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JEE CHEMISTRY MCQ: STATES OF MATTER

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Application +4 | -1

Boyle's temperature of helium is, given van der Waals' constant

\mathrm{a}=0.0351 \mathrm{~atm} \mathrm{~L}^2 \mathrm{~mol}^{-2}

and

b=0.024 \mathrm{~L} \mathrm{~mol}^{-1}

View answer and explanation

Correct option: A

18 \mathrm{~K}

Option A (

18 \mathrm{~K}

) is correct because it is calculated using the formula

T_b = \frac{a}{Rb}

with the given van der Waals constants.

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