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vander Waal's constants a and b for ${CO}_{2}$ gas are
3.592 atm $L^{2} {mol}^{- 2}$ and $0.0427 L {mol}^{- 2}$ respectively.
The Boyle's temperature for the gas will be:

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$1000 K$

$84.12 K$

$1024.6 K$

$373 K$

answer is D.

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Detailed Solution

Boyle's temperature,

$\begin{array}{l} T_{b} = \frac{a}{bR} \\ = \frac{3.592 {atmL}^{2} {mol}^{- 2}}{0.0427 L {mol}^{-} \times 0.0821 {LatmK}^{-} {mol}^{-}} \\ or T_{b} = 1024.6 K \\ [∵ R = 0.0821 {LatmK}^{-} mol] \end{array}$

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