A cylindrical vessel fitted with a light piston contains an amount...

Since the piston is light and the process is slow, the pressure of the gas remains constant at atmospheric pressure $p_a$. Thus, the process is isobaric.

For a monatomic gas, the molar heat capacity at constant pressure is $C_P = \dfrac{5}{2}R$.

The rate of heat conduction into the gas is given by:

$\dfrac{dQ}{dt} = \dfrac{KA(T_s - T)}{x}$

In an isobaric process, the heat absorbed is $dQ = nC_P dT$. Therefore:

$n \left(\dfrac{5}{2}R\right) \dfrac{dT}{dt} = \dfrac{KA(T_s - T)}{x}$

Rearranging the variables to integrate:

$\dfrac{dT}{T_s - T} = \dfrac{2KA}{5nRx} dt$

Integrating both sides from $t = 0$ (where $T = T_0$) to $t$ (where $T = T$):

$\int_{T_0}^{T} \dfrac{dT}{T_s - T} = \int_{0}^{t} \dfrac{2KA}{5nRx} dt$

$-\ln\left(\dfrac{T_s - T}{T_s - T_0}\right) = \dfrac{2KAt}{5nRx}$

$T_s - T = (T_s - T_0)e^{-\dfrac{2KAt}{5nRx}}$

The change in temperature of the gas is:

\Delta T = T - T_0 = (T_s - T_0) \left(1 -e^{-\dfrac{2KAt}{5nRx}}\right)

Let $y$ be the distance moved by the piston. The change in volume is $\Delta V = Ay$.

Using the ideal gas equation at constant pressure, $p_a \Delta V = nR\Delta T$:

$p_a A y = nR \Delta T$

$y =$\dfrac{nR}{p_a A} \Delta$T$

Substituting the value of $\Delta T$:

$y =$\dfrac{nR}{p_a A}(T_s - T_0) \left(1 - e^{-\dfrac{2KAt}{5nRx}}\right)