Show that

a$)$

$(\frac{\text{d}\text{e}\text{l}\text{U}}{\text{d}\text{e}\text{l}\text{T}}{)}_P=C_P-{}_{P}PV$

b$)$

$(\frac{\text{d}\text{e}\text{l}\text{U}}{\text{d}\text{e}\text{l}\text{T}}{)}_P=C_V+\frac{{}_P}{{}_T}({}_{P}T-{}_{T}P)V$

Note $-$ these are two different "ending" points, which may require two different

"starting" points.

Hint $-$ remember the definitions of $C_P,{}_P$ and ${}_T.$ You might want to work backwards!

c$)$ What does the equation a reduce to for an ideal gas? $($Hint $-$ you might want to

substitute the derivative definition of ${}_P$ back in$.)$