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Liquefaction of Gases and Critical Constants

The van der W...

Question

The van der Waals’ constants for a gas are; $a = 3.6 {atm L}^{2} {mol}^{2}$ , $b = 0.6 {L mol}^{- 1}$ . If $R = 0.08 {L atm K}^{- 1}$ . If the Boyle's temperature $(K)$ of this gas is $T_{B}$ , then what is the value of $\frac{T_{B}}{15}$ ?

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Solution

$T_{B} = \frac{a}{R b} \Rightarrow \frac{T_{B}}{15} = \frac{a}{15 R b} = \frac{3.6}{15 \times 0.08 \times 0.6} = 5$

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