The Ideal Gas and the First Law of Thermodynamics

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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 $V _ { 1 } = 1.5 \mathrm {~m} ^ { 3 }$ and its temperature is $T _ { 1 } = 300 \mathrm {~K}$, while the temperature of state B is $T _ { 2 } = 400 \mathrm {~K}$, the volume of state B is ( ).
   • A．$1.5 \mathrm {~m} ^ { 3 }$
   • B．$4.5 m ^ { 3 }$
   • C．$2 m ^ { 3 }$
   • D．$6 m ^ { 3 }$
   정답 및 풀이 보기

   정답: C

   For a fixed mass of ideal gas at constant pressure, $V_1/T_1 = V_2/T_2$. Thus $V_2 = 1.5\times(400/300) = 2\,\mathrm{m^3}$ exactly. Option C is correct.
   CSCA Physics December 2025 Past Paper 전체에서 보기
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
   정답 및 풀이 보기

   정답: 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.
   CSCA Physics March 2026 Past Paper 전체 풀기 — 시간 측정 및 채점
3. CSCA Physics April 2026 Past Paper
   A fixed mass of gas undergoes an isochoric (constant-volume) process. Its temperature rises from $27 ^ { \circ } \mathrm { C }$ to $54 ^ { \circ } \mathrm { C }$. The pressure becomes ( ) times its original value.
   • A．1.09
   • B．1
   • C．2
   • D．1.5
   정답 및 풀이 보기

   정답: A

   For an isochoric process, $\frac { P } { T } =$ constant ( $T$ in kelvin). $T _ { 1 } = 27 + 273 = 300 \mathrm {~K}$, $T _ { 2 } = 54 + 273 = 327 \mathrm {~K}$. $\frac { P _ { 2 } } { P _ { 1 } } = \frac { T _ { 2 } } { T _ { 1 } } = \frac { 327 } { 300 } = 1.09$.
   CSCA Physics April 2026 Past Paper 전체 풀기 — 시간 측정 및 채점