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Equation of the gas state processes

Problem 2.23

One mole of a certain gas is contained in a vessel of volume V=0.250V=0.250 l. At a temperature T1=300 KT_{1}=300 \mathrm{~K} the gas pressure is p1p_{1} =90=90 atm, and at a temperature T2=350 KT_{2}=350 \mathrm{~K} the pressure is p2p_{2} =110=110 atm. Find the Van der Waals parameters for this gas.

Reveal Answer
a=V2(T1p2−T2p1)/(T2−T1)=185 atm. 12/mol2,b==V−R(T2−T1)/(p2−p1)=0.042l/mol\begin{aligned} &\text {} a=V^{2}\left(T_{1} p_{2}-T_{2} p_{1}\right) /\left(T_{2}-T_{1}\right)=185 \text { atm. } 1^{2} / \mathrm{mol}^{2}, b=\\ &=V-R\left(T_{2}-T_{1}\right) /\left(p_{2}-p_{1}\right)=0.042 \mathrm{l} / \mathrm{mol} \end{aligned}
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