Molar volume of gas in function of temperature and pressure

1. Feb 19, 2013

damabo

1. The problem statement, all variables and given/known data

Given are two relations for the molar volume. Are they possible? If so, give the formula for v in function of P and T.
a) dv =R/P dT - RT/P² dP
b) dv = 2R/P dT - RT/2P² dP

2. Relevant equations

3. The attempt at a solution

If I integrate dv I get ∫R/P dT - ∫RT/P² dP= RT/P + RT/P (in case a) and 5/4 * RT/P (in case b).
does this mean they are both 'possible'?

intuitively I would say only a is possible

another thing that struck me - probably resulting from some kind of error I made- was the following discrepancy:
say v=RT/P then dv=dv/dT dT + dv/dP dP = R/P dT - RT/P² dP.
So according to this v=RT/P might well be the solution to the integral ∫dv (in the case of a).

2. Feb 20, 2013

ehild

The molar volume v is function of P and T: v(T,P). If its first and second partial derivatives exist and continuous its differential is

dv=∂v/∂T dT + ∂v/∂P dp.

an the mixed second partial derivatives are equal:

∂(∂v/∂T)∂P=∂(∂v/∂P)∂T.

The integral of dv is independent of he path taken, v is a "potential", only if that condition holds.

In case of the first example, ∂v/∂T=R/P and ∂v/∂P=-RT/P2. The mixed derivatives are equal.

Now you have ∂v/∂T=R/P, and integrate with respect to T: V=RT/P + integration constant. But that constant can depend on P, so v(T,P)=R/P+f(P). You can find f(P) from the condition that the derivative of v with respect to P has to be -RT/P2:
∂v/∂P= -RT/P2+df/dP=-R/P2, so f=constant.

Check if the other dv can be the perfect differential of a potential function.

ehild