$\begin{array}{1 1}17.6kJ\\18.6kJ\\19.6kJ\\20.6kJ\end{array} $

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Answer : 17.6kJ

$W_{1-2}=\int\limits_1^2 PdV$

$\qquad\;\;=C\large\frac{V_2^{-n+1}-V_1^{-n+1}}{-n+1}$

$\qquad\;\;=\large\frac{P_2V_2-P_1V_1}{1-n}$

The pressure at state 2 can be found using

$P_2V_2^n=P_1V_1^n$

$P_2=P_1\big(\large\frac{V_1}{V_2}\big)^n$

$\;\;\;\;\;=(3\times 10^5Pa)\big(\large\frac{0.1}{0.2}\big)^{1.5}$

$\;\;\;\;\;=1.06\times 10^5$Pa

Hence,$W=\big(\large\frac{(1.06\times 10^5Pa)(0.2m^3)-(3\times 10^5Pa)(0.1m^3)}{1-1.5}\big)$

$\Rightarrow 17.6kJ$

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