Question
Assuming $\mathrm{CO}_2$ to be van der Waal's gas, calculate its Boyle temperature.

Answer

$\text{T}_\text{b}=\frac{\text{a}}{\text{Rb}}$
$=\frac{3.59\text{L}^2\text{atm mol}^{-2}}{(0.082\text{L atm K}^{-1}\text{mol}^{-1})(0.0427\text{L mol}^{-1})}$
$=1025.3\text{K}$

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