- #1

kntsy

- 82

- 0

i actually posted it in physics section but no one knows so i think engineering experts would be stronger to this kind of equations.

1.

[tex]\color{blue}dq=dU+PdV[/tex]

so

[tex]C_V = \left(\frac {\partial q}{\partial T}\right)_V= \left(\frac {\partial U}{\partial T}\right)_V[/tex]

leads to

[tex]dU=C_{V}dT[/tex]

so

[tex]\color{red}\left(\frac {\partial U}{\partial V}\right)_T = 0[/tex]

meaning that internal energy of NONideal gas is a sole function of T?

2.

As

[tex]\color{blue}dq=dU+PdV[/tex]

so

[tex]\left(\frac {\partial U}{\partial V}\right)_T= \left(\frac {\partial V}{\partial T}\right)_P \left(C_P-C_V\right) - P[/tex]

why? even for ideal gas the change in volume results in change in internal energy:

[tex]\color{red}\left(\frac {\partial U}{\partial V}\right)_T \not= 0??[/tex]

thanks for answering.