$p_{1} v_{1}^{\gamma}=p_{2} v_{2}^{\gamma}$
$(200)(1200)^{1.5}=P^{2}(300)^{1.5}$
$P_{2}=200[4]^{3 / 2}=1600 {kPa}$
$\mid$ W.D. $\mid=\frac{{p}_{2} {v}_{2}-{p}_{1} {v}_{1}}{v-1}=\left(\frac{480-240}{0.5}\right)=480 {J}$



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The correct option ($s$) is (are)
$(A)$ $q_{A C}=\Delta U_{B C}$ and $W_{A B}=P_2\left(V_2-V_1\right)$
$(B)$ $\mathrm{W}_{\mathrm{BC}}=\mathrm{P}_2\left(\mathrm{~V}_2-\mathrm{V}_1\right)$ and $\mathrm{q}_{\mathrm{BC}}=\mathrm{H}_{\mathrm{AC}}$
$(C)$ $\Delta \mathrm{H}_{\mathrm{CA}}<\Delta \mathrm{U}_{\mathrm{CA}}$ and $\mathrm{q}_{\mathrm{AC}}=\Delta \mathrm{U}_{\mathrm{BC}}$
$(D)$ $\mathrm{q}_{\mathrm{BC}}=\Delta \mathrm{H}_{\mathrm{AC}}$ and $\Delta \mathrm{H}_{\mathrm{CA}}>\Delta \mathrm{U}_{\mathrm{CA}}$