Question

The virial equation for 1 mole of a real gas is written as:
PV=RT

[1+$\frac{A}{V}+\frac{B}{V^2}+\frac{C}{V^3}+$................to the higher power of n]

Where A, B and C are known as virial coefficients. If Vander Waal's equation is written in virial form, then what will be the value of B?

• A. $a-\frac{b}{RT}$
• B. $b^3$
• C. $b-\frac{a}{RT}$
• D. $b^2$

Answer 1

Answer: (C)

$b-\frac{a}{RT}$

A virial equation is an equation of state of gases that has additional terms beyond that for an ideal gas, which account for the interactions between the molecules.

The equation of state for a van der Waals gas is:

$(p+\frac{a}{V_m^2})(V_m-b)=RT$

This leads to the following virial expansion:

$\frac{p{V}_m}{RT}+1+\frac{B}{V_m}+\frac{C}{V_m^2}+-----\left(1\right)$

The van der Waals equations of state can be rewritten as:

$pV=\frac{RT}{V-b}+\frac{a}{V^2}=\frac{RT}{V}{(1-\frac{b}{V})}^{-1}+\frac{a}{V^2}$

and using the binomial expansion, the terms in brackets can be expanded into a series, resulting in

$\frac{pV}{RT}=1+\frac{1}{V}(b-\frac{a}{RT})+{\left(\frac{b}{V}\right)}^2+{\left(\frac{b}{V}\right)}^3+----,$

which is in the same form as the virial expansion is equation (1) with

$B\left(T\right)=b-\frac{a}{RT}$

If the Van der Waal's equation is written in the virial form then the value of $B=b-\frac{a}{RT}$.

Hence, option C is correct.