The pressure of an ideal gas varies with volu | vedclass

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Question

(A) Given the pressure-volume relationship: $P = \alpha V$.The work done $W$ by the gas during expansion from initial volume $V_i = V$ to final volume

Vedclass · Accepted Answer

$\frac{\alpha V^2}{2}(m^2 - 1)$

Vedclass · Answer

(A) Given the pressure-volume relationship: $P = \alpha V$.The work done $W$ by the gas during expansion from initial volume $V_i = V$ to final volume $V_f = mV$ is given by the integral:$W = \int_{V_i}^{V_f} P \, dV$Substituting $P = \alpha V$:$W = \int_{V}^{mV} \alpha V \, dV$Integrating with respect to $V$:$W = \alpha \left[ \frac{V^2}{2} \right]_{V}^{mV}$$W = \frac{\alpha}{2} [(mV)^2 - V^2]$$W = \frac{\alpha V^2}{2} (m^2 - 1)$

The pressure of an ideal gas varies with volume as $P = \alpha V,$ where $\alpha$ is a constant. One mole of the gas is allowed to undergo expansion such that its volume becomes $m$ times its initial volume. The work done by the gas in the process is

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