Thermodynamics — Class 11 MCQs with Answers

Thermodynamics — important questions & MCQs with answers (Class 11 Physics)

13 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 53 are timed and scored when you sign in.

1. Q1Easy

   Zeroth law of thermodynamics defines:

   A.Temperature✓ Correct

   B.Pressure

   C.Volume

   D.Entropy

   Answer: A. Temperature

   Explanation: Zeroth law: if A is in thermal equilibrium with C and B is also in equilibrium with C, then A and B share the same value of temperature — this is what the law defines, not pressure or entropy.

2. Q2Easy

   First law of thermodynamics: ΔU = ?

   A.W − Q

   B.Q + W

   C.Q − W (heat added − work done by system)✓ Correct

   D.Cannot determine

   Answer: C. Q − W (heat added − work done by system)

   Explanation: First law of thermodynamics: ΔU = Q − W, heat added to the system minus work done BY the system — confusing this with W − Q reverses the sign of the work term, the usual slip.

3. Q3Easy

   Isothermal process: T = ?

   A.Constant✓ Correct

   B.Variable

   C.Zero

   D.Maximum

   Answer: A. Constant

   Explanation: Isothermal process: temperature T stays constant throughout, e.g. a gas expanding slowly in contact with a 300 K reservoir stays at 300 K — heat and work still flow even though T doesn't change.

4. Q4Easy

   Second law of thermodynamics (Clausius statement):

   A.Variable

   B.No restriction

   C.Always conserved

   D.Heat cannot flow from cold to hot without external work✓ Correct

   Answer: D. Heat cannot flow from cold to hot without external work

   Explanation: Clausius statement of the second law: heat cannot flow spontaneously from a colder body to a hotter one — moving it that way (as in a refrigerator) always needs external work input.

5. Q5Easy

   Heat engine converts:

   A.Mass into energy

   B.Work into heat

   C.Heat into work (partial)✓ Correct

   D.Cannot define

   Answer: C. Heat into work (partial)

   Explanation: A heat engine converts only part of the absorbed heat Q_h into useful work W, rejecting the rest as Q_c to a cold reservoir — full conversion (100%) is forbidden by the second law.

6. Q6Medium

   In an isothermal process, ΔT = 0, so for ideal gas ΔU =

   A.Q

   B.0✓ Correct

   C.W

   D.PV

   Answer: B. 0

   Explanation: Isothermal process keeps ΔT = 0, and for an ideal gas internal energy depends only on temperature, so ΔU = 0 even though heat Q and work W are both nonzero — don't assume ΔU equals Q or W.

7. Q7Medium

   Gas absorbs 1500 J heat and does 700 J work. ΔU:

   A.2200 J

   B.800 J✓ Correct

   C.700 J

   D.1500 J

   Answer: B. 800 J

   Explanation: First law: ΔU = Q − W = 1500 J − 700 J = 800 J for a gas absorbing 1500 J of heat and doing 700 J of work — adding Q and W instead (2200 J) is the tempting but wrong shortcut.

8. Q8Medium

   Ideal gas in isothermal process. Work done in expansion is:

   A.Positive (gas does work)✓ Correct

   B.Negative

   C.Zero

   D.Cannot say

   Answer: A. Positive (gas does work)

   Explanation: Isothermal expansion: as volume increases the gas pushes outward on the surroundings, doing positive work (W > 0) even though ΔU = 0 — the heat absorbed exactly equals this work, Q = W.

9. Q9Medium

   Refrigerator (heat pump) takes heat from cold and rejects to hot. This requires:

   A.Variable

   B.No work

   C.Heat only

   D.Work input✓ Correct

   Answer: D. Work input

   Explanation: Second law: heat naturally flows hot to cold, so moving it from cold to hot in a refrigerator is non-spontaneous and needs external work input — assuming it runs on heat alone violates Clausius's statement.

10. Q10Medium

    Engine absorbs 500 J and delivers 200 J work. Efficiency:

    A.40%✓ Correct

    B.30%

    C.20%

    D.100%

    Answer: A. 40%

    Explanation: Heat engine efficiency η = W/Q_h = 200 J / 500 J = 0.4 = 40%, dividing work output by heat input — mixing up numerator and denominator would wrongly give 250%, an impossible result.

11. Q11Hard

    Second law (Kelvin–Planck) says:

    A.Energy disappears

    B.Heat flows from cold to hot spontaneously

    C.Entropy decreases

    D.No engine can convert all heat to work in a cyclic process✓ Correct

    Answer: D. No engine can convert all heat to work in a cyclic process

    Explanation: Kelvin–Planck statement: no cyclic engine can convert all absorbed heat into work — some heat must always be rejected to a cold reservoir, so a claimed 100% efficient cyclic engine is impossible.

12. Q12Hard

    For isothermal expansion of ideal gas, Q = W. This is because:

    A.Q is always W

    B.ΔU = 0 for isothermal process✓ Correct

    C.Energy conserved

    D.Cannot relate

    Answer: B. ΔU = 0 for isothermal process

    Explanation: Isothermal process: since ideal-gas internal energy depends only on T and ΔT = 0, ΔU = 0; the first law ΔU = Q − W then forces Q = W — it follows from ΔU = 0, it isn't just an assumed equality.

13. Q13Hard

    In cyclic process, total work done equals:

    A.Always negative

    B.Always positive

    C.Total heat absorbed (Σ Q)✓ Correct

    D.Variable

    Answer: C. Total heat absorbed (Σ Q)

    Explanation: Over a complete cycle ΔU = 0 since internal energy is a state function, so by the first law the total work done equals the net heat absorbed, ΣW = ΣQ — not a fixed sign or an unrelated quantity.

Thermodynamics — FAQs

More Class 11 Physics chapters

• 01Physical World
• 02Units and Measurements
• 03Motion in a Straight Line
• 04Motion in a Plane
• 05Laws of Motion
• 06Work, Energy and Power

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• Class 11 Biology chapters
• Class 11 Chemistry chapters
• Class 11 Computer Science chapters
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• Class 11 Mathematics chapters