3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
A gas mixture contains 1 mol of helium (monatomic) and 1 mol of nitrogen (diatomic, rigid molecules, no vibration), both ideal, at pressure 1.0×105 Pa. The pressure of the mixture is increased suddenly by 0.30%; the compression is fast enough to be adiabatic and slow enough to be quasi-static (reversible). The fractional decrease in the volume of the mixture is ______. Exactly one option is correct.
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Answer: B
Molar heat capacities of the mixture: CV=21⋅23R+1⋅25R=2R, CP=CV+R=3R, so γmix=3/2.\nFor a reversible adiabatic change PVγ= const, so the adiabatic bulk modulus is Bad=−VdVdP=γP.\nHence VΔV=γPΔP=1.50.30%=0.20%.\nTraps: 0.30% uses the isothermal modulus B=P; 0.18% uses γ=5/3 (helium alone); 0.21% uses γ=1.4 (nitrogen alone). For a mixture, γ is NOT the average of the two γ values; average the heat capacities instead.\nDifficulty: HardQuestion 2Level 2 · ExamSingle correct
One mole of a monatomic ideal gas is in a thermally insulated cylinder fitted with a massless, frictionless, insulating piston. Initially the gas is in equilibrium at temperature T0 and pressure P0 (equal to the outside pressure). The outside pressure on the piston is suddenly raised to 2P0 and then held constant until the gas reaches a new equilibrium. The final temperature of the gas is ______. Exactly one option is correct.
A1.40T0
B22/5T0≈1.32T0
C1.25T0
D1.60T0
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Answer: A
Sign convention: Q is heat supplied to the gas, W is work done by the gas, ΔU=Q−W.\nNo heat flows (Q=0). The work done on the gas by the constant outside pressure is Pext(V0−Vf)=2P0(V0−Vf), so the first law gives 23R(Tf−T0)=2P0V0−2P0Vf.\nInitial state: P0V0=RT0, so 2P0V0=2RT0. Final equilibrium pressure is 2P0, so 2P0Vf=RTf.\nThus 23Tf−23T0=2T0−Tf⇒25Tf=27T0⇒Tf=57T0=1.40T0.\nTrap: 22/5T0 is the reversible adiabatic result (TP(1−γ)/γ= const), which does not apply to a sudden, irreversible compression. The irreversible process ends hotter because more work is done on the gas for the same final pressure.\nDifficulty: HardQuestion 3Level 3 · AdvancedSingle correct
Two moles of a diatomic ideal gas (rigid molecules, CV=25R) are taken quasi-statically from state A (3P0, V0) to state B (P0, 3V0) along the straight line joining A and B on the P-V diagram (pressure decreases linearly as volume increases). During this expansion the gas first absorbs heat and then rejects heat. The volume at which it stops absorbing heat and starts rejecting heat is ______. Exactly one option is correct.
A2V0
B49V0
C25V0
D37V0
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Answer: D
Sign convention: Q is heat supplied to the gas, W is work done by the gas, ΔU=Q−W.\nWrite v=V/V0. The line is P=P0(4−v), so nRT=PV=P0V0(4v−v2).\nFor a small step: dQ=nCVdT+PdV=25d(PV)+PdV=P0V0[25(4−2v)+(4−v)]dv=P0V0(14−6v)dv.\nSo dQ>0 for v<7/3 and dQ<0 for v>7/3: the switch is at V=37V0.\nInsight: the switch is where the line is tangent to an adiabat, not where the temperature is highest. The temperature is maximum at V=2V0 (trap), where dU=0 but the gas is still doing work and still absorbing heat. 25V0 is the answer for a monatomic gas (23(4−2v)+(4−v)=10−4v).\nCheck: Q absorbed from v=1 to 7/3 is P0V0∫17/3(14−6v)dv=316P0V0>0, and the total QAB=ΔU+W=0+4P0V0 (since PAVA=PBVB), consistent with 316−34=4.\nDifficulty: Very hard