QJEEM16Ch9a9b: JEE Main 2016 Chemistry - Class 11 JEEM16Ch9a9b Solution: At very high pressures, the compressibility factor of one mole…

According to van der Waals' equation, for one mole of a gas

$\left( {p + {a \over {{V^2}}}} \right)(V - b) = RT$ ...... (1)

At high pressure, ${a \over {{V^2}}}$ can be neglected

So, $p + {a \over {{V^2}}} \approx p$ ...... (2)

From Eqs. (1) and (2), we get

$p(V - b) = RT$

$pV - pb = RT$

$pV = RT + pb$

Dividing both the sides by RT, we get

${{pV} \over {RT}} = 1 + {{pb} \over {RT}}$

$Z = 1 + {{pb} \over {RT}}$