As shown in the figure, an insulated container is fitted with a...

Let $T_1$ and $T_2$ be the temperatures of the gases in sections $S_1$ and $S_2$ respectively. Assume $T_1 > T_2$, so heat flows from $S_1$ to $S_2$.

The rate of heat transfer through the partition $P_1$ is given by Fourier's law of heat conduction:
$\dfrac{dQ}{dt} = \dfrac{KA}{x} (T_1 - T_2)$

For section $S_1$, the volume is constant (since $P_1$ is immovable). Thus, the process is isochoric. The heat lost by $S_1$ is:
$dQ = -n_1 C_V dT_1$
Since the gas is monoatomic and $n_1 = 1$ mole, $C_V = \dfrac{3}{2}R$.
$\Rightarrow dT_1 = -\dfrac{2 dQ}{3R}$

For section $S_2$, the pressure is constant (since $P_2$ is freely movable and exposed to atmospheric pressure). Thus, the process is isobaric. The heat gained by $S_2$ is:
$dQ = n_2 C_P dT_2$
Since the gas is monoatomic and $n_2 = 1$ mole, $C_P = \dfrac{5}{2}R$.
$\Rightarrow dT_2 = \dfrac{2 dQ}{5R}$

The change in the temperature difference $d(\Delta T) = d(T_1 - T_2)$ is:
$d(T_1 - T_2) = dT_1 - dT_2 = -\dfrac{2 dQ}{3R} - \dfrac{2 dQ}{5R} = -\dfrac{16 dQ}{15R}$

Substituting the rate of heat transfer $\dfrac{dQ}{dt}$ into the equation:
$\dfrac{d(T_1 - T_2)}{dt} = -\dfrac{16}{15R} \left( \dfrac{KA}{x} (T_1 - T_2) \right)$

Let $\Delta T = T_1 - T_2$. The differential equation becomes:
$\dfrac{d(\Delta T)}{\Delta T} = -\dfrac{16 KA}{15 R x} dt$

Integrating both sides from $t = 0$ to $t$, where the temperature difference goes from $\Delta T_0$ to $\dfrac{\Delta T_0}{2}$:
$\int_{\Delta T_0}^{\Delta T_0 / 2} \dfrac{d(\Delta T)}{\Delta T} = -\dfrac{16 KA}{15 R x} \int_0^t dt$

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

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

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

Given that $\ln 2 \approx 0.7$:
$t = \dfrac{15 \times 0.7}{16} \dfrac{x R}{KA} = \dfrac{10.5}{16} \dfrac{x R}{KA} = \dfrac{21}{32} \dfrac{x R}{KA} = 0.65625 \dfrac{x R}{KA}$

Comparing this with the given expression $t = n \dfrac{x R}{K A}$, we get:
$n = 0.65625$

Answer: $0.66$