- #1
RJLiberator
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Homework Statement
An imperfect gas obeys the equation
[tex](p+\frac{a}{V^2_m})(V_m-b)=RT[/tex]
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
[tex]W = -\int_a^b pdV\[/tex]
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:
[tex]p=\frac{RT-\frac{a}{V_m}+\frac{ab}{V^2_m}}{(V_m-b)}[/tex]
or
[tex]p=\frac{RT}{(V_m-b)}-\frac{a}{V^2_m}[/tex]
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.
[tex] -\int_{5*10^{-3}}^{2*10^{-5}} \frac{RT}{(V_m-b)}-\frac{a}{V^2_m} dx \ [/tex]
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?