One mole of ideal monoatomic gas is taken round the cyclic process $ABCA$ . The following is a schematic $PV$ graph of a process $ABCA$ .

Find the heat exchange in the process $AB$ .

$\dfrac{9P_0V_0}{2}$
$3P_0V_0$
$-3P_0V_0$
$\dfrac{3P_0V_0}{2}$

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In an isochoric process,

$dQ$ = $dU$ $+$ $dW$

( here d is short for delta not differential)

But $dU$ = $u C_v dT$ = $C_v dT$

(u is short for mu - no. of moles =1 )

Also $dW$ = $PdV$ = $0$ since $dV$ = $0$ for an isochoric process.

Hence $dQ$ = $dU$ = $C_v dT$

From Ideal gas equation, we get,

$dT$ = $dPV$ $/$ $R$ ( since u =1 )

Substituting,

$dQ$ = $C_v dPV$ $/$ $R$ = $3/2 R * 2P * V$ $/$ $R$ = $3PV$

( because for monoatomic ideal gases gamma = $C_p$ $/$ $C_v$ = $5/3$ and also $C_p$ - $C_v$ = $R$ from which we get $C_p$ = $3/2$ )