Heat and Thermodynamics

Concept:The problem combines heat transfer through a conducting partition with thermodynamics of ideal gases undergoing different processes.Explanation:The container has two sections $S_1$ and $S_2$, each with 1 mole of monoatomic gas.$S_1$ is rigid (constant volume) → isochoric process; molar heat capacity $C_v = \frac{3}{2}R$.$S_2$ has a movable piston maintaining constant atmospheric pressure → isobaric process; $C_p = \frac{5}{2}R$.Let $T_1$ and $T_2$ be instantaneous temperatures of $S_1$ and $S_2$, with $T_1 > T_2$.Heat $dQ$ flows from $S_1$ to $S_2$ through partition of conductivity $K$, area $A$, thickness $x$.For $S_1$ (isochoric cooling): $dQ = -n C_v dT_1 = -\frac{3}{2}R dT_1$ → $dT_1 = -\frac{2}{3R}dQ$.For $S_2$ (isobaric heating): $dQ = n C_p dT_2 = \frac{5}{2}R dT_2$ → $dT_2 = \frac{2}{5R}dQ$.Change in temperature difference: $d(\Delta T) = dT_1 - dT_2 = -\frac{16}{15R}dQ$.Heat transfer rate: $\frac{dQ}{dt} = \frac{KA}{x}(T_1 - T_2) = \frac{KA}{x}\Delta T$ → $dQ = \frac{KA}{x}\Delta T dt$.Substitute: $d(\Delta T) = -\frac{16KA}{15xR}\Delta T dt$ → $\frac{d(\Delta T)}{\Delta T} = -\frac{16KA}{15xR}dt$.Integrate from initial $\Delta T_0$ to $\frac{\Delta T_0}{2}$ over time $\tau$:$\int_{\Delta T_0}^{\Delta T_0/2} \frac{d(\Delta T)}{\Delta T} = -\frac{16KA}{15xR}\int_0^\tau dt$ → $\ln\frac{1}{2} = -\frac{16KA}{15xR}\tau$.Thus $-\ln2 = -\frac{16KA}{15xR}\tau$ → $\tau = \frac{15}{16}\ln2 \cdot \frac{xR}{KA}$.Given $\ln2 \approx 0.7$, $n = \frac{15}{16}\times 0.7 = 0.65625 \approx 0.66$.Answer:$n \approx 0.66$.