Isothermal Work Calculation for an Imperfect Gas

[RJLiberator]

Homework Statement

An imperfect gas obeys the equation
(p+\frac{a}{V^2_m})(V_m-b)=RT
where a = 8*10^(-4)Nm^4mol^(-2) and b=3*10^(-5)m^3mol^(-1). Calculate the work required to compress 0.3 mol of this gas isothermally from a volume of 5*10^(-3)m^3 to 2*10^(-5)m^3 at 300K.

Homework Equations

This is an isothermal work calculation so I should be able to use

W = -\int_a^b pdV\

The Attempt at a Solution

I feel like I have a good solution, however, when I plug the integral calculation into Wolfram alpha, I get arctanh and imaginary numbers meaning I must have done something wrong.

1. I take the equation giving and set it equal to p
I tried both of these for p:

p=\frac{RT-\frac{a}{V_m}+\frac{ab}{V^2_m}}{(V_m-b)}
or
p=\frac{RT}{(V_m-b)}-\frac{a}{V^2_m}

These two should equal the same thing, I just did different algebraic manipulations to get there. Either way, both do not work.

2. I put p into the integral from the desired volumes and compute using wolfram alpha.

-\int_{5*10^{-3}}^{2*10^{-5}} \frac{RT}{(V_m-b)}-\frac{a}{V^2_m} dx \

Now, I use R = 8.314, T = 300. I get imaginary numbers everytime. Cauchy Principal value turned out to be -15445.6 if that means anything.

Any idea what I'm doing wrong?

 

[TSny]

-\int_{5*10^{-3}}^{2*10^{-5}} \frac{RT}{(V_m-b)}-\frac{a}{V^2_m} dx \

What does x represent in dx? Note that $V_m$ is molar volume (not volume of the gas).

 

[RJLiberator]

Wopps, that is meant to be dV.

Ah, molar volume, to handle this I simply multiply by 0.3 everywhere I see a V_m?

V_m = V/n where n= 0.3mol in this case, so we can have 0.3V_m everywhere in the integral.

 

[TSny]

You could replace each $V_m$ in the integral by $V/n$ so that you can then integrate with respect to the volume V.

Or, you could leave the $V_m$'s in place and let $V_m$ be the integration variable by expressing $dV$ in terms of $dV_m$ and changing your limits to molar volumes.

 

[RJLiberator]

BOOM, that did it. So my mess up was the molar volume... ah. Thank you kindly for your help, @TSny.