Question

Pressure exerted by 1 mole of methane in a 0.25 litre container at 300K using vander Waal's equation (given a = 2.253 $atm{l}^{2}mo{l}^{-2}$, b = 0.0428 lit $mo{l}^{-1}$) is

• A. 82.82 atm
• B. 152.51 atm
• C. 190.52 atm
• D. 70.52 atm

Answer 1

The correct option is A 82.82 atm

$(P+\frac{n^{2}a}{V^2})$(V - nb) = nRT
$(P+\frac{2.253}{0.25\times 0.25})$(0.25 - 0.0428) = 0.0821 $\times$ 300
or (P + 36.048)(0.2072) = 24.63
$\Rightarrow$P + 36.08 = 118.87
$\Rightarrow$P = 82.82 atm