Answer :

Correct answer is : 408.92

P₁ = 110 KPa, V₁ = 5 m³ , V₂ = 2.5 m³, P₂ = ?, W = ?

$P_1{V}_1^n=P_2{V}_2^n$

$110\times 5^{1.2}=P_2\times {2.5}^{1.2}$ $\Rightarrow$ P₂ = 252.7136 KPa

Work done during the process is

$W=\frac{P_1{V}_1^{}-P_2{V}_2^{}}{n-1}=\frac{110\times 5^{}-252.7136\times {2.5}^{}}{1.2-1}$

W = - 408.92 kJ

The magnitude of Work done is 408.92 kJ