Differential of X [ dX = U/V dV + U/p dp ] Internal Energy?

[says]

Homework Statement

dX = U/V dV + U/p dp

1. Write the differential of X in terms of the independent variables.
2. Prove that this is an exact differential.
3. Use the ideal gas equation of state to verify that X is actually the internal energy and that it satisfies the above equation. Would you expect the internal energy to be a state variable for the van der Waals gas?
   

Homework Equations

Internal Energy
dU = TdS - PdV

The Attempt at a Solution

dX = U/V dV + U/p dp ... I'm a bit too embarrassed to type anything that I've tried. I'm not really sure where to start.

dX = U / -V² + U / -p²
= - U / V² - U / p²

Partial Derivative Attempt

dX = U/V dV + U/p dp

dX = [ U / V dV ]_p + [ U / p dp ]_V

Any help to put me in the right direction would be much appreciated. Thanks

 

[Chestermiller]

If dX is going to be an exact differential, then a good place to start on this is to write down the chain rule expression for dX in terms of dP and dV. Then compare that equation with your starting differential equation.

Chet

 

[says]

dU / dX = dU/dV dU/dp - Chain Rule expression for dX in terms of dP and dV?

M =(V,U) dV N= (U,p) dp

( ∂U / ∂V = dV ) + ( ∂U / ∂p dp ) = dX

X (V,p)
∴ X = U (Internal Energy)

 

[Chestermiller]

dU / dX = dU/dV dU/dp - Chain Rule expression for dX in terms of dP and dV?

M =(V,U) dV N= (U,p) dp

( ∂U / ∂V = dV ) + ( ∂U / ∂p dp ) = dX

X (V,p)
∴ X = U (Internal Energy)

What I had in mind was:
$$dX=\left(\frac{\partial X}{\partial v}\right)_pdv+\left(\frac{\partial X}{\partial p}\right)_vdp$$
and setting U = U(p,v), so, if the equation represents an exact differential, then
$$\left(\frac{\partial X}{\partial v}\right)_p=\frac{U(p,v)}{v}$$
$$\left(\frac{\partial X}{\partial p}\right)_v=\frac{U(p,v)}{p}$$

Chet

 

[says]

Ok. Exact and partial differentials are still pretty new to me so I think I'll do a bit more reading and come back to this. Thank you for your reply!

 

[Chestermiller]

Ok. Exact and partial differentials are still pretty new to me so I think I'll do a bit more reading and come back to this. Thank you for your reply!

Yes. You need to be able to work with them to do many thermo analyses.

Chet

 

[says]

What I had in mind was:
$$dX=\left(\frac{\partial X}{\partial v}\right)_pdv+\left(\frac{\partial X}{\partial p}\right)_vdp$$
and setting U = U(p,v), so, if the equation represents an exact differential, then
$$\left(\frac{\partial X}{\partial v}\right)_p=\frac{U(p,v)}{v}$$
$$\left(\frac{\partial X}{\partial p}\right)_v=\frac{U(p,v)}{p}$$

Chet

Doesn't this mean that internal energy is not a state variable for the van der Waals gas?

van der Waals equation of state:

Ideal Gas Equation of State:

pV = nRT

 

[Chestermiller]

Internal energy is a state variable for all materials.

Chet