How Does Compression Work Differ for Van der Waals Gas vs. Ideal Gas?

[corr0105]

One mole of a van der Waals gas is compressed quasi-satically and isothermally from volume V1 to V2. For a van der Waals gas, the pressure is:
p= RT/(V-b)-a/V^2
where a and b are material constants, V is the volume and RT is the gas constant x temperature.

For the first part of the problem I was supposed to write an expression for the work done. According to the equation ∂w=-pdV (where w=work, p=pressure, and V=volume) we can solve the equation for work by integrating the pressure equation from V1 to V2. Doing this, we get:
w=RTln((V_1-b)/(V_2-b))+a(1/V_1 -1/V_2 )

The second part of the question asks: Is more or less work required than for an ideal gas in the low-density limit? What about the high-density limit? Why?
Basically, I don't understand what the second part of the question is asking. Any help would be much appreciated! Thanks.

(Sorry, I tired to format the equations in Microsoft equation editor first so they'd look normal, but it didn't work.. I don't know how to do that on here :-/ )

 

[member 553083]

Answer: In the low-density limit, a van der Waals gas requires less work than an ideal gas due to the attraction between molecules. At high densities, a van der Waals gas requires more work than an ideal gas because of the repulsion between molecules. This is due to the terms 'a' and 'b' in the equation for pressure, which represent the attraction and repulsion between molecules in the van der Waals gas.