Question

An amount n (in moles) of a monatomic gas at an initial temperature T₀ is enclosed in a cylindrical vessel fitted with a light piston. The surrounding air has a temperature T_s (> T₀) and the atmospheric pressure is P_α. Heat may be conducted between the surrounding and the gas through the bottom of the cylinder. The bottom has a surface area A, thickness x and thermal conductivity K. Assuming all changes to be slow, find the distance moved by the piston in time t.

Answer 1

In time dt, heat transfer through the bottom of the cylinder is given by
$\frac{dQ}{dt}=\frac{KA\left(T_s-T_0\right)}{x}$

For a monoatomic gas, pressure remains constant.

$$\begin{gathered} \therefore dQ=n{C}_{p}dT \\ \frac{n{\mathrm{C}}_{p}dT}{dt}=\frac{KA\left(T_s-T_0\right)}{x} \end{gathered}$$

For a monoatomic gas,
$C_p=\frac{5}{2}R$
$\Rightarrow \frac{n5RdT}{2dt}=\frac{KA\left(T_s\mathit{-}{T}_{\mathit{0}}\right)}{x}$
$$\begin{gathered} \Rightarrow \frac{5nR}{2}\frac{dT}{dt}=\frac{KA\left(T_s-T_0\right)}{x} \\ \Rightarrow \frac{dT}{\left(T_s-T_0\right)}=\frac{-2KAdt}{5nRx} \end{gathered}$$

Integrating both the sides,
$$\begin{gathered} {\left(T_{\mathrm{s}}-T_0\right)}_{T_0}^T=\frac{-2KAt}{5nRx} \\ \Rightarrow \ln \left(\frac{T_s-T}{T_s-T_0}\right)=-\frac{-2KAt}{5nRx} \\ \Rightarrow T_{\mathrm{s}}-T=\left(T_s-T_0\right)e^{{}^{-\frac{2KAt}{5nRx}}} \\ \Rightarrow T=T_s-\left(T_s-T_0\right)e^{\frac{-2KAt}{5nRx}} \\ \Rightarrow T-T_0=(T_s-T_0)-\left(T_s-T_0\right)e^{\frac{-2KAt}{5nRx}} \\ \Rightarrow T-T_0=(T_s-T_0)[1-e^{\frac{-2KAt}{5nRx}}] \end{gathered}$$

From the gas equation,
$$\begin{gathered} \frac{P_{a}Al}{nR}=T-T_0 \\ \therefore \frac{P_{\mathrm{a}}Al}{nR}=\left(T_s-T_0\right)[1-e^{\frac{-2KAt}{5nRx}}] \\ \Rightarrow l=\frac{nR}{P_{\mathrm{a}}A}\left(T_s-T_0\right)[1-e^{\frac{-2KAt}{5nRx}}] \end{gathered}$$