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Q.
If in an isothermal process the volume of ideal gas is halved, then we can say that
ManipalManipal 2011Thermodynamics
Solution:
In an isothermal process, work done is
$d W =p d V$
$\therefore d W =p\left(V_{2}-V\right)$
$=p\left(\frac{V}{2}-V\right)$
$\left(\because V_{1}=V,\, V_{2}=\frac{V}{2}\right)$
$=-\frac{p V}{2}$
So, the work done by the gas is negative.