Proving a thermodynamic relationship

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Potatochip911
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Homework Statement


Prove that ##TdS = C_vdT + \alpha T / \kappa dV##

Homework Equations


##T dS = dU - pdV##
##\alpha = \frac{1}{v}\left(\frac{\partial v}{\partial T}\right )_P##
##\kappa = -\frac{1}{v}\left(\frac{\partial v}{\partial P}\right)_T##

The Attempt at a Solution



The ##C_vdT## part is quite easy since for a constant volume process ##dU = C_vdT## but I can't seem to figure out how to get the second part of the expression. After multiplying by forms of 1 I end up with $$-pdV = \frac{\alpha\left(\frac{\partial v}{\partial P}\right)_T}{\kappa \left(\frac{\partial v}{\partial T}\right)_P}PdV$$, now using the cyclical rule here doesn't seem logical since that would introduce a negative so it seems like I need to replace the pressure P with something although I'm not sure what relation I can use to do that.
 
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You've probably figured it out over the past few days, but for one thing, you've got a sign problem: ##T\,dS=dU+p\,dV## because ##p## is compressive stress.
 
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Your mistake is that dU is not equal to ##C_vdT##. That is only correct for an ideal gas. In general, $$dS=\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV=\frac{C_vdT}{T}+\left(\frac{\partial S}{\partial V}\right)_TdV$$
From one of the Maxwell relationships, $$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$$Therefore, $$dS=\frac{C_vdT}{T}+\left(\frac{\partial P}{\partial T}\right)_VdV$$So, $$TdS=C_vdT+T\left(\frac{\partial P}{\partial T}\right)_VdV$$
 
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Chestermiller said:
Your mistake is that dU is not equal to ##C_vdT##. That is only correct for an ideal gas.

At constant volume, ##dU=C_V\,dT## for all materials, as Potatochip911 noted.
 
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