Thermodynamics MCQ With Answers for JKSSB Junior Engineer (Mechanical)

Technical · 2026-10-21 · 6 min read

Thermodynamics is the first subject in most mechanical engineering papers. Questions are mostly direct: a law, a cycle, a formula, or a short calculation with Carnot efficiency, constant-volume heating or the gas laws. This page gives a revision list and then solved MCQs with explanations.

Quick revision list

  • Laws: the zeroth law defines temperature, the first law is energy conservation, and the second law gives the direction of processes (Kelvin–Planck and Clausius statements)
  • Cycles: Carnot (two isothermal, two adiabatic), Otto (heat added at constant volume), Diesel (constant pressure), Rankine (steam) and Brayton (gas turbine)
  • Ideal gas: Cp − Cv = R, γ for air is about 1.4, and for a reversible adiabatic process pV^γ is constant
  • Steam: dryness fraction, the triple and critical points, and throttling (constant enthalpy)

Worked numerical problems

  • First law: a gas receives 500 kJ of heat and does 200 kJ of work. The change in internal energy is Q − W = 500 − 200 = 300 kJ
  • Ideal gas: the volume of 2 kg of air at 300 K and 1 bar (100,000 Pa), with R = 287 J/kg·K, is mRT/p = 2 × 287 × 300 ÷ 100,000 = 1.72 m³
  • Otto cycle: the efficiency is 1 − 1/r^(γ − 1). For a compression ratio of 8 and γ = 1.4, it is 1 − 8^(−0.4) = 1 − 0.435 = 56.5%
  • Refrigerator: a Carnot refrigerator working between 270 K and 300 K has COP = T_L ÷ (T_H − T_L) = 270 ÷ 30 = 9. As a heat pump the COP is 10
  • Entropy: when 600 kJ of heat is transferred reversibly at 300 K, the entropy change is Q ÷ T = 2 kJ/K
  • Wet steam: with h_f = 419 kJ/kg, h_fg = 2,257 kJ/kg and a dryness fraction of 0.9, the enthalpy is 419 + 0.9 × 2,257 = 2,450 kJ/kg

Process equations to remember

  • Constant volume: work is zero, and Q = m·Cv·ΔT
  • Constant pressure: work is p(V₂ − V₁), and Q = m·Cp·ΔT
  • Isothermal: pV is constant, the work is p₁V₁ ln(V₂/V₁), and ΔU is zero for an ideal gas
  • Adiabatic (reversible): pV^γ is constant, there is no heat transfer, and the work is (p₁V₁ − p₂V₂) ÷ (γ − 1)
  • Polytropic: pV^n is constant, and n takes a value between 0 and γ for most real processes

Sign conventions and common traps

Heat added to a system is positive, and work done by a system is positive. Many wrong answers come from mixing these signs. Temperatures in efficiency formulas must always be in kelvin, not Celsius. A Carnot efficiency cannot be 100%, and a heat engine's efficiency cannot exceed the Carnot efficiency between the same two temperatures. The COP of a refrigerator can be greater than 1, which is not a violation of any law, because it compares heat moved to work supplied.

How to revise this topic

  • Learn the two statements of the second law (Kelvin–Planck and Clausius) and what each forbids
  • Memorise the efficiency formulas for Carnot, Otto, Diesel and Rankine cycles
  • Practise ten numerical problems a week, and check your units every time
  • Draw the p–V and T–s diagrams of each cycle until you can sketch them without help

Questions candidates ask

Which cycles matter most? Carnot, Otto, Diesel, Rankine and Brayton, and the vapour compression refrigeration cycle.

Why must I use kelvin? Efficiency and gas law formulas depend on absolute temperature, so using Celsius gives wrong results.

What is the difference between a heat engine and a refrigerator? A heat engine converts heat into work, and a refrigerator uses work to move heat from a cold space to a warm one.

One more point to remember

Draw a table of the four basic processes with the work, heat and the relation between pressure and volume for each. Fill it in from memory every few days.

Practise

Solve the MCQs below, then try the full-length mocks and read the exam guide. More technical topics are on the JE Mechanical guide.

By Shivam · Written with AI assistance and fact-checked · Questions on this page are original · Editorial policy · Report an error

Practice questions

Q1. The SI unit of pressure is:

  1. Joule
  2. Pascal (N/m²)
  3. Watt
  4. Newton
Show answer

Pascal (N/m²) — One pascal is one newton per square metre.

Q2. The zeroth law of thermodynamics is the basis for the measurement of:

  1. Temperature
  2. Pressure
  3. Work
  4. Entropy
Show answer

Temperature — If two bodies are each in thermal equilibrium with a third, they are in equilibrium with each other, which lets us define temperature.

Q3. The first law of thermodynamics is a statement of the conservation of:

  1. Energy
  2. Momentum
  3. Mass only
  4. Entropy
Show answer

Energy — Energy can change form but cannot be created or destroyed.

Q4. The Kelvin–Planck statement of the second law says that it is impossible to build a heat engine that:

  1. Works without a boiler
  2. Produces work from a single cold body
  3. Runs without friction
  4. Converts all the heat it receives into work, in a cycle
Show answer

Converts all the heat it receives into work, in a cycle — A cyclic engine must reject some heat to a sink, so its efficiency is always below 100%.

Q5. A Carnot engine works between 600 K and 300 K. Its efficiency is:

  1. 75%
  2. 50%
  3. 100%
  4. 25%
Show answer

50% — Efficiency = 1 − T2/T1 = 1 − 300/600 = 0.5.

Q6. The Carnot cycle consists of:

  1. Two isothermal and two adiabatic processes
  2. Two isobaric and two isochoric processes
  3. Two adiabatic and two isobaric processes
  4. Two isothermal and two isochoric processes
Show answer

Two isothermal and two adiabatic processes — Heat is added and rejected isothermally, and the temperature changes adiabatically (isentropically).

Q7. In the ideal Otto cycle, heat is added at constant:

  1. Entropy
  2. Temperature
  3. Pressure
  4. Volume
Show answer

Volume — The Otto cycle is the ideal cycle for spark-ignition engines, with heat added at constant volume.

Q8. In the ideal Diesel cycle, heat is added at constant:

  1. Temperature
  2. Entropy
  3. Pressure
  4. Volume
Show answer

Pressure — The Diesel cycle adds heat at constant pressure.

Q9. For the same compression ratio, the thermal efficiency of the Otto cycle is:

  1. Zero
  2. Higher than that of the Diesel cycle
  3. Lower than that of the Diesel cycle
  4. Equal to that of the Diesel cycle
Show answer

Higher than that of the Diesel cycle — At equal compression ratio the Otto cycle is more efficient, but diesel engines run at much higher compression ratios.

Q10. The entropy of an isolated system during an irreversible process:

  1. Decreases
  2. Stays constant
  3. Increases
  4. Becomes zero
Show answer

Increases — Entropy of an isolated system never decreases.

Q11. The ratio of specific heats (γ) for air is about:

  1. 2.0
  2. 1.67
  3. 1.0
  4. 1.4
Show answer

1.4 — Air is mostly diatomic gas, for which γ is about 1.4.

Q12. For an ideal gas, Cp − Cv equals:

  1. γ
  2. The universal gas constant divided by γ
  3. The characteristic gas constant R
  4. Zero
Show answer

The characteristic gas constant R — Mayer's relation: Cp − Cv = R.

Q13. In a reversible isothermal process of an ideal gas, the change in internal energy is:

  1. Zero
  2. Equal to the heat added
  3. Equal to the work done
  4. Always positive
Show answer

Zero — Internal energy of an ideal gas depends only on temperature.

Q14. For a reversible adiabatic process of an ideal gas, which relation holds?

  1. p/T = constant
  2. pV = constant
  3. V/T = constant
  4. pV^γ = constant
Show answer

pV^γ = constant — The adiabatic relation is pV^γ = constant.

Q15. The work done by a gas in a constant-pressure process is:

  1. mCvΔT
  2. p(V2 − V1)
  3. Zero
  4. V(p2 − p1)
Show answer

p(V2 − V1) — Work at constant pressure equals pressure times the change in volume.

Q16. The triple point of water is at:

  1. 0 K
  2. 273.15 K
  3. 273.16 K
  4. 373.15 K
Show answer

273.16 K — At the triple point, solid, liquid and vapour exist together, at 273.16 K and 0.611 kPa.

Q17. At the critical point of a substance:

  1. The liquid and vapour phases cannot be told apart
  2. Only solid exists
  3. Only vapour exists
  4. Latent heat of vaporization is at its maximum
Show answer

The liquid and vapour phases cannot be told apart — At the critical point the latent heat of vaporization becomes zero and the two phases merge.

Q18. The coefficient of performance (COP) of a refrigerator is:

  1. Heat rejected divided by refrigerating effect
  2. Always less than 1
  3. Work input divided by refrigerating effect
  4. Refrigerating effect divided by work input
Show answer

Refrigerating effect divided by work input — COP = Q_L / W, the cooling obtained per unit of work.

Q19. For the same machine, the COP of a heat pump is related to the COP of a refrigerator as:

  1. COP(heat pump) = COP(refrigerator) − 1
  2. COP(heat pump) = COP(refrigerator) + 1
  3. COP(heat pump) = 1 / COP(refrigerator)
  4. They are equal
Show answer

COP(heat pump) = COP(refrigerator) + 1 — Q_H = Q_L + W, so dividing by W gives the relation.

Q20. The dryness fraction of wet steam is:

  1. Volume of steam divided by volume of water
  2. Mass of dry steam divided by total mass of the mixture
  3. Mass of water divided by total mass
  4. Zero for dry saturated steam
Show answer

Mass of dry steam divided by total mass of the mixture — Dry saturated steam has a dryness fraction of 1.

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