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Next exam on November 14 – 15, 2026:
All concepts

The Ideal Gas and the First Law of Thermodynamics

理想气体与热力学第一定律

The ideal gas equation of state and state-change problems, the microscopic meaning of pressure, kinetic theory and Brownian motion, and applications of the first law of thermodynamics.

7 real-exam questions · 4 sittings14 in mock papers124 practice

Formulas

  • Ideal gas equation of state

    Law
    p1V1T1=p2V2T2\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}

    Applies when:A fixed mass of ideal gas; T must be the absolute (kelvin) temperature

  • First law of thermodynamics

    Law
    ΔU=W+Q\Delta U=W+Q

    Applies when:W is positive when work is done on the gas, Q positive when heat is absorbed

How the real exam tests it

Questions from real CSCA papers that test this concept. Work each one out before you open the answer.

  1. CSCA Physics December 2025 Past Paper
    An ideal gas of a certain mass undergoes an isobaric process (constant pressure) from state A to state B. Given that the volume of state A is V1=1.5 m3V _ { 1 } = 1.5 \mathrm {~m} ^ { 3 } and its temperature is T1=300 KT _ { 1 } = 300 \mathrm {~K}, while the temperature of state B is T2=400 KT _ { 2 } = 400 \mathrm {~K}, the volume of state B is ( ).
    • A.1.5 m31.5 \mathrm {~m} ^ { 3 }
    • B.4.5m34.5 m ^ { 3 }
    • C.2m32 m ^ { 3 }
    • D.6m36 m ^ { 3 }
    Show answer & solution

    Answer: C

    For a fixed mass of ideal gas at constant pressure, V1/T1=V2/T2V_1/T_1 = V_2/T_2. Thus V2=1.5×(400/300)=2 m3V_2 = 1.5\times(400/300) = 2\,\mathrm{m^3} exactly. Option C is correct.
    See it in the full CSCA Physics December 2025 Past Paper
  2. CSCA Physics March 2026 Past Paper
    For an ideal gas at constant pressure, if volume increases, then
    • A.Temperature increases
    • B.Temperature decreases
    • C.Internal energy unchanged
    • D.Releases heat
    Show answer & solution

    Answer: A

    PV = nRT: P constant, V up → T up. Higher T = greater molecular kinetic energy = more internal energy (not C). Isobaric expansion: gas does work and absorbs heat (not D). Answer: A.
    Take the full CSCA Physics March 2026 Past Paper — timed & graded
  3. CSCA Physics April 2026 Past Paper
    A fixed mass of gas undergoes an isochoric (constant-volume) process. Its temperature rises from 27∘C27 ^ { \circ } \mathrm { C } to 54∘C54 ^ { \circ } \mathrm { C }. The pressure becomes ( ) times its original value.
    • A.1.09
    • B.1
    • C.2
    • D.1.5
    Show answer & solution

    Answer: A

    For an isochoric process, PT=\frac { P } { T } = constant ( TT in kelvin). T1=27+273=300 KT _ { 1 } = 27 + 273 = 300 \mathrm {~K}, T2=54+273=327 KT _ { 2 } = 54 + 273 = 327 \mathrm {~K} . P2P1=T2T1=327300=1.09\frac { P _ { 2 } } { P _ { 1 } } = \frac { T _ { 2 } } { T _ { 1 } } = \frac { 327 } { 300 } = 1.09.
    Take the full CSCA Physics April 2026 Past Paper — timed & graded

Real-exam evidence

Recorded past-paper appearances are listed below; online verification is labelled separately.

7 questions across 4 sittings1 of 4 sittings available to verify free online

14 questions across 6 mock papers — our own papers, so a coverage signal rather than real-exam evidence

124 related practice questions availablePractise free: Ideal Gas LawPractise free: Kinetic Theory of GasesPractise free: First Law of Thermodynamics

Sitting evidence comes from the CSCA past papers on record; questions whose original paper is missing are excluded, and sprint papers are deliberately not counted.