The process $AB$ is shown in the diagram. As the gas is taken from $A$ to $B$, its temperature
Diffcult
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Let us first find the equation of the straight line.
$\frac{P-2 p}{V-v}=\frac{2 p-p}{v-2 v}$
$P-2 p=V \times\left(\frac{-p}{v}\right)$
$\therefore$ equation of the line : $P=V \times\left(\frac{-p}{v}\right)+2 p$
We know that internal Energy $(U)=P V$
$U=P \times V=V \times\left(V \times\left(\frac{-p}{v}\right)+2 p\right)$
$U=-V^{2} \times\left(\frac{p}{v}\right)+2 p V$
As we go from point $A$ to point $B$ volume increases.
As volume increases internal energy initially increases and then decreases, since the internal energy as a function of volume is a downward parabola.
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