Free Thermodynamics MCQs with Answers
714 Thermodynamics MCQs from Physics, each with the correct answer and a written explanation of why it is correct. Free and unlimited, with no account needed.
Thermal equilibrium and heat explain how temperature and energy transfer are related, while molar specific heats describe the heat required by a gas at constant volume or constant pressure. The first law connects heat supplied, work done and change in internal energy, with sign conventions kept distinct from temperature changes.
Last updated
Read the Thermodynamics notesFree MDCAT chapter notes with key terms714 questions · page 6 of 36
- A. Work done by the system = increase in internal energy of the system
- B. Work done on the system = increase in internal energy of the system
- C. Work done on the system = decrease in internal energy of the system
- D. Work done on the system = decrease in internal energy of the system + heat released
Explanation: In adiabatic process, Q=0, so ΔU = -W. This means that if work is done on the system (W is negative), internal energy increases (ΔU is…
Correct answer: Work done on the system = decrease in internal energy of the system- A. Pressure
- B. Volume
- C. Temperature
- D. All of these
Explanation: For an ideal gas, there are no intermolecular forces, so its internal energy consists only of the kinetic energy of its molecules.
Correct answer: Temperature- A. 8200 J
- B. 5600 J
- C. 7900 J
- D. 6400 J
Explanation: First, convert the heat absorbed to Joules: Q = 2 kcal × 4200 J/kcal = 8400 J (using 1 cal ≈ 4.2 J).
Correct answer: 7900 J- A. Temperature increases
- B. Mechanical energy comes into the system
- C. ΔQ = Zero
- D. All of these
Explanation: During an adiabatic contraction, work is done on the system, which is an input of mechanical energy.
Correct answer: All of these- A. Pressure is constant
- B. Volume is constant
- C. Temperature is constant
- D. Entropy is constant
Explanation: In isochoric process, volume is constant (ΔV = 0), work done (W=PΔV) is zero, regardless change in pressure or temp.
Correct answer: Volume is constant- A. nRT ln(V₂/V₁)
- B. nRT (V₂/V₁)
- C. nR ln(T₂/T₁)
- D. nRT (T₂/T₁)
Explanation: For an isothermal process, the work done by an ideal gas is W = nRT ln(V₂/V₁), where n is moles, R is the gas constant, T is constant…
Correct answer: nRT ln(V₂/V₁)- A. Zero
- B. Positive
- C. Negative
- D. Equal to heat supplied
Explanation: In an adiabatic expansion, the gas does work on the surroundings, so work done by the gas is positive, and its internal energy decreases…
Correct answer: Positive- A. Pressure only
- B. Volume only
- C. Temperature only
- D. Pressure and volume
Explanation: For an ideal gas, internal energy is a function of temperature only, as U = (f/2)nRT, where f is degrees of freedom, n is moles, R is the…
Correct answer: Temperature only- A. Positive
- B. Negative
- C. Zero
- D. Equal to work done
Explanation: In isothermal process, temperature of ideal gas remains constant, its internal energy, depend only on temperature (U =(f/2)nRT), does not…
Correct answer: Zero- A. Change in internal energy
- B. Work done
- C. Change in pressure
- D. Change in temperature
Explanation: In an isochoric process, volume is constant, so no work is done (W = PΔV = 0).
Correct answer: Change in internal energy- A. PV = constant
- B. P/V = constant
- C. PVᵞ = constant
- D. PᵞV = constant
Explanation: In an adiabatic process, the relation for an ideal gas is PV^γ = constant, where γ is the ratio of specific heats (Cp/Cv), distinguishing…
Correct answer: PVᵞ = constant- A. 200 J
- B. 800 J
- C. -200 J
- D. 500 J
Explanation: By the first law of thermodynamics, ΔU = Q - W, where Q = 500 J (heat absorbed) and W = 300 J (work done by the system).
Correct answer: 200 J- A. Heat is added to the system
- B. Work is done on the gas
- C. Volume remains constant
- D. Pressure decreases
Explanation: In adiabatic compression, no heat is exchanged (Q = 0), and work is done on the gas, increasing its internal energy (ΔU = - W), which…
Correct answer: Work is done on the gas- A. TVᵞ = constant
- B. TᵞV = constant
- C. TV⁽ᵞ⁻¹⁾ = constant
- D. T/V = constant
Explanation: For an adiabatic process, PV^γ = constant. Using the ideal gas law (PV = nRT), this becomes TV^(γ-1) = constant, where γ is the ratio of…
Correct answer: TV⁽ᵞ⁻¹⁾ = constant- A. UII > ΔUI
- B. UI = ΔUII
- C. UII < ΔUI
- D. Relation between ΔUI and ΔUII cannot be determined
Explanation: As internal energy is a state function therefore change in internal energy does not depends upon the path followed i.e. ΔUI=Δ UII
Correct answer: UI = ΔUII- A. Q = U
- B. U = W
- C. W = U/W
- D. Q = W
Explanation: Since it is an isothermal process, there is no change in temperature hence the internal energy is constant.
Correct answer: Q = W- A. Isothermal process
- B. Adiabatic compression
- C. Isobaric expansion
- D. Isochoric cooling
Explanation: In adiabatic compression, work done on gas with no heat exchange (Q=0), increasing its internal energy and its temperature.
Correct answer: Adiabatic compression- A. Zero
- B. Equal to net heat transfer
- C. Equal to internal energy change
- D. Always positive
Explanation: In a cyclic process, the system returns to its initial state, so ΔU = 0.
Correct answer: Equal to net heat transfer- A. nRΔT
- B. (P₁V₁ - P₂V₂)/(γ - 1)
- C. nRT ln(V₂/V₁)
- D. PΔV
Explanation: For an adiabatic process, work done is W = (P₁V₁ - P₂V₂)/(γ - 1), derived from the pressure-volume relationship and the first law, where γ…
Correct answer: (P₁V₁ - P₂V₂)/(γ - 1)- A. The temperature of the cold reservoir increases
- B. The temperature of the hot reservoir decreases
- C. The temperature difference between reservoirs increases
- D. The pressure of the gas increases
Explanation: Carnot efficiency is η = 1 - (Tcold/Thot). Increasing the temperature difference (higher Thot or lower Tcold) increases efficiency.
Correct answer: The temperature difference between reservoirs increases