1 , text{mole} of an ideal gas at temperature | vedclass

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(B) The work done by an ideal gas is given by $W = \int_{V_1}^{V_2} P \, dV$.Given the process law $P/V = K$ (where $K$ is a constant), we have $P = K

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$\frac{R}{2}(T_2 - T_1)$

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(B) The work done by an ideal gas is given by $W = \int_{V_1}^{V_2} P \, dV$.Given the process law $P/V = K$ (where $K$ is a constant), we have $P = KV$.Substituting this into the work integral: $W = \int_{V_1}^{V_2} KV \, dV = \frac{1}{2} K(V_2^2 - V_1^2)$.From the ideal gas equation $PV = RT$, and substituting $P = KV$, we get $(KV)V = RT$, which implies $KV^2 = RT$.Therefore, $K V_2^2 = RT_2$ and $K V_1^2 = RT_1$.Substituting these into the work expression: $W = \frac{1}{2} (KV_2^2 - KV_1^2) = \frac{1}{2} (RT_2 - RT_1) = \frac{R}{2}(T_2 - T_1)$.

$1 \, \text{mole}$ of an ideal gas at temperature $T_1$ expands according to the law $P/V = \text{constant}$. Find the work done when the final temperature becomes $T_2$.

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