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Assuming CO(2) to be van der Waals gas, ...

Assuming `CO_(2)` to be van der Waals gas, calculate its Boyle temperature. Given `a=3.59" L"^(2)" atm " mol^(-2)` and `b=0.0427" L "mol^(-1)`.

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The correct Answer is:
1024
Boyle temperature in terms of van der Waals constants is : `T_(B) = a/(Rb)` . Hence ,
`T_(B) = (3.59L^(2)"atm mol"^(-2))/((0.0821 " L atm L"^(-1)mol^(-1))(0.0427" L mol"^(-1)))=1024 K `.
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