A monatomic ideal gas undergoes a process where its pressure P is related to its volume V by the equation P = P0 exp(alpha V), where P0 and alpha are constants. If the volume changes from V1 to V2, what is the work done by the gas?

A. P_0 (e^ V_2 - e^ V_1) — The work done by a gas during a volume change is calculated by integrating the pressure function with respect to volume.

D. P_0 (V_2 e^ V_2 - V_1 e^ V_1)

JEE Advanced · PhysicsHard

$A monatomic ideal gas undergoes a process where its pressure P is related to its volume V by the equation P = P_{0} e^{α V}, where P_{0} and α are constants. If the volume changes from V_{1} to V_{2}, what is the work done by the gas?$

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Correct answer: A — $\frac{P _{0}}{α} (e^{α V_{2}} - e^{α V_{1}})$

$The work done W by a gas during a process is given by the integral W = \int_{V_{1}}^{V_{2}} P d V .$
$Substitute the given pressure function P = P_{0} e^{α V} into the integral: W = \int_{V_{1}}^{V_{2}} P_{0} e^{α V} d V .$
$Evaluate the integral using the rule \int e^{a x} d x = \frac{1}{a} e^{a x}, which yields W = \frac{P _{0}}{α} [e^{α V}]_{V_{1}}^{V_{2}} .$
$Apply the limits of integration to obtain W = \frac{P _{0}}{α} (e^{α V_{2}} - e^{α V_{1}}) .$
$Hence the answer is (A).$