Deriving the TdS Equation for Thermal Physics

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Elzair
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Homework Statement


Derive the following equation


Homework Equations


[tex]TdS = C_{V} \left( \frac{\partial T}{\partial P} \right)_{V}dP + C_{P} \left( \frac{\partial T}{\partial V} \right)_{P}dV[/tex]


The Attempt at a Solution



[tex]dU = \delta Q - \delta W[/tex]

[tex]\delta Q = TdS[/tex] for a closed system

[tex]C_{P} = T \left( \frac{\partial S}{\partial T} \right)_{P}[/tex]

[tex]C_{V} = T \left( \frac{\partial S}{\partial T} \right)_{V}[/tex]

I am not sure where to go from here.
 
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I think the trick here is playing around with the relations.

If you consider S as a function of p and v as independent variables, then
[tex]dS = \left(\frac{\partial S}{\partial P}\right)_V dP + \left(\frac{\partial S}{\partial V}\right)_P dV[/tex]

But, [tex]\frac{\partial S}{\partial P}_V = \left(\frac{\partial S}{\partial T}\right)_V \left(\frac{\partial T}{\partial P}\right)_V[/tex], and so on.