The heat transferred in the process.

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Solution

Given:-
$\Rightarrow P = a V \to (1)$ a : constant
$\Rightarrow V_{i} = V_{o}$ and $V_{f} = {3 V}_{o}$
Heat transfer $(Q) : -$
$\Rightarrow d Q = d U + P d V \to (2)$
$\Rightarrow d U = C_{V} d T$
$= \frac{3}{2} R d T \to (3) [C_{V} = \frac{3}{2} R f o r M o n o a t o m i c]$
Now Ideal gas equation,
$\Rightarrow P V = R T$
$\Rightarrow {a V}^{2} = R T$ [using $(1)$ ]
$\Rightarrow 2 a V = \frac{R d T}{d V}$
$\Rightarrow R d T = 2 a V d V$
Put in $(3)$
$\Rightarrow d U = \frac{3}{2} (2 a V d V) = 3 a V d V$
$\Rightarrow d Q = d U + P d V$
$= 3 a V d V + a V d V$ [using $(1)$ & $(2)$ ]
$\Rightarrow \int d Q = \int 4 a V d V$
$\Rightarrow Q = 4 a {∣ ∣ ∣ \frac{V^{2}}{2} ∣ ∣ ∣}_{V_{o}}^{{3 V}_{o}} = 2 a [{(3 V_{o})}^{2} - {V_{o}}^{2}]$
$\Rightarrow$ Heat transfer; $Q = 16 a {V_{o}}^{2}$
Hence, the answer is $Q = 16 a {V_{o}}^{2} .$

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