6. One mole of an ideal gas was expanded so that during the process, the pressure of the gas turned out to be directly proportional to its volume. In this process, the gas heated up by $\Delta T=100{ }^{\circ} \mathrm{C}$. Determine the work done by the gas in this process. The gas constant $R=8.31$ J/mol$\cdot$K. (15 points)
Answer: 415.5 J

Solution. From the condition: $p=\alpha V$ (3 points). The work of the gas is equal to the area under the graph of the given process, constructed in the coordinates $p-V$:
$A=\frac{p_{1}+p_{2}}{2} \cdot\left(V_{2}-V_{1}\right)=\frac{\alpha V_{1}+\alpha V_{2}}{2} \cdot\left(V_{2}-V_{1}\right)=\frac{\alpha}{2} \cdot\left(V_{2}^{2}-V_{1}^{2}\right) \quad(4$ points). From the equation of state of an ideal gas: $p V=\alpha V V=\alpha V^{2}=v R T \quad$ (4 points). As a result, we get the final answer:

$A=\frac{\alpha}{2} \cdot\left(V_{2}^{2}-V_{1}^{2}\right)=\frac{1}{2} v R \Delta T=\frac{1}{2} \cdot 1 \cdot 8.31 \cdot 100=415.5$ Joules (4 points).