68 free MCQs on Thermodynamics with worked answers and explanations. Laws of thermodynamics, heat engines, entropy, and gas processes.
Take the timed Thermodynamics chapterwise test →Below are 68 practice questions on Thermodynamics, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Thermodynamics notes.
The Carnot cycle alternates two isothermal steps (heat absorbed at TH, released at TC) with two adiabatic steps (no heat exchange); the enclosed loop area equals the net work output, and η = 1−TC/TH is the maximum possible efficiency between those two temperatures.
Q1.
The SI unit of temperature is:
Answer: C. Kelvin
Why: The SI unit of temperature is Kelvin (K). Absolute zero = 0 K = -273.15 degrees Celsius.
Q2.
Conversion formula from Celsius to Kelvin is:
Answer: B. K = C + 273
Why: T(K) = T(C) + 273.15. For most purposes T(K) = T(C) + 273.
Q3.
The first law of thermodynamics is basically:
Answer: B. Conservation of energy
Why: The first law is conservation of energy: delta U = Q - W (internal energy change = heat added - work done by system).
Q4.
During an isothermal process:
Answer: B. Temperature is constant
Why: Isothermal = constant temperature. For ideal gas: PV = constant (Boyle's law applies).
Q5.
In an adiabatic process:
Answer: C. No heat is exchanged
Why: Adiabatic process: Q = 0 (no heat exchange with surroundings). First law: delta U = -W.
Q6.
Ideal gas equation is:
Answer: A. PV = nRT
Why: PV = nRT is the ideal gas equation. n = number of moles, R = 8.314 J/(mol K).
Q7.
Zeroth law of thermodynamics defines:
Answer: B. Temperature scale
Why: Zeroth law: if A is in equilibrium with B, and B with C, then A and C are in equilibrium. This defines temperature.
Q8.
A Carnot engine operates between temperatures 300 K and 600 K. Its efficiency is:
Answer: B. 50%
Why: Carnot efficiency = 1 - T<sub>cold</sub>/T<sub>hot</sub> = 1 - 300/600 = 1 - 0.5 = 50%.
Q9.
The second law of thermodynamics states that entropy of an isolated system:
Answer: C. Always increases or stays constant
Why: Second law: entropy of an isolated system always increases or stays constant. Spontaneous processes increase total entropy.
Q10.
Boyle's law states that at constant temperature:
Answer: B. P is proportional to 1/V
Why: Boyle's law: P proportional to 1/V (at constant T). Or PV = constant.
Q11.
Q = mcT formula is used to calculate:
Answer: C. Sensible heat exchange
Why: Q = mc delta T calculates heat exchanged during temperature change. c is specific heat capacity.
Q12.
Specific heat of water is approximately:
Answer: C. 4200 J/(kg K)
Why: Specific heat of water = 4200 J/(kg K) = 4.2 J/(g K). Water has an unusually high specific heat.
Q13.
A refrigerator works on which principle?
Answer: B. Moves heat from cold to hot body using work input
Why: Refrigerator: uses work input to transfer heat from cold body (inside) to hot body (surroundings). It is a heat pump.
Q14.
Internal energy of an ideal gas depends only on:
Answer: C. Temperature
Why: For an ideal gas, internal energy depends only on temperature (kinetic energy of molecules).
Q15.
Charles law states that at constant pressure:
Answer: B. V/T = constant
Why: Charles law: volume is proportional to absolute temperature at constant pressure. V/T = constant.
Q16.
Latent heat is the heat exchanged during:
Answer: B. Phase change
Why: Latent heat is exchanged during phase change (e.g., melting or boiling) with NO temperature change.
Q17.
Thermal efficiency of a heat engine is defined as:
Answer: A. Work output / Heat input
Why: Efficiency = W/Q<sub>in</sub> = work output / heat absorbed from hot source.
Q18.
In an isochoric (constant volume) process, work done by the gas is:
Answer: C. Zero
Why: W = P delta V. At constant volume, delta V = 0, so W = 0. All heat added goes to internal energy.
Q19.
Avogadro's number is:
Answer: A. 6.022 x 10<sup>23</sup>
Why: Avogadro's number N<sub>A</sub> = 6.022 x 10<sup>23</sup> mol<sup>-1</sup> (particles per mole).
Q20.
Heat flows spontaneously from:
Answer: B. Hot to cold body
Why: Second law: heat flows spontaneously from higher temperature (hot) to lower temperature (cold).
Q21.
1 mole of ideal gas at 27 degrees C has internal energy (Cv = 3R/2 for monatomic):
Answer: C. 3741 J
Why: U = n Cv T = 1 x (3/2 x 8.314) x 300 = 3741 J. (T = 27+273 = 300 K)
Q22.
An ideal gas expands isothermally. Work done by gas:
Answer: B. Is positive
Why: Isothermal expansion: gas does positive work on surroundings (W > 0). Temperature stays constant so delta U = 0, Q = W.
Q23.
A Carnot engine has 60% efficiency. If cold reservoir is at 300 K, hot reservoir temperature:
Answer: C. 750 K
Why: eta = 1 - T<sub>cold</sub>/T<sub>hot</sub>. 0.6 = 1 - 300/T<sub>hot</sub>. 300/T<sub>hot</sub> = 0.4. T<sub>hot</sub> = 750 K.
Q24.
For a gas expanding adiabatically, gamma = Cp/Cv = 5/3. If pressure halves, volume changes by factor:
Answer: C. 2<sup>5/3</sup>
Why: PV<sup>gamma</sup> = constant. P<sub>1</sub> V<sub>1</sub><sup>5/3</sup> = P<sub>2</sub> V<sub>2</sub><sup>5/3</sup>. V<sub>2</sub><sup>5/3</sup> = (P<sub>1</sub>/P<sub>2</sub>) V<sub>1</sub><sup>5/3</sup> = 2 V<sub>1</sub><sup>5/3</sup>. V<sub>2</sub> = 2<sup>3/5</sup> V<sub>1</sub>. Answer: 2<sup>3/5</sup>.
Q25.
Entropy change when 100 J of heat is added to a system at 400 K:
Answer: B. 0.25 J/K
Why: delta S = Q<sub>rev</sub> / T = 100/400 = 0.25 J/K.
Q26.
Which process has the steepest slope on a P-V diagram?
Answer: B. Adiabatic
Why: Adiabatic process: PV<sup>gamma</sup> = constant. Since gamma > 1, adiabatic curve is steeper than isothermal on P-V diagram.
Q27.
1 kg of water at 100 degrees C converts to steam. Latent heat of vaporization = 2.26 x 10<sup>6</sup> J/kg. Heat required:
Answer: D. 2260000 J
Why: Q = m x L = 1 x 2.26 x 10<sup>6</sup> = 2.26 x 10<sup>6</sup> J = 2,260,000 J.
Q28.
Heat engine takes 1000 J from hot source, does 400 J of work. Heat rejected to cold reservoir:
Answer: C. 600 J
Why: Energy conservation: Q<sub>hot</sub> = W + Q<sub>cold</sub>. Q<sub>cold</sub> = 1000 - 400 = 600 J.
Q29.
Mean kinetic energy of a gas molecule at temperature T:
Answer: C. (3/2)kT
Why: Average KE per molecule = (3/2)kT for a monatomic ideal gas. k = 1.38 x 10<sup>-23</sup> J/K (Boltzmann constant).
Q30.
RMS speed of gas molecules is proportional to:
Answer: B. sqrt(T)
Why: v<sub>rms</sub> = sqrt(3RT/M) or sqrt(3kT/m). RMS speed is proportional to sqrt(T) (absolute temperature).
Q31.
For an ideal gas, relation between Cp and Cv is:
Answer: A. Cp = Cv + R
Why: Mayer's relation: Cp - Cv = R (for 1 mole of ideal gas). R = 8.314 J/(mol K).
Q32.
A process in which PV<sup>n</sup> = constant is called:
Answer: C. Polytropic
Why: Polytropic process: PV<sup>n</sup> = constant. Special cases: isothermal (n=1), adiabatic (n=gamma), isobaric (n=0), isochoric (n=infinity).
Q33.
Work done by an ideal gas in an isobaric expansion from V<sub>1</sub> to V<sub>2</sub> at pressure P:
Answer: A. P(V<sub>2</sub>-V<sub>1</sub>)
Why: W = P x delta V = P(V<sub>2</sub> - V<sub>1</sub>) for isobaric (constant pressure) process.
Q34.
Coefficient of performance (COP) of a refrigerator is:
Answer: B. Q<sub>cold</sub>/W
Why: COP = Q<sub>cold</sub>/W (heat removed from cold reservoir per unit work done). A good refrigerator has high COP.
Q35.
At absolute zero (0 K), all molecular motion:
Answer: B. Stops completely
Why: At absolute zero, all thermal motion stops. This is the lowest possible temperature (third law of thermodynamics).
Q36.
Cp for monatomic ideal gas (in terms of R):
Answer: C. 5R/2
Why: Monatomic: Cv = 3R/2. Cp = Cv + R = 3R/2 + R = 5R/2.
Q37.
Ideal gas at constant temperature has pressure 4 atm and volume 2 L. If pressure becomes 2 atm, volume:
Answer: C. 4 L
Why: PV = constant. P<sub>1</sub>V<sub>1</sub> = P<sub>2</sub>V<sub>2</sub>. 4 x 2 = 2 x V<sub>2</sub>. V<sub>2</sub> = 4 L.
Q38.
In a reversible process, entropy of the universe:
Answer: C. Remains constant
Why: Reversible process: entropy of universe remains constant (delta S<sub>universe</sub> = 0). Irreversible processes increase it.
Q39.
Speed of sound in a gas depends on:
Answer: C. gamma and T
Why: v = sqrt(gamma P/rho) = sqrt(gamma RT/M). Speed depends on gamma (ratio of specific heats) and temperature.
Q40.
If temperature of a gas doubles (at constant pressure), its volume:
Answer: C. Doubles
Why: Charles law: V/T = constant. If T doubles (in Kelvin), V doubles at constant pressure.
Q41.
One mole of ideal gas undergoes cycle: isothermal expansion at T<sub>1</sub>, then cooling to T<sub>2</sub>, then isothermal compression, then heating. This is a:
Answer: A. Carnot cycle
Why: Carnot cycle consists of two isothermal processes (at T<sub>1</sub> and T<sub>2</sub>) and two adiabatic processes.
Q42.
1 mole of diatomic gas (Cv = 5R/2) undergoes isochoric heating from 300K to 600K. Heat added:
Answer: C. 6235 J
Why: Q = n Cv delta T = 1 x (5/2 x 8.314) x 300 = 5/2 x 8.314 x 300 = 6235.5 J.
Q43.
Carnot efficiency = 80%. If cold reservoir is at 400 K, hot reservoir is at:
Answer: B. 2000 K
Why: eta = 1 - T<sub>c</sub>/T<sub>h</sub>. 0.8 = 1 - 400/T<sub>h</sub>. 400/T<sub>h</sub> = 0.2. T<sub>h</sub> = 2000 K.
Q44.
An ideal gas undergoes free expansion into vacuum (Q=0, W=0). Temperature change:
Answer: C. Remains same
Why: In free expansion, no work done (W=0) and no heat exchange (Q=0). So delta U = 0. For ideal gas, U depends on T only, so T stays constant.
Q45.
Work done during isothermal expansion of n moles from V<sub>1</sub> to V<sub>2</sub> at temperature T:
Answer: A. nRT ln(V<sub>2</sub>/V<sub>1</sub>)
Why: W = nRT ln(V<sub>2</sub>/V<sub>1</sub>) for isothermal expansion of ideal gas.
Q46.
Entropy change when 0.5 kg of ice melts at 0 degrees C. Latent heat = 336 kJ/kg:
Answer: C. 616 J/K
Why: Q = mL = 0.5 x 336000 = 168000 J. T = 273 K. delta S = Q/T = 168000/273 = 615.4 J/K approximately 616 J/K.
Q47.
For 2 moles of monatomic ideal gas, ratio of work done to heat exchanged in isothermal expansion:
Answer: B. 1
Why: In isothermal process for ideal gas: delta U = 0. Q = W. Ratio W/Q = 1.
Q48.
COP of a Carnot refrigerator operating between 200 K and 300 K:
Answer: B. 2
Why: COP = T<sub>cold</sub>/(T<sub>hot</sub> - T<sub>cold</sub>) = 200/(300-200) = 200/100 = 2.
Q49.
Mean free path of gas molecules increases when:
Answer: D. Pressure decreases or temperature increases
Why: Mean free path lambda = kT/(sqrt(2) pi d<sup>2</sup> P). Increases when T increases or P decreases.
Q50.
A heat engine performs 1000 J of work and rejects 1500 J to the cold reservoir. Efficiency:
Answer: C. 40%
Why: Q<sub>hot</sub> = W + Q<sub>cold</sub> = 1000 + 1500 = 2500 J. eta = W/Q<sub>hot</sub> = 1000/2500 = 0.4 = 40%.
Q51.
Maxwell-Boltzmann distribution of molecular speeds: most probable speed v<sub>p</sub> compared to rms speed v<sub>rms</sub>:
Answer: B. v<sub>p</sub> < v<sub>rms</sub>
Why: v<sub>p</sub> = sqrt(2RT/M), v<sub>rms</sub> = sqrt(3RT/M). Ratio v<sub>p</sub>/v<sub>rms</sub> = sqrt(2/3) < 1. So v<sub>p</sub> < v<sub>rms</sub>.
Q52.
In an Otto cycle, heat is added at:
Answer: B. Constant volume
Why: Otto cycle (gasoline engine): heat is added and rejected at constant volume. Compression and expansion are adiabatic.
Q53.
For Van der Waals gas (a, b are constants): (P + an<sup>2</sup>/V<sup>2</sup>)(V - nb) = nRT. At high pressure compared to ideal gas, the gas:
Answer: A. Is less compressible (harder to compress)
Why: At high pressures, the (V-nb) term dominates. The effective volume is reduced by b (molecular volume), making it less compressible.
Q54.
Triple point of water is at:
Answer: B. 273.16 K and 611.7 Pa
Why: Triple point of water: T = 273.16 K (0.01 degrees C), P = 611.7 Pa. All three phases coexist.
Q55.
Joule-Thomson effect: when a real gas expands through a porous plug at constant enthalpy, temperature:
Answer: C. Can increase or decrease depending on inversion temperature
Why: Below the inversion temperature, gas cools on expansion (most gases at room T). Above inversion T, gas heats up. Temperature change depends on inversion temperature.
Q56.
Dulong-Petit law states that molar heat capacity of most metals at room temperature is:
Answer: C. 3R
Why: Dulong-Petit law: Cv = 3R per mole for most solid elements at room temperature (6 degrees of freedom, each contributing R/2).
Q57.
Entropy can be expressed as S = k ln(omega) where omega is:
Answer: B. Number of microstates
Why: Boltzmann's entropy formula: S = k ln(omega), where omega is the number of microscopic states (microstates) for the macrostate.
Q58.
A gas expands adiabatically from (P<sub>1</sub>, V<sub>1</sub>) to (P<sub>2</sub>, V<sub>2</sub>). Work done by gas:
Answer: D. Both B and C
Why: W = -delta U = nCv(T<sub>1</sub>-T<sub>2</sub>) = (P<sub>1</sub>V<sub>1</sub>-P<sub>2</sub>V<sub>2</sub>)/(gamma-1). These are equivalent. Both B and C are correct.
Q59.
If a system undergoes a cyclic process, net change in internal energy is:
Answer: D. Zero
Why: After a complete cycle, system returns to initial state. State function U depends only on state, so delta U = 0 for any cycle.
Q60.
Clausius inequality states: for any process, cyclic integral of dQ/T:
Answer: D. Is <= 0
Why: Clausius inequality: contour integral of dQ/T <= 0. Equality holds for reversible processes; strict inequality for irreversible.
Q61.
A Carnot engine operates between a source at 500 K and a sink at 300 K. Its efficiency is:
Answer: C. 40%
Why: η = 1 − T<sub>sink</sub>/T<sub>source</sub> = 1 − 300/500 = 0.4 = 40%.
Q62.
One mole of an ideal gas (γ = 1.4) is adiabatically compressed to half its volume. The ratio of final to initial temperature is approximately:
Answer: B. 1.32
Why: T₂/T₁ = (V₁/V₂)<sup>γ−1</sup> = 2<sup>0.4</sup> ≈ 1.32.
Q63.
One mole of an ideal gas expands isothermally from V to 2V at 300 K (R = 8.31 J/mol·K). The work done by the gas is about:
Answer: B. 1.73 kJ
Why: W = nRT ln2 = 1·8.31·300·0.693 ≈ 1730 J ≈ 1.73 kJ.
Q64.
A gas absorbs 200 J of heat and does 150 J of work on its surroundings. The change in its internal energy is:
Answer: A. 50 J
Why: ΔU = Q − W = 200 − 150 = 50 J.
Q65.
In a complete cyclic process a gas absorbs 100 J of heat. The net work done by the gas over the cycle is:
Answer: C. 100 J
Why: Over a cycle ΔU = 0, so by the first law Q = W = 100 J.
Q66.
For a diatomic ideal gas (rigid molecules) with Cv = 5R/2, the ratio γ = Cp/Cv is:
Answer: C. 1.4
Why: Cp = Cv + R = 7R/2, so γ = (7R/2)/(5R/2) = 7/5 = 1.4.
Q67.
An ideal gas undergoes free expansion into a vacuum inside an insulated container. Its temperature:
Answer: C. remains unchanged
Why: No work is done and no heat exchanged, so ΔU = 0 and the temperature of an ideal gas stays constant.
Q68.
A refrigerator with coefficient of performance 4 extracts 400 J of heat from the cold reservoir per cycle. The work input required is:
Answer: B. 100 J
Why: COP = Q<sub>cold</sub>/W → W = 400/4 = 100 J.