# Problem Understanding The virial equation Volume Terms

recently i found virial equation for finding P and V for a gas,but i just don't understand what those V's are. $PV/RT=1+B/V+...\\$

are the V's in both sides of equation the same ? or they are different? our professor did this:

$PV/RT=1+BP/RT\\$

and then said $V-V(ig)=B\\$

the question is why he replaced V in the right side by $P/RT\\$

that V is for ideal gas?

or for calculating the work on it, he said we can use:

$dW=RT/V(1+B/V+C/V^2)dV\\$ as $dW=PdV\\$

are the dV and the V on the left side of the virial equation from a same source?

i appreciate your help,I'm a little confused.

An ideal gas has the equation of state:

$$\frac{PV}{nRT} = 1 \tag{1}$$

A real gas has an equation of state that is different to this, but since most gases are quite close to ideal we can write their equation of state by starting with the ideal gas equation and adding extra terms. Actually we usually rewrite equation (1) as :

$$\frac{P}{RT} = \rho$$

where $\rho = n/V$ is the molar density i.e. the number of moles per unit volume. Then the virial series is written as:

$$\frac{P}{RT} = \rho + B_2\rho^2 + B_3\rho^3 + \, ...$$

So for an ideal gas the virial coefficients $B_2$, $B_3$, etc are zero, while for real gases we expect them to be small. Hopefully the terms on the right hand side decrease rapidly so $1 \gg B_2\rho^2 \gg B_3\rho^3 \gg B_4\rho^4$ and so on, so we can get a good approximation by ignoring the cubic and higher terms, and that's what your professor has done. In that case the virial series simplifies to:

$$\frac{P}{RT} = \rho + B_2\rho^2 = \frac{n}{V} + B_2\left(\frac{n}{V}\right)^2$$

or multiplying both sides by $V$ and taking one mole of gas so $n = 1$ we get the equation you give:

$$\frac{PV}{RT} = 1 + B_2\frac{1}{V}$$

The $V$ in this equation is just the molar volume i.e. it's the volume of a mole of gas at the pressure $P$ and temperature $T$. It's the same $V$ on both sides of the equation.

When we're using the virial series we assume the gas is close to ideal, so the molar volume $V$ is close to the molar volume of an ideal gas:

$$V \approx \frac{RT}{P}$$

That's what your professor is doing when he substitutes for $V$ on the right hand side. This is an approximation, but then truncating the virial expansion at the $B_2$ term is also an approximation so the extra degree of approximation doesn't make a lot of difference.