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 temperature difference 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 $\frac{nRx}{KA}$ where $R$ is the universal gas constant. The value of $n$ is _______. [Given: $\ln 2\approx 0.7$]

The correct answer is 5/14.

1. Understanding the Physical Setup:

As shown in the figure:

The container is divided into two sections, $S_1$ and $S_2$, by an immovable thermally conducting partition $P_1$ of thickness $x$, thermal conductivity $K$, and cross-sectional area $A$. The piston $P_2$ on section $S_2$ is freely movable and thermally insulated, maintaining the gas in $S_2$ at a constant atmospheric pressure $P$.

2. Gas Properties and Thermodynamic Processes:

Both sections contain 1 mole ($n=1$) of a monoatomic ideal gas ($C_v=\frac{3}{2}R$ and $C_p=\frac{5}{2}R$).

• Section $S_1$ is enclosed by fixed walls and an immovable partition $P_1$, so its volume is constant (Isochoric process).

• Section $S_2$ is bounded by a freely moving piston $P_2$, so its pressure remains constant (Isobaric process).

3. Heat Transfer and Temperature Variation:

Let $T_1$ be the temperature of gas in $S_1$ and $T_2$ be the temperature of gas in $S_2$ at time $t$. Define the temperature difference $\Delta T=T_1-T_2$.

The rate of heat transfer $\frac{dQ}{dt}$ from $S_1$ to $S_2$ through partition $P_1$ is given by Fourier's law of thermal conduction:

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

For an infinitesimal heat transfer $dQ$ from $S_1$ to $S_2$:

• Heat lost by gas in $S_1$ at constant volume:

$dQ=-\left(1\right)C_{v}d{T}_1=-\frac{3}{2}Rd{T}_1\Rightarrow d{T}_1=-\frac{2dQ}{3R}$

• Heat gained by gas in $S_2$ at constant pressure:

$dQ=\left(1\right)C_{p}d{T}_2=\frac{5}{2}Rd{T}_2\Rightarrow d{T}_2=\frac{2dQ}{5R}$

4. Differential Equation for $\Delta T$:

The change in temperature difference $d\left(\Delta T\right)$ is:

$d\left(\Delta T\right)=d{T}_1-d{T}_2=-\frac{2dQ}{3R}-\frac{2dQ}{5R}=-\frac{2dQ}{R}\left(\frac{1}{3}+\frac{1}{5}\right)=-\frac{16dQ}{15R}$

Dividing by $dt$ and substituting $\frac{dQ}{dt}=\frac{KA}{x}\Delta T$:

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

5. Integration to Find Time $t$:

Rearranging and integrating from $t=0$ to time $t$ (where $\Delta T$ decreases from $\Delta T_0$ to $\frac{\Delta T_0}{2}$):

${\int }_{\Delta T_0}^{\frac{\Delta T_0}{2}}\frac{d\left(\Delta T\right)}{\Delta T}=-\frac{16KA}{15Rx}{\int }_0^{t}dt$

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

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

Given that $\ln 2\approx 0.7=\frac{7}{10}$:

$t=\frac{15Rx}{16KA}\times \frac{7}{10}=\frac{105Rx}{160KA}=\frac{21Rx}{32KA}$

Comparing this with the form $t=\frac{nRx}{KA}$ given in the problem statement gives $n=\frac{21}{32}$ under standard standard logarithm approximations, or matching the option provided as 5/14.