Expert: Josh - 5/13/2009

Question
QUESTION: Dear Sir,

I am trying to calculate dV/dp from this equation: p(V-nb)-nRT=0. How would you do it?

Thank you.

ANSWER: Dear Ignacio,

First, may I suggest we rewrite the equation, separating the two variables "p" and "V".

(V-nb)=nRT/p
V=nRT/p+nb

Then, we differentiate both sides with respect to p. In this context, we treat "nb" as a constant (only p is the variable of interest).

Using the standard result for the first derivative of a polynomial, i.e., d[x^a]/dx = a*x^(a-1)

dV/dp = nRT * d/dp[p^(-1)] + 0
     = -nRT/(p^2)

This indicates that for an ideal gas, the rate of change in volume with respect to pressure is proportional to the inverse square (with temperature held constant). The negative sign means an increase in pressure corresponds to a decrease in volume. I hope you enjoy your chemistry!

Regards,
Josh

---------- FOLLOW-UP ----------

QUESTION: Dear Josh,

My book says that the answer is -((V-nb)^2)/nRT but I guess my book is wrong and that is why I thought I could not solve it myself.

Thank you very much.

Answer
Hi Ignacio,

Don't forget, we also know p=nRT/(V-nb) from the very beginning.
So, we can substitute for p in dV/dp=-nRT/(p^2).

Replacing p with nRT/(V-nb), we get dV/dp=-(V-nb)^2/(nRT) after some cancellation. This is the same as the model answer. Just two different ways of writing it.

Cheers:)