JEE Main 2019 — Thermodynamics (C) Question with Solution

$\mathrm{PV}=$ constant (isothermal expansion)

$\mathrm{P}\propto \frac{1}{\mathrm{V}}$. The graph will be a straight line, $\mathrm{y}=\mathrm{mx}+\mathrm{c}$. So, the above graph is correct.



$\mathrm{PV}=$ constant (isothermal expansion)

On increasing or decreasing $\mathrm{P}$, $\mathrm{PV}$ becomes constant. So, the above graph is correct.

$\left(\mathrm{A},\mathrm{ }\mathrm{C}\right)$ graphs are correct.



$\mathrm{PV}=$ constant (isothermal expansion)

$\mathrm{P}\propto \frac{1}{\mathrm{V}}$. The graph will be a straight line, $\mathrm{y}=\mathrm{mx}+\mathrm{c}$. So, the above graph is incorrect.



$\mathrm{\Delta U}={\mathrm{nC}}_{\mathrm{v}}\mathrm{dT}$

and for isothermal process, $\mathrm{\Delta U}=0.$ So, the above graph is incorrect.

Options $\mathrm{B}, \mathrm{D}$ are incorrect.