JEE Advanced · PhysicsMedium

$An ideal gas undergoes a process where its pressure P is proportional to its volume V . If the gas expands from volume V_{0} to 2 V_{0}, what is the work done by the gas, given P = k V ?$

A. $0.5 k V_{0}^{2}$
B. $1.5 k V_{0}^{2}$
C. $2 k V_{0}^{2}$
D. $3 k V_{0}^{2}$

Show correct answer & step-by-step solution

Correct answer: B — $1.5 k V_{0}^{2}$

Solution

$The work done W by a gas is given by the integral W = \int_{V_{0}}^{2 V_{0}} P d V .$
$Substitute the given relation P = k V into the integral: W = \int_{V_{0}}^{2 V_{0}} k V d V .$
$Evaluate the integral: W = k [\frac{V ^{2}}{2}]_{V_{0}}^{2 V_{0}} = \frac{k}{2} ((2 V_{0})^{2} - V_{0}^{2}) .$
$Simplify the expression: W = \frac{k}{2} (4 V_{0}^{2} - V_{0}^{2}) = \frac{k}{2} (3 V_{0}^{2}) = 1.5 k V_{0}^{2} .$
$Hence the answer is (B).$

Attempt this question & track your score

Sign up free to answer, get instant scoring, and let SolveGini track which Physics topics you need to revise.

Attempt & Track Free →

More Thermodynamics practice questions

View all Thermodynamics questions →