Question 11

Let $T_1$ be the temperature in $S_1$ and $T_2$ be the temperature in $S_2$.

Since the partition $P_1$ is immovable, section $S_1$ undergoes an isochoric process.

The heat gained by $S_1$ is $dQ = n C_v dT_1 = 1 \cdot \frac{3}{2} R dT_1$.

Since the piston $P_2$ is freely movable, section $S_2$ undergoes an isobaric process. The heat lost by $S_2$ is $-dQ = n C_p dT_2 = 1 \cdot \frac{5}{2} R dT_2$.

The rate of heat transfer through the partition is:

$\frac{dQ}{dt} = \frac{KA(T_2 - T_1)}{x}$

Let $\Delta T = T_2 - T_1$. Then:

$dT_1 = \frac{2 dQ}{3R}$

$dT_2 = -\frac{2 dQ}{5R}$

$d(\Delta T) = dT_2 - dT_1 = -\frac{2 dQ}{5R} - \frac{2 dQ}{3R} = -\frac{16 dQ}{15R}$

Substituting for $dQ$:

$d(\Delta T) = -\frac{16}{15R} \left( \frac{KA \Delta T}{x} dt \right)$

$\frac{d(\Delta T)}{\Delta T} = -\frac{16 KA}{15 x R} dt$

Integrating from $t=0$ (where $\Delta T = \Delta T_0$) to $t$ (where $\Delta T = \frac{\Delta T_0}{2}$):

$\int_{\Delta T_0}^{\Delta T_0/2} \frac{d(\Delta T)}{\Delta T} = -\frac{16 KA}{15 x R} \int_{0}^{t} dt$

$\ln\left(\frac{1}{2}\right) = -\frac{16 KA t}{15 x R}$

$t = \frac{15 x R \ln 2}{16 KA}$

Comparing with $t = \frac{n x R}{KA}$, we get:

$n = \frac{15 \ln 2}{16} = \frac{15 \times 0.7}{16} = \frac{10.5}{16} = 0.65625 \approx 0.66$