JEE Main · PhysicsMedium

$An ideal gas undergoes a process where its pressure $P$ is proportional to its volume $V$. If the gas expands from volume $V_0$ to $2V_0$, what is the work done by the gas, given $P = kV$?$

Show correct answer & step-by-step solution

Correct answer: B — $$1.5 k V_0^2$$

Solution

The work done is

W = \int_{V_{0}}^{2 V_{0}} P d V

. Substituting

P = k V

, we get

W = \int_{V_{0}}^{2 V_{0}} k V d V = \frac{k}{2} [V^{2}]_{V_{0}}^{2 V_{0}} = \frac{k}{2} (4 V_{0}^{2} - V_{0}^{2}) = 1.5 k V_{0}^{2}