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The volume of an ideal gas is doubled in an isothermal process. Then, which of the following is true?
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The correct answer is:
Work done by the gas is positive
For isothermal process,
$d U=0$
$\begin{aligned} & \text { Work done }=d W=p\left(V_2-V_1\right) \\ & \therefore \quad d W=+p V \quad\left[\because V_2=2 V_1=2 V\right] \\ & \end{aligned}$
$d U=0$
$\begin{aligned} & \text { Work done }=d W=p\left(V_2-V_1\right) \\ & \therefore \quad d W=+p V \quad\left[\because V_2=2 V_1=2 V\right] \\ & \end{aligned}$
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