One mole of a monatomic ideal gas undergoes a cyclic process as shown in the figure (where V is the volume…

1. Process $I$ is an isochoric process
2. In process $II$ , gas absorbs heat
3. In process $IV$ , gas releases heat
4. Processes $I \mathrm{a}\mathrm{n}\mathrm{d} III$ are not isobaric

Solution

$\left(A\right)$ Process- $I$ is not isochoric, V is decreasing
$\left(B\right)$ Process- $II$ is isothermal expansion
$∆U=0, W>0$
$∆Q>0$
$\left(C\right)$ Process- $IV$ is isothermal compression
$∆U=0, W<0$
$∆Q<0$
$\left(D\right)$ Process- $I \mathrm{a}\mathrm{n}\mathrm{d} III$ are NOT isobaric because in isobaric process $T \propto V$ hence isobaric $T–V$ graph will be linear

Asked in: JEE Advanced 2018 (Paper 1)

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