a) Do diagrama da figura B encontramos para os pontos $i$ e $f$:\n$\left\{\begin{array}{l} P_{i}=10,0 \times 10^{4} \mathrm{~Pa} \\ V_{i}=12,0 \times 10^{4} \mathrm{~m}^{3} \\ T_{i}=\left(\theta_{i}\left({ }^{\circ} \mathrm{C}\right)+273\right) \mathrm{K}=(7+273) \mathrm{K}=280 \mathrm{~K} \end{array}\right.$ e $\left\{\begin{array}{l} P_{f}=7,0 \times 10^{4} \mathrm{~Pa} \\ V_{f}=15,0 \times 10^{4} \mathrm{~m}^{3} \end{array}\right.$ \nUsando a Lei dos gases perfeitos:\n$\frac{P V}{T}=n R=c t e \Rightarrow \frac{P_{i} V_{i}}{T_{i}}=\frac{P_{f} V_{f}}{T_{f}} \Rightarrow T_{f}=\left(\frac{P_{f}}{P_{i}}\right)\left(\frac{V_{f}}{V_{i}}\right) T_{i}=\left(\frac{7}{10}\right) \times \left(\frac{15}{12}\right) \times 280 \mathrm{~K}=\frac{7}{2} \times \frac{280}{4} \mathrm{~K}=245 \mathrm{~K}$\nAssim: $\theta_{2}\left({ }^{\circ} \mathrm{C}\right)=245-273 \Rightarrow \theta_{2}=-28 { }^{\circ} \mathrm{C}$.\n$W_{i f}$ é numericamente igual a $A$.\n$$ W_{i f}=(42+8) \times 0,5 \times 10^{4} \times 1,0 \times 10^{4}=2,5 \times 10^{9} \mathrm{~J} $$