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 $S_2$ is $\Delta T_0$. The time it takes for the temperature difference to become $\frac{\Delta T_0}{2}$ is $nxR/KA$, where $R$ is the universal gas constant. The value of $n$ is: [Given: $\ln 2 \approx 0.7$]

Step 1: Identify the thermodynamic processes for each section.

Section $S_1$ has a constant volume, while section $S_2$ remains at constant atmospheric pressure.

$$C_{V} = \frac{3}{2}R$$ $$C_{P} = \frac{5}{2}R$$

Step 2: Express the rate of heat transfer through the partition.

The heat flow rate from $S_1$ to $S_2$ depends on the temperature difference $\Delta T = T_1 - T_2$.

$$\frac{dq}{dt} = \frac{KA}{x}(T_1 - T_2)$$ $$\frac{dq}{dt} = \frac{KA}{x}\Delta T$$

Step 3: Relate the heat transfer to the temperature changes.

For section $S_1$, the heat lost reduces its internal energy at constant volume.

$$\frac{dT_1}{dt} = -\frac{2}{3R}\frac{dq}{dt}$$

For section $S_2$, the heat gained increases its enthalpy at constant pressure.

$$\frac{dT_2}{dt} = \frac{2}{5R}\frac{dq}{dt}$$

Step 4: Set up the differential equation for the temperature difference.

Subtract the temperature change rates to find the rate of change of $\Delta T$.

$$\frac{d(\Delta T)}{dt} = \frac{dT_1}{dt} - \frac{dT_2}{dt}$$ $$\frac{d(\Delta T)}{dt} = -\left(\frac{2}{3R} + \frac{2}{5R}\right)\frac{dq}{dt}$$ $$\frac{d(\Delta T)}{dt} = -\frac{16}{15R}\left(\frac{KA}{x}\Delta T\right)$$ $$\frac{d(\Delta T)}{dt} = -\frac{16KA}{15xR}\Delta T$$

Step 5: Solve the differential equation for the time elapsed.

Integrate the equation from the initial state to the state where the temperature difference is halved.

$$\int_{\Delta T_0}^{\frac{\Delta T_0}{2}} \frac{d(\Delta T)}{\Delta T} = -\frac{16KA}{15xR}\int_0^t dt$$ $$\ln\left(\frac{1}{2}\right) = -\frac{16KA}{15xR}t$$ $$t = \frac{15}{16}\ln 2 \frac{xR}{KA}$$

Step 6: Calculate the value of $n$.

Compare the expression for $t$ to the given form $t = \frac{nxR}{KA}$.

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

Answer:

The value of $n$ is approximately $0.66$ (or $\frac{15}{16}\ln 2$).