A text problem

1. Jun 3, 2003

PrudensOptimus

I tried to solve it by knowing that a, b, n, and R are constants, so only V, T are variables.

So I did this:

P = nR(dT/d(V-nb)) - ((an^2)*(-2V^-3))

but I still didn't get the correct answer. I believe I did something wrong, could someone help me out?

2. Jun 3, 2003

KLscilevothma

The question says "If gas in a cylinder is maintained at a constant temperature T". So I don't think T is a variable. Does the answer contain somthing like dT/dV? I don't think so because T isn't a variable.

Last edited: Jun 3, 2003
3. Jun 3, 2003

PrudensOptimus

not only does the answer including T, it has a, n, in it too.

4. Jun 3, 2003

KLscilevothma

-nRTV/(V-nb)2 + (2an2)/V3 ?

5. Jun 3, 2003

KLscilevothma

If T isn't a constant but a variable, I would expect (dT/dV) as part of the answer. (chain rule)

By the way, remember you need to use quotient rule when differentiate (nRT/(V - nb)) with respect to V as V is in the denominator

Last edited: Jun 3, 2003
6. Jun 3, 2003

plus

dT/dV will not be in the answer, as T is assumed to be constant, so therefore does not depend upon V.

7. Jun 3, 2003

PrudensOptimus

Yep how did you get that?

8. Jun 3, 2003

KLscilevothma

T is constant in this question

dP/dV
=d/dV [nRT/(V - nb) - an2/V2]
=d/dV [(nRT/(V - nb)] - d/dV (an2/V2)
now take all the constants out to the left hand side of d/dV
=nRT*d/dV [1/(V-nb)] - an2* d/dV (1/V2) .......................(1)

The blue part:
[1/(V-nb)] = (V-nb)-1
d/dV [1/(V-nb)] = -1*(V-nb)-2 = - 1/(V-nb)2
(the power rule)

Alternately,
d/dV [1/(V-nb)]
= [(V-nb)d/dV (1) - 1*d/dV (V-nb)]/(V-nb)2
(the quotient rule)
= (0-1)/(V-nb)2
= - 1/(V-nb)2

the green part
d/dV (1/V2)
= -2V-3

I think you can do it because you got it right in your first post

Substitute the blue part and green part back to (1), then you'll get the answer.

9. Jun 4, 2003

PrudensOptimus

awesome!!!:)

10. Jun 4, 2003

heisenberg

Is that yoda guy smart or what? WOW