/mole.

2. A piston/cylinder with a cross-sectional area of 0.01 m^2 is resting on the stops. With an outside pressure of 100 kPa, what should be the water pressure to lift the piston?
a) 178kPa
b) 188kPa
c) 198kPa
d) 208kPa
Answer: c
Clarification: Pw = Po + mg/A = 100000 + (100*9.8/0.01) = 198kPa.

3. A large exhaust fan in a lab room keeps the pressure inside at 10 cm water relative vacuum to the hallway? What is the net force acting on the door measuring 1.9 m by 1.1 m?
a) 2020 N
b) 2030 N
c) 2040 N
d) 2050 N
Answer: d
Clarification: Net force acting on the door = Gauge pressure*area
= 0.1cm water*1.9*1.1 = 2050 N.

4. A 5 m long vertical tube having cross sectional area 200 cm^2 is placed in a water. It is filled with 15°C water, with the bottom closed and the top open to 100 kPa atmosphere. How much water is present in tube?
a) 99.9 kg
b) 109.9 kg
c) 89.9 kg
d) 79.9 kg
Answer: a
Clarification: m = ρ V = V/v = AH/v
= 200 × 10^(−4) × 5/0.001001 = 99.9 kg.

5. A 5 m long vertical tube having cross sectional area 200 cm2 is placed in a water. It is filled with 15°C water, with the bottom closed and the top open to 100 kPa atmosphere. What is the pressure at the bottom of tube ?
a) 119 kPa
b) 129 kPa
c) 139 kPa
d) 149 kPa
Answer: d
Clarification: ∆P = ρ gH = gH/v = 9.80665 × 5/0.001001
= 48.98 kPa
P(total) = P(top) + ∆P = 149 kPa.

6. Find the pressure of water at 200°C and having specific volume of 1.5 m3/kg.
a) 0.9578 m3/kg
b) 0.8578 m3/kg
c) 0.7578 m3/kg
d) 0.6578 m3/kg
Answer: a
Clarification: The state is superheated vapour between 100 and 150 kPa.
v = 1.3136 + (0.8689 – 1.3136)(140 − 100)/(150 – 100)
= 1.3136 + ( − 0.4447) × 0.8 = 0.9578 m3/kg.

7. Find the pressure of water at 200°C and having specific volume of 1.5 m^3/kg.
a) 141.6 kPa
b) 111.6 kPa
c) 121.6 kPa
d) 161.6 kPa
Answer: d
Clarification: v > vg so that it is superheated vapour.
Between 100 kPa and 200 kPa,
P = 100 + (200 – 100)(1.5 − 2.17226)/(1.08034 − 2.17226)
= 161.6 kPa.

8. A 5m^3 container is filled with 840 kg of granite (density is 2400 kg/m^3) and the rest of the volume is air (density is 1.15 kg/m^3). Find the mass of air present in the container.
a) 9.3475 kg
b) 8.3475 kg
c) 6.3475 kg
d) 5.3475 kg
Answer: d
Clarification: Mass of the air (mair) = ρairVair = ρair (Vtotal – Vgranite)
= ρair (Vtotal – (m/ρ)granite) = 1.15*(5 – 840/2400) = 5.3475 kg.

9. A 100 m tall building receives superheated steam at 200 kPa at ground and leaves saturated vapour from the top at 125 kPa by losing 110 kJ/kg of heat. What should be the minimum inlet temperature at the ground of the building so that no steam will condense inside the pipe at steady state?
a) 363.54°C
b) 263.54°C
c) 163.54°C
d) none of the mentioned
Answer: c
Clarification: FLOT for steam flow results: q + h(ground) = h(top) + gZtop.
h(top) = [email protected] kPa = 2685.35 kJ/kg
H(ground) = 2685.35 + (9.80665*100)/1000 – (–110) = 2796.33 kJ/kg
Minimum temperature at the ground of the building T(ground)
= 163.54°C.

10. The pressure gauge on an air tank shows 60 kPa when the diver is 8 m down in the ocean. At what depth will the gauge pressure be zero?
a) 34.118 m
b) 24.118 m
c) 14.118 m
d) none of the mentioned
Answer: b
Clarification: Pressure at 10 m depth = Patm + ρgh = 101.325 + 1000*9.80665*8/1000 = 179.778 kPa
Absolute pressure of the air in the tank = 179.778 + 60 = 239.778 kPa
Depth at which gauge pressure is zero (H) = (239.778 – 101.325)*1000/(1000*9.80665) = 14.118 m.

11. A piston-cylinder device initially contains air at 150 kPa and 27°C. At this state, the volume is 400 litre. The mass of the piston is such that a 350 kPa pressure is required to move it. The air is now heated until its volume has doubled. Determine the final temperature.
a) 1400 K
b) 400 K
c) 500 K
d) 1500 K
Answer: a
Clarification: Final temperature (T3)= 1400 K.–> P1V1/T1=P3V3/T3.

12. A piston-cylinder device initially contains air at 150 kPa and 27°C. At this state, the volume is 400 litre. The mass of the piston is such that a 350 kPa pressure is required to move it. The air is now heated until its volume has doubled. Determine work done by the air.
a) 120 kJ
b) 130 kJ
c) 100 kJ
d) 140 kJ
Answer: d
Clarification: Work done = P3(V3- V1) = 140 kJ.

13. Find the change in u for carbon dioxide between 600 K and 1200 K for a constant Cv0 value.
a) 291.8 kJ/kg
b) 391.8 kJ/kg
c) 491.8 kJ/kg
d) 591.8 kJ/kg
Answer: b
Clarification: Δu = Cv0 ΔT = 0.653 ×(1200–600) = 391.8 kJ/kg.

14. Calculate the change in enthalpy of carbon dioxide from 30 to 1500°C at 100 kPa at constant specific heat.
a) 2237.7 kJ/kg
b) 1637.7 kJ/kg
c) 1237.7 kJ/kg
d) 2337.7 kJ/kg
Answer: c
Clarification: Δh = CpΔT = 0.842 (1500 – 30) = 1237.7 kJ/kg.

15. A sealed rigid vessel has volume of 1 m3 and contains 2 kg of water at 100°C. The vessel is now heated. If a safety pressure valve is installed, at what pressure should the valve be set to have a maximum temperature of 200°C ?
a) 431.3 kPa
b) 531.3 kPa
c) 631.3 kPa
d) 731.3 kPa
Answer: a
Clarification: Initial specific volume (v1) = 1 m3/2 kg = 0.5 m3/kg
Interpolating, pressure for the same specific volume at 200°C
= 400 + {(0.53422-0.5)/(0.53422-0.42492)}*(500-400) = 431.3 kPa.

16. A system undergoing change in state from A to B along path ‘X’ receives 100 J heat and does 40 J work. It returns to state A from B along path ‘Y’ with work input of 30 J. Calculate the heat transfer involved along the path ‘Y’.
a) – 60 J
b) 60 J
c) – 90 J
d) 90 J
Answer: c
Clarification: ∆UX = AQB – AWB = 100 – 40 = 60 J
Heat transfer involved along the ‘Y’ (BQA) = ∆UY + BWA = – ∆UX + BWA
= – 60 – 30 = – 90 J.

250+ TOP MCQs on Polytropic Process-2 and Answers

Thermodynamics Inteview Questions and Answers for freshers on “Polytropic Process-2”.

1. A cylinder/piston contains air at 100 kPa and 20°C with a V=0.3 m^3. The air is compressed to 800 kPa in a reversible polytropic process with n = 1.2, after which it is expanded back to 100 kPa in a reversible adiabatic process. Find the net work.
a) -174.6 kJ/kg
b) -154.6 kJ/kg
c) -124.6 kJ/kg
d) -194.6 kJ/kg
Answer: a
Clarification: m = P1V1/RT1 = (100 × 0.3)/(0.287 × 293.2) = 0.3565 kg
T2/T1 = (P2/P1)^[(n-1)/n] = 293.2(800/100)^(0.167) = 414.9 K
W = (P2 v2 – P1 v1)/(1-n) = R(T2-T1)/(1-n)
= 0.287(414.9-293.2)/(1-1.20) = -174.6 kJ/kg.

2. A piston-cylinder contains carbon dioxide at 2MPa with V=50 L. The device has a mass of 4 kg. Everything is initially at 200°C. By heat transfer the whole system cools to 25°C, at which point the gas pressure is 1.5 MPa. Find the work done.
a) -10.0 kJ
b) -12.0 kJ
c) -14.0 kJ
d) -16.0 kJ
Answer: c
Clarification: CO2: m = P1V1/RT1 = 2000 × 0.05/(0.18892 × 473.2) = 1.1186 kg
V2 = V1(P1/P2)(T2/T1) = 0.05(2/1.5)(298.2/473.2) = 0.042 m^3
Work = ⌠PdV = (P1 + P2)(V2 – V1)/2 = (2000 + 1500)(0.042 – 0.050)/2
= -14.0 kJ.

3. A gas initially at 500°C, 1 MPa is contained in a piston-cylinder arrangement with an initial volume of 0.1 m^3. It is then slowly expanded according to the relation PV = constant until a final pressure of 100 kPa is attained. Determine the work for this process.
a) 200.3 kJ
b) 210.3 kJ
c) 220.3 kJ
d) 230.3 kJ
Answer: d
Clarification: Process: PV = C ⇒ V2 = P1V1/P2 = 1000 × 0.1/100 = 1 m^3
1W2 = ∫ P dV = ⌠ CV^(-1)dV = C ln(V2/V1)
1W2 = P1V1 ln(V2/V1) = 1000 × 0.1 ln (1/0.1) = 230.3 kJ.

4. Helium gas expands from 350 K, 125 kPa and 0.25 m^3 to 100 kPa in a polytropic process with n = 1.667. How much work does it give out?
a) 3.09 kJ
b) 4.09 kJ
c) 5.09 kJ
d) 6.09 kJ
Answer: b
Clarification: Process: PV^n = constant = P1(V1)^n = P2(V2)^n
V2 = V1 (P1/P2)^(1/n) = 0.25 × (125/100)^(0.6) = 0.2852 m^3
Work = (P2V2 – P1V1)/(1-n) = (100× 0.2852 – 125× 0.25)/(1 – 1.667)
= 4.09 kJ.

5. Air goes through a polytropic process from 325 K, 125 kPa to 500 K, 300 kPa. Find the specific work in the process.
a) -51.8 kJ/kg
b) -61.8 kJ/kg
c) -71.8 kJ/kg
d) -81.8 kJ/kg
Answer: a
Clarification: Process: Pv^(n) = Const = P1(v1)^n = P2(v2)^n
Ideal gas Pv = RT hence v1 = RT/P = 0.287 × 325/125 = 0.7462 m^3/kg
v1 = RT/P = 0.287 × 500/300 = 0.47833 m^3/kg
n = ln(P2/P1) / ln(v1/v2) = ln 2.4 / ln 1.56 = 1.969
Work = (P2v2 – P1v1)/(1-n) = R(T2-T1)/(1-n) = 0.287(500 – 325)/(1-1.969)
= -51.8 kJ/kg.

6. A piston-cylinder contains 0.1 kg air at 400 K, 100 kPa which goes through a polytropic compression process (n = 1.3) to a pressure of 300 kPa. How much work has been done by air in the process?
a) -277 kJ
b) -377 kJ
c) -477 kJ
d) -577 kJ
Answer: c
Clarification: Process: Pv^(n) = Const;
T2 = T1 ( P2 V2 / P1V1) = T1 ( P2 / P1)(P1 / P2 )^(1/n)
= 400 × (300/100)^(1 – 1/1.3) = 515.4 K
Work = (P2V2 – P1V1)/(1-n) = mR(T2-T1)/(1-n)
= (0.2 × 0.287)(515.4-400)/(1 − 1.3) = -477 kJ.

7. A balloon behaves according to the equation P = (C2)V^(1/3), C2 = 100 kPa/m. The balloon is blown up with air from a volume of 1 m^3 to a volume of 3 m^3. Find the work done by the air assuming it is at 25°C.
a) 219.5 kJ
b) 229.5 kJ
c) 239.5 kJ
d) 249.5 kJ
Answer: d
Clarification: The process is polytropic with exponent n = -1/3.
P1 = (C2)V^(1/3) = 100 × 1^(1/3) = 100 kPa
P1 = (C2)V^(1/3) = 100 × 3^(1/3) = 144.22 kPa
Work = ⌠PdV = (P2V2 – P1V1)/(1-n) = (144.22 × 3 – 100 × 1)/(1 – (-1/3))
= 249.5 kJ.

8. A balloon behaves such that pressure inside it is proportional to the diameter squared. It contains 2kg of ammonia at 0°C, 60% quality. They are now heated so that the final pressure is 600 kPa. Find the work done in the process.
a) 117.5 kJ
b) 127.5 kJ
c) 137.5 kJ
d) 147.5 kJ
Answer: a
Clarification: Process : P∝D^2, with V ∝ D^3 this implies P∝D^2 ∝ V^(2/3)
so PV^(-2/3) = constant, hence n = −2/3
V1 = mv1 = 2(0.001566 + 0.6 × 0.28783) = 0.3485 m^3
V2 = V1*(P2/P1)^(3/2) = 0.3485(600/429.3)^(3/2) = 0.5758 m^3
Work = ⌠PdV = (P2V2 – P1V1)/(1-n) = (600 × 0.5758 – 429.3 × 0.3485)/[1 – (-2/3)]
= 117.5 kJ.

9. Consider a piston-cylinder with 0.5 kg of R-134a as saturated vapour at -10°C. It is compressed to a pressure of 500 kPa in a polytropic process with n = 1.5. Determine the work done during the process.
a) -6.07 kJ
b) -7.07 kJ
c) -8.07 kJ
d) -9.07 kJ
Answer: b
Clarification: Pv^(1.5) = constant until P = 500 kPa
1: v1 = 0.09921 m3/kg, P = Psat = 201.7 kPa
2: v2 = v1(P1/P2)^(1/1.5) = 0.09921×(201.7/500)^(2/3) = 0.05416
hence it is superheated vapour at T2 = 79°C
Work = ⌠PdV = m(P2 v2 – P1 v1)/(1-1.5) = 2*(500 × 0.05416 – 201.7 × 0.09921)/(-0.5)
= -7.07 kJ.

10. R-12 in a piston-cylinder arrangement is initially at 50°C, x = 1. It is then expanded in a process so that P = Cv^(−1) to a pressure of 100 kPa. Find the work.
a) 23.2 kJ/kg
b) 33.2 kJ/kg
c) 43.2 kJ/kg
d) 53.2 kJ/kg
Answer: c
Clarification: State 1: 50°C, x=1, P1 = 1219.3 kPa, v1 = 0.01417 m^3/kg
Process: P = Cv^(-1) ⇒ Work = ∫ P dv = C ln(v2/v1)
State 2: 100 kPa thus v2 = (v1)(P1)/P2 = 0.1728 m^3/kg hence T = – 13.2°C
Work = P1v1[ln(v2/v1)] = 1219.3 × 0.01417 × ln(0.1728/0.01417)
= 43.2 kJ/kg.

11. A piston-cylinder contains water at 3 MPa, 500°C. It is cooled in a polytropic process to 1 MPa, 200°C. Find the specific work in the process.
a) 155.2 kJ
b) 165.2 kJ
c) 175.2 kJ
d) 185.2 kJ
Answer: a
Clarification: Pv^(n) = C thus (P1/P2) = (v2/v1)^n
n= ln(P1/P2) / ln(v2/v1) = 1.0986/0.57246 = 1.919
Work = ⌠PdV = (P2 v2 – P1 v1)/(1-n) = (1000 × 0.20596 – 3000 × 0.11619)/(1 – 1.919)
= 155.2 kJ.

12. A piston/cylinder contains carbon dioxide at 300 kPa, 100°C with a volume of 0.2 m^3. Weights are added at such a rate that the gas compresses according to the relation PV^1.2 = constant to a final temperature of 200°C. Determine the work done during the process.
a) 70.4 kJ
b) -70.4 kJ
c) 80.4 kJ
d) -80.4 kJ
Answer: d
Clarification: For the Polytropic process PV^n = constant
1W2 = ∫PdV = (P2V2 – P1V1)/(1 – n )
Assuming ideal gas, PV = mRT
But mR = P1V1/T1 = 300 × 0.2/373.15 = 0.1608 kJ/K
1W2 = 0.1608(473.15 – 373.15)/(1 – 1.2) = -80.4 kJ.

Thermodynamics for interviews,

250+ TOP MCQs on First and Second Laws Combined and Answers

Thermodynamics Multiple Choice Questions on “First and Second Laws Combined”.

1. The equation TdS=dU+pdV is obtained from which law?
a) first law
b) second law
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: By first law, dQ=dU+pdV and from second law, dQ=TdS.

2. Which of the following equation is true?
a) TdS=dH+Vdp
b) TdS=dH-Vdp
c) TdS=-dH-Vdp
d) TdS=-dH+Vdp
Answer: b
Clarification: It comes from TdS=dU+pdV and H=U+pV.

3. The equation dQ=dE+dW holds good for
a) any process, reversible or irreversible
b) only reversible process
c) only irreversible process
d) none of the mentioned
Answer: a
Clarification: This equation holds good for any process and for any system.

4. The equation dQ=dU+pdW holds good for any process undergone by a closed stationary system.
a) true
b) false
Answer: a
Clarification: When a closed stationary system undergoes a process, this equation holds true.

5. The equation dQ=dU+pdV holds good for
a) open system
b) closed system
c) both of the mentioned
d) none of the mentioned
Answer: b
Clarification: This equation is true only for a reversible(quasi-static) process.

6. The equation TdS=dU+pdV holds good for
a) reversible process
b) reversible process
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: This equation holds good for any process undergone by a closed system since it is a relation among properties which are independent of the path.

7. The equation dQ=TdS is true only for a reversible process.
a) true
b) false
Answer: a
Clarification: This comes from the second law.

8. The equation TdS=dH-Vdp
a) relates only the properties of a system
b) there is no path function term in the equation
c) the equation holds good for any process
d) all of the mentioned
Answer: d
Clarification: Since there is no path function in the equation hence the equation holds good for any process.

9. The entropy change of a system between two identifiable equilibrium state is ___ when the intervening process is reversible or change of state is irreversible.
a) different
b) same
c) depends on the process
d) none of the mentioned
Answer: b
Clarification: To determine the change in entropy, a known reversible path is made to connect the two end states and integration is performed on this path.

10. It is better to state that “the change of state is irreversible, rather than say it is an irreversible process”.
a) true
b) false
Answer: a
Clarification: This is because no irreversible path or process can be plotted on thermodynamic coordinates.

250+ TOP MCQs on Comments on Exergy and Answers

Thermodynamics Multiple Choice Questions on “Comments on Exergy”.

1. ____ is conserved but ____ is not conserved.
a) exergy, energy
b) energy, exergy
c) both exergy and energy are conserved
d) neither exergy nor energy is conserved
Answer: b
Clarification: Energy is conserved but once the exergy is wasted, it can never be recovered.

2. Exergy is a composite property.
a) true
b) false
Answer: a
Clarification: It depends on the state of the system and the surroundings.

3. A dead state
a) is in equilibrium with its surroundings
b) has zero exergy
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: This is the basic fact about dead state which we have seen earlier.

4. Which of the following statement is true?
a) KE is entirely available energy
b) PE is entirely exergy
c) The exergy of thermal energy of reservoirs is equivalent to the work output of a Carnot engine operating between the reservoir at temperature T and environment To.
d) all of the mentioned
Answer: d
Clarification: W=Q(1-To/T).

5. Useful work is given by
a) difference between the actual work and the surrounding work
b) W – W for surroundings
c) W – p(V2-V1)
d) all of the mentioned
Answer: d
Clarification: Some work is lost to the surroundings and hence the useful work is reduced.

6. The surrounding work is zero for
a) cyclic devices
b) steady flow devices
c) system with fixed boundaries
d) all of the mentioned
Answer: d
Clarification: There is no work being done on surroundings in all these cases.

7. The maximum amount of useful work that can be obtained from a system as it undergoes a process between two specified states is called
a) adiabatic work
b) reversible work
c) irreversible work
d) none of the mentioned
Answer: b
Clarification: Reversible work is the maximum work that we can get in a process.

8. If the final state of the system is the dead state, the ____ and the ____ become identical.
a) reversible work, exergy
b) irreversible work, exergy
c) reversible work, irreversible work
d) none of the mentioned
Answer: a
Clarification: For a dead state, reversible work and exergy are same.

9. The difference between the reversible work and the useful work for a process is called irreversibility.
a) true
b) false
Answer: a
Clarification: Irreversibility is given by the product of To and rate of entropy generation, where To is the environment temperature.

10. For a total reversible process,
a) W reversible = useful work
b) irreversibility = 0
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: I = W in reversible – useful work = To * rate of entropy generation.

11. Two engines having same thermal efficiency but supplied with heat from source at different temperatures,
a) will convert same fraction of heat they receive into work
b) from second law, one will perform better than other
c) this can be considered as a deficiency of the first law
d) all of the mentioned
Answer: d
Clarification: This proves that we need second law efficiency.

12. Second law efficiency is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency.
a) true
b) false
Answer: a
Clarification: Second law efficiency = first law efficiency / efficiency of a reversible process.

13. Which of the following is true?
a) for work producing devices, second law efficiency = useful work / reversible work
b) for work absorbing devices, second law efficiency = reversible work / useful work
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: Both of the above statements are true and comes from the formula of second law efficiency.

14. In the context of first law efficiency and second law efficiency, which of the following statement is true?
a) first law efficiency = energy output / energy input and their difference is the energy loss
b) second law efficiency = exergy output / exergy input and their difference is irreversibility
c) by reducing energy loss, first law efficiency can be reduced and by reducing irreversibilities, second law efficiency can be reduced
d) all of the mentioned
Answer: d
Clarification: All of these statements just give a summary of what we have learned in first law efficiency and second law efficiency.

250+ TOP MCQs on Difference in Heat Capacities and their Ratio and Answers

Basic Thermodynamics questions and answers on “Difference in Heat Capacities and their Ratio”.

1. What do we get on equating the first and second TdS equations?
a) Cp-Cv = T*(∂T/∂p)*(∂V/∂T)
b) Cp-Cv = T*(∂p/∂T)*(∂V/∂T)
c) Cp+Cv = T*(∂p/∂T)*(∂V/∂T)
d) none of the mentioned
Answer: b
Clarification: This is the relation we get on equating first and second TdS equations.

2. Consider the equation Cp-Cv = -T*(∂p/∂V)(∂V/∂T)2 , which of the following is correct?
a) (∂V/∂T)2 is always positive
b) (∂p/∂V) for any substance is negative
c) (Cp-Cv) is always positive
d) all of the mentioned
Answer: d
Clarification: From this we can conclude that, Cp is always greater than Cv.

3. When do we have the condition Cp=Cv?
a) as T approaches 0K, Cp tends to approach Cv
b) when (∂V/∂T)=0, Cp=Cv
c) both of the mentioned are correct
d) none of the mentioned are correct
Answer: c
Clarification: These facts come from the equation Cp-Cv = -T*(∂p/∂V)(∂V/∂T)2.

4. For an ideal gas,
a) Cp-Cv = R
b) Cp-Cv = mR
c) Cp=Cv
d) all of the mentioned
Answer: b
Clarification: This comes from the ideal gas equation, pV=mRT.

5. The volume expansivity and isothermal compressibility is defined as
a) volume expansivity = (1/V)*(∂V/∂T) at p and isothermal compressibility = (-1/V)*(∂V/∂T) at T
b) volume expansivity = (1/V)*(∂V/∂T) at p and isothermal compressibility = (-1/V)*(∂V/∂T) at T
c) volume expansivity = (1/V)*(∂V/∂T) at p and isothermal compressibility = (-1/V)*(∂V/∂T) at T
d) none of the mentioned
Answer: a
Clarification: These two terms are used for better representation of the original equation.

6. The equation Cp-Cv = -T*(∂p/∂V)(∂V/∂T)2 can also be expressed as
a) Cp-Cv = T*V*(isothermal compressibility)2 / (volume expansivity)
b) Cp-Cv = T*V*(isothermal compressibility) / (volume expansivity)
c) Cp-Cv = T*V*(volume expansivity)2 / (isothermal compressibility)
d) Cp-Cv = T*V*(volume expansivity) / (isothermal compressibility)
Answer: c
Clarification: This comes from the equation Cp-Cv = -T*(∂p/∂V)(∂V/∂T)2 when we use volume expansivity and isothermal compressibility in it.

7. At constant entropy, the two TdS equations give us the relation
a) Cp+Cv = 0
b) Cp=Cv
c) Cp-Cv = mR
d) Cp/Cv = ɣ
Answer: d
Clarification: This relation is obtained on dividing the two TdS equations.

8. The slope of an isentrope is ____ the slope of an isotherm on p-v diagram.
a) less than
b) greater than
c) equal to
d) less than or equal to
Answer: b
Clarification: This comes from the fact that ɣ>1.

9. Work done in reversible and isothermal compression is ____ the work done in reversible and adiabatic compression.
a) equal to
b) greater than
c) less than
d) less than or equal to
Answer: c
Clarification: We get this from the p-v diagram for compression work in different reversible processes.

10. Isothermal compression requires minimum work.
a) true
b) false
Answer: a
Clarification: This is because work in isothermal is less than the work in adiabatic process and that of polytropic process lies in between these values.

11. Which of the following relation gives ɣ .
a) 1/(isothermal compressibility *adiabatic compressibility)
b) isothermal compressibility * adiabatic compressibility
c) adiabatic compressibility / isothermal compressibility
d) isothermal compressibility / adiabatic compressibility
Answer: d
Clarification: The adiabatic compressibility is defined as (-1/V)*(∂V/∂p).

To practice basic questions on all areas of Thermodynamics,

250+ TOP MCQs on Thermodynamics of Coupled Cycles and Answers

Thermodynamics Multiple Choice Questions on “Thermodynamics of Coupled Cycles”.

1. The efficiency of a binary cycle is given by(here E=efficiency)
a) E = (1-E1)(1-E2)
b) E = 1 – (1-E1)(1-E2)
c) E = 1 / (E1-1)(E2-1)
d) none of the mentioned
Answer: b
Clarification: Here E1 is the efficiency of topping cycle and E2 is the efficiency of bottoming cycle.

2. For n cycles, overall efficiency would be
a) E = (1-E1)(1-E2)(1-E3)……(1-En) – 1
b) E = 1 / (1-E1)(1-E2)(1-E3)……(1-En)
c) E = (1-E1)(1-E2)(1-E3)……(1-En)
d) E = 1 – (1-E1)(1-E2)(1-E3)……(1-En)
Answer: d
Clarification: Hence we can say total loss = product of losses in all cycles.

3. By combining two cycles in series, we can get high combined efficiency even if the individual efficiencies are low.
a) true
b) false
Answer: a
Clarification: Such a high efficiency cannot be achieved by a single cycle.

4. How can we generate required power and required quantity of steam in a single process?
a) by modifying initial steam pressure
b) by modifying exhaust pressure
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: We require modification in both the pressures.

5. In a back pressure turbine,
a) exhaust steam from turbine is used for process heating
b) the process heater replaces the condenser of Rankine cycle
c) exhaust pressure from turbine is the desired saturation pressure
d) all of the mentioned
Answer: d
Clarification: In a back pressure turbine we modify both initial steam pressure and exhaust pressure.

6. A cogeneration plant produces,
a) power
b) process heat
c) both power and process heat
d) none of the mentioned
Answer: c
Clarification: In cogeneration plant, both power and process heat are produced.

7. In a by-product power cycle,
a) the basic need is power produced and process steam is a by-product
b) the basic need is process steam and power produced is a by-product
c) both process steam and power is the basic need
d) both process steam and power is a by-product
Answer: b
Clarification: Here, power produced is a by-product and the basic need is process steam.

8. In a by-product power cycle, condenser losses is ____
a) high
b) low
c) zero
d) infinity
Answer: c
Clarification: In a normal steam plant, this is the biggest loss but in a by-product power cycle it is zero.

9. The fraction of energy utilized in a by-product power cycle is ____
a) very high
b) very low
c) zero
d) infinity
Answer: a
Clarification: The reason being the condenser losses is zero.

10. The power available from back pressure turbine through which the heating steam flows is very less.
a) true
b) false
Answer: a
Clarification: The reason being relatively high back pressure or maybe small heating requirement.