Question

The 'z' for 1 mole of van der Waals gas at $0^{o}C$ and $100$ atm pressure is found to be $0.5$. Assuming vol of a gas molecule is negligible find the value of 'a'.

Answer 1

Solution:-

As we know that, compressibility factor $\left(Z\right)$ of a gas is given by-

$Z=\frac{PV}{RT}$

whereas,

$Z=0.5\left(\text{Given}\right)$

$P=$ Pressure of the gas $=100atm$

$V=$ Volume occupied by the gas $=V\left(\text{say}\right)=?$

$R=$ Universal gas constant $=0.0821Latm{K}^{-1}{mol}^{-1}$

$T=$ Temperature of the gas $=0℃=273K$

$\therefore 0.5=\frac{100\times V}{0.0821\times 273}$

$\Rightarrow V=\frac{0.5\times 22.413}{100}=0.112L$

Further when volume of a gas molecule is negligible, van der Waal’s equation becomes-

$(P+\frac{na}{V^2})(V-0)=nRT$

Since given that no. of moles is 1.

$\therefore (100+\frac{1\times a}{{\left(0.112\right)}^2})\times 0.112=1\times 0.0821\times 273$

$\Rightarrow \frac{1.254+a}{{\left(0.112\right)}^2}\times 0.112=22.413$

$\Rightarrow 1.254+a=22.413\times 0.112$

$\Rightarrow a=2.51-1.254$

$\Rightarrow a=1.25atm{L}^2{mol}^{-2}$

Hence, $a=1.25atm{L}^2{mol}^{-2}$.