Deriving Maxwell's Relation with Thermodynamic Variables

[mdwerner]

Homework Statement

Derive the following general relation : $$\left(\frac{\partial T}{\partial V}\right)_{S} = - \frac{1}{C_{V}} T \left(\frac{\partial p}{\partial T}\right)_{V}$$

Homework Equations

Maxwell's Relation : $$\left(\frac{\partial T}{\partial V}\right)_{S} = - \left(\frac{\partial p}{\partial S}\right)_{V}$$

The Attempt at a Solution

The similarity of the question to the maxwell's relation is the only thing I could recognize - but I don't see what to do? Any suggestions will be appreciated.

 

[Mindscrape]

First try writing out the differentials for T and 1/Cv to see if that gives you any inspiration.

 

[mdwerner]

I don't understand, what do you mean by the differentials of T and 1 / C_V ? A derivative? if so, with respect to what?
One thing I had considered would be to multiply this function by $$\frac{\partial S}{\partial T}$$ but that seems to jumble the left side of the equation...

 

[Mindscrape]

T is $$\frac{\partial U}{\partial S}$$

and

1/Cv is $$\frac{\partial T}{\partial U}$$

 

[paranormal]

I would approach this problem by first understanding the given relation and the variables involved. The given relation is a form of Maxwell's relation, which describes the relationship between the partial derivatives of temperature and volume at constant entropy, and the partial derivative of pressure with respect to entropy at constant volume. This relation is important in thermodynamics as it allows us to relate different thermodynamic variables and properties.

To derive the given relation, we can start with the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In mathematical form, this can be written as:

dU = dQ - pdV

Where dU is the change in internal energy, dQ is the heat added, p is the pressure, and dV is the change in volume.

We can also express the internal energy in terms of temperature and entropy using the fundamental thermodynamic relation:

dU = TdS - pdV

Combining these two equations, we get:

dQ = TdS + pdV

Now, we can take the partial derivative of this equation with respect to volume at constant entropy:

\left(\frac{\partial Q}{\partial V}\right)_{S} = T \left(\frac{\partial S}{\partial V}\right)_{S} + p

We can also express the heat added in terms of temperature and pressure using the definition of heat capacity at constant volume, C_V:

dQ = C_VdT

Substituting this into the previous equation, we get:

C_VdT = T \left(\frac{\partial S}{\partial V}\right)_{S} + p

Rearranging this equation and using the definition of entropy, S = \frac{Q}{T}, we get:

T \left(\frac{\partial S}{\partial V}\right)_{S} = -C_VdT + p

Finally, we can use the definition of pressure in terms of temperature and volume, p = \left(\frac{\partial U}{\partial V}\right)_{S}, to substitute for p in the equation above:

T \left(\frac{\partial S}{\partial V}\right)_{S} = -C_VdT + \left(\frac{\partial U}{\partial V}\right)_{S}

Now, we can take the partial derivative of this