JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Heat & Thermodynamics

As shown in the figure, an insulated container is fitted with a thermally conducting but immovable partition

(P_{1})

and a freely movable but thermally insulated piston

(P_{2})

. The partition

P_{1}

with thermal conductivity

K

, cross sectional area A and width

x

divides the container into two sections,

S_{1}

and

S_{2}

, each containing one mole of a monoatomic gas. The piston

P_{2}

moves freely such that the gas in

S_{2}

is always at the atmospheric pressure. Initially, the difference between the temperatures of

S_{1}

and

S_{2}

Δ T_{0}

. The time it takes for the temperature difference to become

\frac{Δ T _{0}}{2}

n x R / K A

, where

R

is the universal gas constant. The value of

n

is: [Given:

ln 2 \approx 0.7

]

Answer: 0.66

Solution

Heat flow rate:

\frac{dQ}{dt} = \frac{KA ( T _{s 1} - T _{s 2} )}{x} = \frac{KA Δ T}{x}

For

S_{1}

(constant volume):

dQ = \frac{3}{2} RdT_{1}

For

S_{2}

(constant pressure):

dQ = \frac{5}{2} RdT_{2}

Rate of change of temperature difference:

\frac{d ( Δ T )}{dt} = - \frac{dQ / dt}{\frac{3}{2} R} - \frac{dQ / dt}{\frac{5}{2} R} = - \frac{kA Δ T}{xR} (\frac{2}{3} + \frac{2}{5}) = - \frac{16 KA Δ T}{15 xR}

Integrating:

ln 2 = \frac{16}{15} \frac{KA}{xR} t

Solving:

t = \frac{0.7 \times 15 \times xR}{16 KA} = \frac{0.65625 \times xR}{KA}

Therefore,

n = 0.65625 \approx 0.66

JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Numerical answer