Question

The volume $V$ of a given mass of monatomic gas changes with temperature $T$ according to the relation $V=K{T}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$. The work done when temperature changes by $90\text{K}$ will be $x\mathrm{R}$. The value of $x$ is ____. [$\mathrm{R}=$universal gas constant]

Answer 1

Step 1: Given data

$V=K{T}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$
where $V$ is the volume of the gas and $T$ is its temperature

Let $T_1$ be the initial temperature and $T_2$ be the final temperature. We are given that,
$T_2-T_1=90\mathrm{K}$

When the temperature is increased by $90\mathrm{K}$, work done is $x\mathrm{R}\dots \left(1\right)$

Step 2: Formulas used

We know that work done by a gas is,
$W=\int PdV$

From ideal gas law,
$\begin{array}{cc}& PV=n\mathrm{R}T\\ \Rightarrow & P=\frac{n\mathrm{R}T}{V}\end{array}$

where $P$ is the pressure, $V$ is the volume $n$ is the number of moles of gas, $T$ is the temperature and $\mathrm{R}$ is the universal gas constant.

Step 3: Calculating the value of x

From the work done equation,
$W=\int \frac{n\mathrm{R}TdV}{V}$

Also, by differentiating the given equation with respect to temperature,
$dV=\frac{2}{3}K{T}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}dT$

Thus, work done from $T_1$ to $T_2$ is,
$\begin{array}{rcl}& \Rightarrow & W={\int }_{T_1}^{T_2}\frac{n\mathrm{R}T}{K{T}^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}·\frac{2}{3}K{T}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}dT\\ & =& \frac{2}{3}{\int }_{T_1}^{T_2}\frac{n\mathrm{R}T}{T^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}dT\\ & =& \frac{2}{3}{\int }_{T_1}^{T_2}n\mathrm{R}dT\\ & =& \frac{2}{3}n\mathrm{R}\left[T_2-T_1\right]\\ & =& \frac{2}{3}n\mathrm{R}·90\\ \therefore W& =& 60n\mathrm{R}\end{array}$

Assuming $n=1$,
$W=60\mathrm{R}$

Comparing with equation $\left(1\right)$,
$\Rightarrow x=60$

Therefore, the value of $x$ is $60$.