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Q.

The van der Waal’s constants for a gas are a = 3.6 atm L²mol^-2,b = 0.6L mol^-1. If R = 0.08L atm K^-1mol^-1 and the Boyle’s temperature (K) is T_B of this gas, then what is the value of $\frac{T_{B}}{15}$ ?

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answer is 5.

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Detailed Solution

$\begin{array}{l} T_{B} = \frac{a}{Rb} = \frac{3 .6}{0 .08 \times 0 .6} = 75 K, \\ \frac{75}{15} = 5 K \end{array}$

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The van der Waal’s constants for a gas are a = 3.6 atm L2mol-2,b = 0.6L mol-1. If R = 0.08L atm K-1mol-1 and the Boyle’s temperature (K) is TB of this gas, then what is the value of TB15 ?