Need help understanding this thermo. derivation

[mdawg467]

Im trying to understand this concept of deriving an expression for ($\frac{\partial T}{\partial P}$)_s in terms of T, V, C_p, $\alpha$, and $\kappa$_T

($\frac{\partial T}{\partial P}$)_s is evaluated by measuring the temperature change and the specific volume change accompanying a small pressure change in a reversible adiabatic process.

I attached the derivation that the book does, but I cannot follow it. Any help would be greatly appreciated!

Thanks

 

[qbert]

There are a lot of steps in that simple looking derivation. Where are you lost?

 

[Philip Wood]

Don't understand the notation in your thumbnail, but append my derivation, expressed in ordinary partial derivative notation.

 

[mdawg467]

Don't understand the notation in your thumbnail, but append my derivation, expressed in ordinary partial derivative notation.

Thanks dude, I was getting lost with the notation as well. This helps out big time.. Not sure what the book was doing lol.

 

[sardar]

for reaching out for help understanding this thermo derivation. I can definitely assist you with this concept. Let's break down the derivation step by step to make it easier to understand.

First, we start with the definition of entropy (S) in terms of temperature (T) and pressure (P), which is given by:

dS = (\frac{C_p}{T})dT - (\frac{\alpha}{\T})dP

Where C_p is the heat capacity at constant pressure, \alpha is the coefficient of thermal expansion, and \kappa_T is the isothermal compressibility.

Next, we use the Maxwell relation to relate the partial derivatives of S with respect to T and P:

(\frac{\partial S}{\partial T})_P = (\frac{\partial S}{\partial P})_T * (\frac{\partial P}{\partial T})_S

Substituting this into our entropy equation, we get:

dS = (\frac{C_p}{T})dT - (\frac{\alpha}{\T})dP = (\frac{\partial S}{\partial T})_P dT + (\frac{\partial S}{\partial P})_T dP

Now, we can equate the coefficients of dT and dP on both sides to get:

(\frac{C_p}{T})dT = (\frac{\partial S}{\partial T})_P dT

and

-(\frac{\alpha}{\T})dP = (\frac{\partial S}{\partial P})_T dP

We can then rearrange these equations to get:

(\frac{\partial S}{\partial T})_P = \frac{C_p}{T}

and

(\frac{\partial S}{\partial P})_T = -\frac{\alpha}{\T}

Next, we use the definition of enthalpy (H) in terms of temperature and pressure:

dH = (\frac{C_p}{T})dT + VdP

where V is specific volume.

We can then use the same Maxwell relation to relate the partial derivatives of H with respect to T and P:

(\frac{\partial H}{\partial T})_P = (\frac{\partial H}{\partial P})_T * (\frac{\partial P}{\partial T})_S

Substituting this into our enthalpy equation, we get:

dH = (\