Pressure exerted by 1 mole of methane in a 0.25 L container at 300 K using van der Waals’ equation is $\left( Given,\ a = 2.253\ atm\ L^{2}\ mol^{- 2},\ b = 0.0428\ L\ {mol}^{- 1} \right)$

Question

Dr. Abubacker · Accepted Answer

(a)
From van der Waals’ equation,
$\left( p + \frac{n^{2}a}{V^{2}} ight)\ (V - nb) = RT$
$\left( p + \frac{2.253}{0.25 	imes 0.25} ight)(0.25 - 0.0428)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 0.0821 	imes 300$
$\ (p + 36.048)(0.2072) = 24.63$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p + 36.048 = 118.87\$
$p = 118.87 - 36.048 = 82.82\ atm$

Q. Pressure exerted by 1 mole of methane in a 0.25 L container at 300 K using van der Waals’ equation is $\left( Given,\ a = 2.253\ atm\ L^{2}\ mol^{- 2},\ b = 0.0428\ L\ {mol}^{- 1} \right)$

Explanation

Reviewed by Dr. Abubacker

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