Question

Consider the reversible isothermal expansion of an ideal gas in a closed system at two different temperatures $T_{1}$ and $T_{2}\left(T_{1} < T_{2}\right)$. The correct graphical depiction of the dependence of work done $(W)$ on the final volume $(V)$ is

• A.
• B.
• C.
• D.

Answer 1

Answer: (C)

For isothermal reversible expansion,
$|W|=n R T \ln \frac{V_{f}}{V_{i}}=n R T \ln \frac{V}{V_{i}}$
where, $V=$ final volume, $V_{i}=$ initial final.
or $|W|=n R T \ln V-n R T \ln V_{i}$
On comparing with equation of straight line, $y=m x+c$, we get
slope $=m=+n R T$
intercept $=-n R T \ln V_{i}$
Thus, plot of $|W|$ with $\ln V$ will give straight line in which slope of $2\left(T_{2}\right)$ is greater than slope of $1\left(T_{1}\right)$ which is given in all options.
Now, if $V_{i}< 1$ then $y$ intercept $\left(-n R T V_{i}\right)$ becomes positive and if it is positive for one case then it is positive for other case also. Thus, it is not possible that one $y$-intercept goes above and other $y$-intercept goes below. Thus, option (b) and (d) are incorrect.
If we extent plot given in option (a) it seems to be merging which is not possible because if they are merging they give same $+ve\, y$-intercept. But they cannot give same $y$-intercept because value of $T$ is different.
Now, if we extent the line of $T_{1}$ and $T_{2}$ given in option (c) it seems to be touching the origin. If they touch the origin then $y$-intercept becomes zero which is not possible. Thus, it is not the exactly correct answer but among the given options it is the most appropriate one.