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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Similar questions

Q. The van der Waals’ constants for a gas are;

a = 3.6 {atm L}^{2} {mol}^{2}

b = 0.6 {L mol}^{- 1}

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T_{B}

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\frac{T_{B}}{15}

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The van der Waals’ constants for a gas are; a=3.6 atm L2 mol2, b=0.6 L mol−1. If R=0.08 L atm K−1. If the Boyle's temperature (K) of this gas is TB, then what is the value of TB15?

TB=aRb⇒TB15=a15Rb=3.615×0.08×0.6=5

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 $\frac{T_{B}}{15}$ ?

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