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Heat through a conducting partition between a gas at constant volume and a gas at constant pressure: the time for the temperature gap to halve

An insulated container: gas S1 at fixed volume and gas S2 held at atmospheric pressure by a free piston, joined by a conducting partition. The gap decays at rate (16/15) KA/xR, so it halves at t = (15/16) ln 2 xR/KA - n = 0.66 with the given ln 2 = 0.7.

Sub-topics covered

• Heat conduction through a slab: H = KA dT/x
• Molar heat capacities of a monoatomic gas: C_V = 3R/2 at fixed volume, C_P = 5R/2 at fixed pressure
• Two bodies exchanging heat: the gap decays as exp(-(KA/x)(1/C1 + 1/C2) t)
• Half time = ln 2 / rate constant

What it teaches

• S1 (rigid) loses heat at constant volume, C_V = 3R/2; S2 (free piston at 1 atm) gains heat at constant pressure, C_P = 5R/2
• d(dT)/dt = -(KA/x)(2/3R + 2/5R) dT = -(16/15)(KA/xR) dT
• t_half = (15/16) ln 2 xR/KA: n = 0.656 with ln 2 = 0.7 (0.650 with the exact log), both inside the key
• With the piston locked the rate would be 4/3 and n = 0.525: part of S2's heat does work on the piston, so the gap closes more slowly

Simulation ID

• ADV-2026-P1-PHY-Q11 · revision 1

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