Derive (∂u/∂T)p = Cp - P(dv/dT)p using enthalpy

[ChronicQuantumAddict]

The question is as follows:

the partial derivative (given as a partial, but i don't know the notation, so letter d is really little delta for the partial)

(du/dT)p = Cp - P(Beta)v​

where Beta = expansivity coefficient = 1/v (dv/dT)p

again, all the "d's" are lowercase delta's for the partial derrivatives, and the "p's" next to the partials and the one with the Cp are to signify that pressure is constant.

I know i need to start with enthalpy, dh, but I am pretty much stuck. if someone would point me in the right direction i would be much obliged. thanks

 

[Astronuc]

It appears that one is trying to show the relationship:

(du/dT)p = Cp - P(Beta)v

or

$$(\frac{\partial u}{\partial T})_p = c_p - p\beta v$$

where

$$\beta = \frac{1}{v} (\frac{\partial v}{\partial T})_p$$


OK, how about starting with $$h = u + pv$$, or

$$u = h - pv$$

differentiating with respect to T at constant P,

$$(\frac{\partial u}{\partial T})_p = (\frac{\partial h}{\partial T})_p - (\frac{\partial (pv)}{\partial T})_p$$

and go from there remembering the definition of $c_p$ is

$$c_p = (\frac{\partial h}{\partial T})_p$$

 

[ChronicQuantumAddict]

thanks

duh, thank a lot. i see it clearly now. much thanks

 

[Astronuc]

I have those moments too.