6. An ideal gas was expanded in such a way that during the process, the pressure of the gas turned out to be directly proportional to its volume. As a result, the gas heated up by $\Delta T=100^{\circ} \mathrm{C}$, and the work done by the gas was $A=831$ J. Determine the amount of substance that participated in this process. The gas constant $R=8.31$ J/mol$\cdot$K. (15 points)
Answer: 2 moles

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 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: $p V=\alpha V V=\alpha V^{2}=\nu R T \quad(4$ points). As a result, we get: $A=\frac{\alpha}{2} \cdot\left(V_{2}^{2}-V_{1}^{2}\right)=\frac{1}{2} v R \Delta T \quad(2$ points $) . \quad$ Final answer: $\quad v=\frac{2 A}{R \Delta T}=\frac{2 \cdot 831}{8.31 \cdot 100}=2$ moles (2 points).