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 is calculated by integrating the pressure with respect to volume, W = integral of P dV from V1 to V2. Substituting P = P0 exp(alpha V) and performing the integration yields the result.

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

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$A monatomic ideal gas undergoes a process where its pressure $P$ is related to its volume $V$ by the equation $P = P_0 e^{\alpha V}$, where $P_0$ and $\alpha$ are constants. If the volume changes from $V_1$ to $V_2$, what is the work done by the gas?$

A. $\frac{P_0}{\alpha} (e^{\alpha V_2} - e^{\alpha V_1})$
B. $\frac{P_0}{\alpha} (e^{\alpha V_1} - e^{\alpha V_2})$
C. $P_0 \alpha (V_2 - V_1)$
D. $\frac{P_0}{\alpha} (V_2 e^{\alpha V_2} - V_1 e^{\alpha V_1})$

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Correct answer: A — $\frac{P_0}{\alpha} (e^{\alpha V_2} - e^{\alpha V_1})$

Solution

The work done is calculated by

W = \int_{V_{1}}^{V_{2}} P d V

. Substituting

P = P_{0} e^{α V}

and integrating gives

W = \int_{V_{1}}^{V_{2}} P_{0} e^{α V} d V = \frac{P _{0}}{α} [e^{α V}]_{V_{1}}^{V_{2}} = \frac{P _{0}}{α} (e^{α V_{2}} - e^{α V_{1}})

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A monatomic ideal gas undergoes a process where its pressure $P$ is related to its volume $V$ by the equation $P = P_0 e^{\alpha V}$, where $P_0$ and $\alpha$ are constants. If the volume changes from $V_1$ to $V_2$, what is the work done by the gas?

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$A monatomic ideal gas undergoes a process where its pressure $P$ is related to its volume $V$ by the equation $P = P_0 e^{\alpha V}$, where $P_0$ and $\alpha$ are constants. If the volume changes from $V_1$ to $V_2$, what is the work done by the gas?$