Question

The ratio of heat absorbed and work done by the gas in a thermodynamic process is $\frac{5}{2}$. Find the parameters representing $X$ and $Y$ axis respectively in the graph shown below.

• A. $T$ and $P$
• B. $T$ and $V$
• C. $V$ and $\frac{1}{P}$
• D. $V$ and $\rho$

Answer 1

The correct option is B $T$ and $V$

Since ratio of heat absorbed and workdone by the gas is,
$\frac{\Delta Q}{W}=\frac{5}{2}$

For option$\left(a\right)$. Let us suppose, parameters are temperature$\left(T\right)$ and pressure$\left(P\right)$, then graph will for $T-P$ will be a straight line for a constant volume process.
$PV=nRT$
$\Rightarrow T\propto P$, if $V=\text{constant}$
For which $W=0\&\frac{\Delta Q}{W}=\infty$.
$\therefore$ option $\left(a\right)$ is wrong

For option$\left(b\right)$. If parameters are temperature $\left(T\right)$ and volume$\left(V\right)$, the straight line graph represent a constant pressure.
$PV=nRT$
for $P=\text{constant}$ hence $T\propto V$
$\Rightarrow \Delta Q=n{C}_P\Delta T$
and $W=nR\Delta T$
$\frac{\Delta Q}{W}=\frac{C_P}{R}...\left(i\right)$
For monoatomic ideal gas, $C_P=\left(\frac{f+2}{2}\right)R=\frac{5}{2}R$
Hence from Eq.$\left(i\right)$ we get,
$\Rightarrow \frac{\Delta Q}{\Delta W}=\frac{5}{2}$
$\therefore$ option $\left(b\right)$ is correct.

For option$\left(c\right)$. If parameters are volume $\left(V\right)$ and $\frac{1}{P}$ then graph between $V$ and $\frac{1}{P}$ will be a straight line passing from origin, for a constant temperature i.e $T=\text{constant}$.
$PV=nRT$
$\Rightarrow V=\frac{1}{P}\left(nRT\right)$
$\therefore$we get $\Delta U=0\Rightarrow \Delta Q=W$
Hence $\frac{\Delta Q}{W}=1$, so option $\left(c\right)$ is incorrect.

For option$\left(d\right)$. If parameters are volume$\left(V\right)$ and density$\left(\rho \right)$ the graph will be always be rectangular hyperbola. Hence option d is not possible for given graph.
So, option $\left(d\right)$ is wrong.