At constant temperature and pressure, the compressbility factor (Z) for one mole of a van der waals gas is 0.5. if the volumes of the gas molecules are considered to negligible, then show that `a=(1)/(2)V_(m)RT`, where `V_(m)` and T are the molar volume and temperature of the gas respectively.

Explanation

We know, `Z=(PV)/(nRT), ` given `Z=0.5` and n=1
`therefore PV=0.5 RT=(1)/(2)RT` . . [1]
The equation of state for 1 mol of a van der waals gas is,
`(P+(a)/(V_(m)^(2)))(V_(m)-b)=RT` [`V_(m)`=molar volume]
If volumes of the molecules are considered to be negligible, as per given condition, then `V_(m)-b~~V_(m)`.
`therefore (P+(a)/(V_(m)^(2)))V_(m)=RT` or, `PV_(m)+(a)/(V_(m))=RT)` . . . [2]
Again, from equation [1] and we get, `PV=(1)/(2)RT`
For 1 mol (n=1) of the gas, `V=V_(m)` (molar volume)
Therefore, `PV_(m)=(1)/(2)RT)` . . [3]
From equation [2] and [3] we have,
`(1)/(2)RT+(a)/(V_(m))=RT` or, `a=(1)/(2)V_(m)RT`