Show that the Internal Energy of an Ideal gas is a Function of T only

[knowlewj01]

Homework Statement

Show that internal energy U = U(T) only for an ideal gas who'se equation of state is:

P(V-b) = RT

(the claussius equation for n moles of gas)

Homework Equations

Thermodynamic Equation of state:

\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_V - P

The Attempt at a Solution

so, basically we need to prove that for this gas the following criteria are met:

\left(\frac{\partial U}{\partial V}\right)_T = \left(\frac{\partial U}{\partial P}\right)_T = 0

internal energy U does not change with respect to P or V.

using the equation of state of the gas, we can differentiate P with respect to T at constant V:

\left(\frac{\partial P}{\partial T}\right)_V = \frac{R}{V-b}

now substitute this into the thermodynamic equation of state:

\left(\frac{\partial U}{\partial V}\right)_T = \frac{RT}{V-b} - P

and from the equation of state of the gas, we can obtain P:

P = \frac{RT}{V-b}

so the result is that \left(\frac{\partial U}{\partial V}\right)_T = 0

thats half the work done, I am not sure how to prove \left(\frac{\partial U}{\partial P}\right)_T = 0

any pointers? Thanks

 

[knowlewj01]

Just read some more notes on this, does the fact that

\left(\frac{\partial U}{\partial V}\right)_T = 0

imply that:

\left(\frac{\partial U}{\partial P}\right)_T = 0

if so, why is this? i can't see why

 

[knowlewj01]

starting from the thermodynamic equation of state:

\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_V - P

rearrange it to get \left(\frac{\partial U}{\partial P}\right)_T

dU = T\left(\frac{\partial P}{\partial T}\right)_V dV - P dV

\left(\frac{\partial U}{\partial P}\right) = T\left(\frac{\partial P}{\partial T}\right)_V \left(\frac{\partial V}{\partial P}\right)_T - P\left(\frac{\partial V}{\partial P}\right)_T

now using the equation of state of the gas find dp/dt and dv/dp

\left(\frac{\partial P}{\partial T}\right)_V = \frac{R}{V-b}

\left(\frac{\partial V}{\partial P}\right)_T = -\frac{RT}{P^2}

substitute back in:

\left(\frac{\partial U}{\partial P}\right)_T = - \left(\frac{RT}{P^2}\cdot \frac{RT}{V-b}\right) - \left(-\frac{RT}{P}\right)

\frac{RT}{V-b} = P

so

\left(\frac{\partial U}{\partial P}\right)_T = -\frac{RT}{P} + \frac{RT}{P} = 0