Question

One mole of monatomic ideal gas is contained in a rigid container of volume $V$ with walls of total surface area $A$, thickness $x$ and thermal conductivity $K$. The gas is at an initial temperature $T_o$ and pressure $P_o$. The temperature of the surrounding air is $T_s$. All the temperatures are in absolute scale. If the temperature of the gas at any time $t$ can be expressed as $T\left(t\right)=T_s-(T_s-T_o)e^{-\lambda t}$, find the value of $\lambda$.

• A. $\lambda =\frac{2KA}{3Rx}$
• B. $\lambda =\frac{4KA}{3Rx}$
• C. $\lambda =\frac{2KA}{3R}$
• D. $\lambda =\frac{5KA}{3Rx}$

Answer 1

The correct option is A $\lambda =\frac{2KA}{3Rx}$

For monoatomic gas,
$C_V=\frac{3R}{2},C_P=\frac{5R}{2},T_s>T_o,$
Assume 1 mole of gas is present in the container.


Let at any time $t$, the temperature of gas be $T$.
The heat transfer to the container is
$\frac{dQ}{dt}=\frac{KA(T_s-T)}{x}$ ... (i)
Heat gained by the gas is
$\frac{dQ}{dt}=n{C}_V\frac{dT}{dt}$ ... (ii)
From equation (i) and (ii),
$n{C}_V\frac{dT}{dt}=\frac{KA(T_s-T)}{x}$
$\Rightarrow \left(\frac{3R}{2}\right)\frac{dT}{dt}=\frac{KA(T_s-T)}{x}$ $(n=1)$
Integrating on both sides,
$\Rightarrow \frac{3R}{2}\underset{T_o}{\overset{T}{\int }}\frac{dT}{T_s-T}=\frac{KA}{x}\underset{0}{\overset{t}{\int }}dt$
$\Rightarrow \frac{3R}{2}{\left[\frac{\ln (T_s-T)}{-1}\right]}_{T_o}^T=\frac{KA}{x}{\left[t\right]}_0^t$
$\Rightarrow \frac{T_s-T}{T_s-T_o}=e^{\frac{-2KAt}{3Rx}}$
$\Rightarrow T\left(t\right)=T_s-(T_s-T_o)e^{-\lambda t}$ where $\lambda =\frac{2KA}{3Rx}$