Question

A 10.64 mol sample of methane gas is maintained in a 0.8073 L container at 301.3 K. What is the pressure in atm calculated using the Van der waal’s equation for CH4 gas under these conditions? For CH4 a = 2.253L ^2 atm/mol^2 and b = 4.278 x 10^-2 L/mol

Answer 1

Ans:-

Given $V=0.8073 L$ $T=301.3 K$ $n=10.64 mol$ $P=?$

$a= 2.253 \dfrac{L^2 atm}{mol^2}$ $b = 4.278\times10^{-2} \dfrac{L}{mol}$

We know that the Vander waal's equation for $CH_4$ gas

$P=\dfrac{n\times{R}\times{T}}{V-n\times{b}}-\dfrac{a\times{n^2}}{V^2}$

$\Rightarrow$ $P=\dfrac{10.64\times{0.083}\times{301.3}}{0.8073-10.64\times{0.04278}}-\dfrac{2.253\times{10.64^2}}{0.8073^2}$

$\Rightarrow$ $P=755.6628-391.3582=364.304atm$

Therefore the pressure of gas $CH_4$ is $364.304$ atm