PRAYOGX
ADV-2026-P1-PHY-Q11

Two-Chamber Heat Lab — how fast does the temperature gap halve?

JEE Advanced 2026 · Paper 1 · Physics · Q.11 · numerical value (+4 / 0) · Thermal properties and thermodynamics

01The question

Q.11 As shown in the figure, an insulated container is fitted with a thermally conducting but immovable partition (P1) and a freely movable but thermally insulated piston (P2). The partition P1 with thermal conductivity K, cross sectional area A and width x divides the container into two sections, S1 and S2, each containing one mole of a monoatomic gas. The piston P2 moves freely such that the gas in S2 is always at the atmospheric pressure. Initially, the difference between the temperatures of S1 and S2 is ΔT0. The time it takes for the temperature difference to become ΔT0/2 is nxR/KA, where R is the universal gas constant. The value of n is:
[ Given: ln 2 ≈ 0.7 ]
S₁S₂P₁P₂x

02The experiment

Answer— Press START: heat leaks through the partition and the run times how long the gap takes to halve.

Ready.

Piston P₂
ΔT₀
Speed

03Detailed solution

Answer: n = — (= (15/16) × 0.7 with the given ln 2; with the exact ln 2 it is 0.650. The accepted range is 0.63 to 0.70.)

  1. Heat through the partition. The rate is H = KA(T1 − T2)/x, and the rest of the container is insulated.
  2. Where it goes. S1 has a fixed volume, so it loses heat at constant volume: CV dT1/dt = −H with CV = (3/2)R. S2 is held at atmospheric pressure by the free piston, so it gains heat at constant pressure and expands: CP dT2/dt = +H with CP = (5/2)R. The partition is thin, so its own heat capacity is neglected.
  3. The gap decays exponentially. With ΔT = T1 − T2:
    dΔT/dt = −(KA/x)(1/CV + 1/CP) ΔT = −(KA/xR)(2/3 + 2/5) ΔT = −(16/15)(KA/xR) ΔT.
  4. Half-time. ΔT = ΔT0 e−16KAt/15xR, so ΔT = ΔT0/2 at
    t = (15/16) ln 2 · xR/KA = (15/16)(0.7) · xR/KA ⇒ n = 0.656 ≈ 0.66.
  5. Why the piston matters. If P2 were locked, S2 would also warm at constant volume, the rate would be 4/3 and n = (3/4)(0.7) = 0.525. With the free piston, part of the heat that reaches S2 goes into pushing the piston out, so S2 warms more slowly and the gap closes more slowly.

Understand the concept

Two bodies exchanging heat through a conductor close their temperature gap exponentially. The rate constant is the conductance KA/x times (1/C1 + 1/C2). The heat capacities here depend on how each gas is held: a rigid chamber uses CV, and a chamber at fixed pressure uses CP = CV + R, because some of its heat does work on the piston. The half-time does not depend on how big the initial gap is.

Key takeaways

  • Conduction: H = KAΔT/x.
  • Constant volume uses CV = 3R/2 and constant pressure uses CP = 5R/2, for a monoatomic gas.
  • dΔT/dt = −(KA/x)(1/C1 + 1/C2)ΔT, so the half-time is ln 2 divided by that rate constant.
  • Here n = (15/16) ln 2 ≈ 0.66.

Verification: an independent derivation, a sympy route, a numerical integration of both gases' energy equations (with the piston's work computed from PdV and the half-time found from the numbers), and the page's own engine, run by tests/verify_p1phyq11.py. The official key (0.63 to 0.70) and the MathonGo solution (0.66) agree. Status: script-verified by the PrayogX simulation factory; no review by a person has been recorded. Time on the page is in units of xR/KA. The starting temperatures are an example, since only the gap matters.

How to use this simulation

  1. Press ▶ START EXPERIMENT. Heat flows through the partition from the hotter S1 to S2.
  2. Watch S1 cool at fixed volume while S2 warms and pushes its piston out, and read both thermometers.
  3. The plot shows the gap falling exponentially. The run marks when it reaches half and reads n from that time.
  4. Try Piston P₂: Locked (the gap halves faster, n = 0.525) or change ΔT₀ (n does not change).
  5. Replay runs it again. Reset clears it. Classroom fills the screen for a projector.