Question

Show that the difference of heat capacities for a substance is given by the relation:

Cp-Cv= -T(¶V/¶T)^2 p x(¶p/¶V)T?

Answer 1

Answer on Question 50672, Physics, Molecular Physics | Thermodynamics

Question:

Show that the difference of heat capacities for a substance is given by the relation:

$$C _ {P} - C _ {V} = - T \left(\frac {\partial V}{\partial T}\right) _ {P} ^ {2} \left(\frac {\partial P}{\partial V}\right) _ {T}$$

Answer:

Let us write the formula for the heat capacities at constant pressure and constant volume:

$$C _ {P} = T \left(\frac {\partial S}{\partial T}\right) _ {P}$$ $$C _ {V} = T \left(\frac {\partial S}{\partial T}\right) _ {V}$$

Let's write the entropy as a function of volume $V$ and temperature $T$:

$$d S = \left(\frac {\partial S}{\partial T}\right) _ {V} d T + \left(\frac {\partial S}{\partial V}\right) _ {T} d V$$

Then we divide equation (1) by $dT$ at constant pressure and obtain:

$$\left(\frac {\partial S}{\partial T}\right) _ {P} - \left(\frac {\partial S}{\partial T}\right) _ {V} = \left(\frac {\partial S}{\partial V}\right) _ {T} \left(\frac {\partial V}{\partial T}\right) _ {P}$$

Then we multiply both sides of this equation by $T$:

$$C _ {P} - C _ {V} = T \left(\frac {\partial S}{\partial V}\right) _ {T} \left(\frac {\partial V}{\partial T}\right) _ {P}$$

From the expression for the Helmholtz free energy $F = U - TS$ we have:

$$d F = d U - T d S - S d T = - P d V - S d T$$

Therefore, $\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V$ and we can rewrite the equation (2):

$$C _ {P} - C _ {V} = T \left(\frac {\partial P}{\partial T}\right) _ {V} \left(\frac {\partial V}{\partial T}\right) _ {P}$$

Let us consider the equation $\left(\frac{\partial P}{\partial T}\right)_V = \left(\frac{\partial S}{\partial V}\right)_T$ . We multiply the left-hand side of the equation by $\left(\frac{\partial S}{\partial V}\right)_T\left(\frac{\partial T}{\partial V}\right)_P\left(\frac{\partial V}{\partial P}\right)_T$ and the right-hand side of one by $(-1)$ :

$$\left(\frac {\partial P}{\partial T}\right) _ {V} \left(\frac {\partial T}{\partial V}\right) _ {P} \left(\frac {\partial V}{\partial P}\right) _ {T} = - 1$$

Therefore, $\left(\frac{\partial P}{\partial T}\right)_V = -\left(\frac{\partial V}{\partial T}\right)_P\left(\frac{\partial P}{\partial V}\right)_T$ (4).

So, we substitute equation (4) into equation (3) and obtain:

$$C _ {P} - C _ {V} = T \left(\frac {\partial P}{\partial T}\right) _ {V} \left(\frac {\partial V}{\partial T}\right) _ {P} = - T \left(\frac {\partial V}{\partial T}\right) _ {P} \left(\frac {\partial V}{\partial T}\right) _ {P} \left(\frac {\partial P}{\partial V}\right) _ {T} = - T \left(\frac {\partial V}{\partial T}\right) _ {P} ^ {2} \left(\frac {\partial P}{\partial V}\right) _ {T}$$

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