In the previous review module, we saw that a function of two independent variables can be expressed using a total differential. Thermodynamic state functions are no different. Because the state of a single-component system can be specified by two independent variables, thermodynamic functions can be written as functions of those variables.
For example, the internal energy may be expressed as
\[ U = U(V,T) \]
indicating that the internal energy is determined by the volume and temperature of the system.
The total differential of the internal energy is therefore
\[ dU = \left( \frac{\partial U} {\partial V} \right)_T dV + \left( \frac{\partial U} {\partial T} \right)_V dT \]
This expression separates changes in the internal energy into contributions arising from changes in volume and changes in temperature.
One of the partial derivatives in the total differential of \(U\) should look familiar. In Chapter 3, the constant-volume heat capacity was defined as
\[ C_V = \left( \frac{\partial U} {\partial T} \right)_V \]
Substituting this definition into the total differential gives
\[ dU = \left( \frac{\partial U} {\partial V} \right)_T dV + C_V\,dT \]
Thus, the constant-volume heat capacity appears naturally as the coefficient describing how the internal energy changes with temperature.
The remaining coefficient,
\[ \left( \frac{\partial U} {\partial V} \right)_T \]
is called the internal pressure, a quantity that will become important later in this chapter. The Joule Experiment was an attempt to measure This quantity directly. Later (Chapter 6) we will show that
\[ \left( \frac{\partial U}{\partial V} \right)_T = T \left( \frac{\partial p}{\partial T} \right)_V - p \]
Show that for an ideal gas, \[ \left( \frac{\partial U}{\partial V} \right)_T = 0 \]
For an ideal gas,
\[ p = \frac{nRT}{V} \]
And so,
\[ \left( \frac{\partial p}{\partial T} \right)_V = \frac{nR}{V} \]
So
\[ \left( \frac{\partial U}{\partial V} \right)_T = T \left( \frac{nR}{V} \right) - p\]
And substituting \( p = \frac{nRT}{V} \) we get
\[ \left( \frac{\partial U}{\partial V} \right)_T = \frac{nRT}{V} - \frac{nRT}{V} = 0\]
Note: It is easy to evaluate a partial derivative if it only contains p, T, and V, because we can assume an equation of state (such as the Ideal Gas Law or the van der Waal Law) and evaluate the derivitive analytically.
Enthalpy can likewise be expressed as a function of pressure and temperature:
\[ H = H(p,T) \]
Its total differential is
\[ dH = \left( \frac{\partial H} {\partial p} \right)_T dp + \left( \frac{\partial H} {\partial T} \right)_p dT \]
The constant-pressure heat capacity was defined in Chapter 3 as
\[ C_P = \left( \frac{\partial H} {\partial T} \right)_p \]
Substituting this definition into the total differential gives
\[ dH = \left( \frac{\partial H} {\partial p} \right)_T dp + C_P\,dT \]
Once again, a measurable thermodynamic property appears naturally as the coefficient of a differential term.
The remaining coefficient,
\[ \left( \frac{\partial H} {\partial p} \right)_T \]
cannot be measured directly. One of the major goals of this chapter is to develop mathematical tools that allow quantities such as this to be expressed in terms of measurable properties.
Big picture: Total differentials provide a powerful way to describe changes in thermodynamic state functions. The coefficients that appear in these expressions are partial derivatives, some of which correspond to experimentally measurable quantities such as \(C_V\) and \(C_P\). The remainder of this chapter focuses on expressing the other partial derivatives in terms of measurable thermodynamic properties.
Evaluate
\[ \left( \frac{\partial U} {\partial V} \right)_T \]
for a gas that obeys the van der Waals equation of state:
\[ p = \frac{nRT} {V-nb} - \frac{an^2} {V^2} \]
We will use the relationship
\[ \left( \frac{\partial U} {\partial V} \right)_T = T \left( \frac{\partial p} {\partial T} \right)_V - p \]
This relationship will be derived later; for now, we will use it to practice evaluating thermodynamic derivatives from an equation of state.
First calculate the partial derivative of pressure with respect to temperature at constant volume:
\[ \left( \frac{\partial p} {\partial T} \right)_V = \left( \frac{\partial} {\partial T} \left[ \frac{nRT} {V-nb} - \frac{an^2} {V^2} \right] \right)_V \]
At constant volume, \(n\), \(R\), \(V\), \(a\), and \(b\) are constants. The attractive-force term does not contain \(T\), so its derivative is zero:
\[ \left( \frac{\partial p} {\partial T} \right)_V = \frac{nR} {V-nb} \]
Substitute this result into the internal-energy derivative:
\[ \left( \frac{\partial U} {\partial V} \right)_T = T \left( \frac{nR} {V-nb} \right) - p \]
Now substitute the van der Waals expression for \(p\):
\[ \left( \frac{\partial U} {\partial V} \right)_T = \frac{nRT} {V-nb} - \left[ \frac{nRT} {V-nb} - \frac{an^2} {V^2} \right] \]
Distribute the negative sign:
\[ \left( \frac{\partial U} {\partial V} \right)_T = \frac{nRT} {V-nb} - \frac{nRT} {V-nb} + \frac{an^2} {V^2} \]
The finite-volume terms cancel, leaving
\[ \boxed{ \left( \frac{\partial U} {\partial V} \right)_T = \frac{an^2} {V^2} } \]
Interpretation: For a van der Waals gas, the internal energy depends on volume at constant temperature because the gas includes attractive intermolecular interactions. Increasing the volume separates the molecules, changing their interaction energy. The parameter \(a\) measures the strength of those attractions, so it appears directly in \(\left(\partial U/\partial V\right)_T\).
Big picture: The total differential is the mathematical tool that allows thermodynamic state functions to be expressed in terms of their independent variables. Many important thermodynamic quantities, including heat capacities, appear naturally as coefficients in these differentials, making total differentials the starting point for much of thermodynamic analysis.