10.4: Expressing Thermodynamic Functions with Independent Variables V and T

If we choose \(V\) and \(T\) as the independent variables, we can express the differential of \(E\) as a function of \(V\) and \(T\). We also have the differential relationship \(dE=TdS-PdV\). These expressions for \(dE\) must be equal:

\[dE = {\left(\frac{\partial E}{\partial V}\right)}_TdV + {\left(\frac{\partial E}{\partial T}\right)}_VdT = TdS + PdV \nonumber \] Rearranging, we find a total differential for \(dS\) with \(V\) and \(T\) as the independent variables:

\[dS = \frac{\mathrm{1}}{T}{\left(\frac{\partial E}{\partial T}\right)}_VdT + \frac{\mathrm{1}}{T}\left[{\left(\frac{\partial E}{\partial V}\right)}_T + P\right]dV \nonumber \]

From the coefficient of \(dT\), we have

\[{\left(\frac{\partial S}{\partial T}\right)}_V = \frac{\mathrm{1}}{T}{\left(\frac{\partial E}{\partial T}\right)}_V = \frac{C_V}{T} \nonumber \]

where we use the definition \({\left({\partial E}/{\partial T}\right)}_V=C_V\). (When we write “\(C_V\),” we usually think of it as a property of a pure substance. The relationship above is valid for any reversible system. When we are describing a system that is not a pure substance, \(C_V\) is just an abbreviation for \({\left({\partial E}/{\partial T}\right)}_V\).) From the coefficient of \(dV\), we have

\[{\left(\frac{\partial S}{\partial V}\right)}_T = \frac{\mathrm{1}}{T}\left[{\left(\frac{\partial E}{\partial V}\right)}_T + P\right] = {\left(\frac{\partial P}{\partial T}\right)}_V \nonumber \]

where we use the relationship \({\left({\partial S}/{\partial V}\right)}_T={\left({\partial P}/{\partial T}\right)}_V\) that we find in §1. Substituting into the expression for \(dS\), we find \[dS = \frac{C_V}{T}dT + {\left(\frac{\partial P}{\partial T}\right)}_VdV \nonumber \]

Now, from \(dE=TdS-PdV\), we have

\[dE = C_VdT + \left[T{\left(\frac{\partial P}{\partial T}\right)}_V + P\right]dV \nonumber \]

From \(H=E+PV\), we have

\[ \begin{align*} dH &= dE + d\left(PV\right) \\[4pt] &= dE + {\left(\frac{\partial \left(PV\right)}{\partial T}\right)}_VdT + {\left(\frac{\partial \left(PV\right)}{\partial V}\right)}_TdV \\[4pt]&= dE + V{\left(\frac{\partial P}{\partial T}\right)}_VdT + \left[P + V{\left(\frac{\partial P}{\partial V}\right)}_T\right]dV \\[4pt]&= \left[C_V + V{\left(\frac{\partial P}{\partial T}\right)}_V\right]dT + \left[T{\left(\frac{\partial P}{\partial T}\right)}_V + V{\left(\frac{\partial P}{\partial V}\right)}_T\right]dV \end{align*} \nonumber \]

Of course, we already have

\[dA\mathrm{=-}SdT + PdV \nonumber \]

From \(G=H-TS\), by an argument that parallels the above derivation of \(dH\), we obtain

\[dG = \left[V{\left(\frac{\partial P}{\partial T}\right)}_V + S\right]dT + V{\left(\frac{\partial P}{\partial V}\right)}_TdV \nonumber \]

Finally, we can write \(P=P\left(T,V\right)\) to find

\[dP = {\left(\frac{\partial P}{\partial T}\right)}_VdT + {\left(\frac{\partial P}{\partial V}\right)}_TdV \nonumber \]

\(P\), \(T\), \(V\), \(C_V\), \({\left({\partial P}/{\partial T}\right)}_V\), and \({\left({\partial P}/{\partial V}\right)}_T\) are all experimentally accessible for any reversible system. If we have this information for a system that undergoes a change from a state specified by \(T_1\) and \(V_1\) to a second state specified by \(T_2\) and \(V_2\), we can use these relationships to calculate \(\Delta E\), \(\Delta S\), and \(\Delta H\). To do so, we calculate the appropriate line integral along a reversible path. One such path is an isothermal reversible change, at \(T_1\), from \(V_1\) to \(V_2\), followed by a constant-volume change, at \(V_2\), from \(T_1\) to \(T_2\). In principle, the same procedure can used to calculate \(\Delta A\) and \(\Delta G\). However, because \(S\) appears in the differentials\(\ dA\) and \(dG\), this requires that we first find \(S\) as a function of \(V\) and\(\ T\).

If the system is a pure substance for which we have an equation of state, we can find \({\left({\partial P}/{\partial T}\right)}_V\), and \({\left({\partial P}/{\partial V}\right)}_T\) by straightforward differentiation. When the substance is a gas, an equation of state may be available in the literature. When the substance is a liquid or a solid, these partial derivatives can still be related to experimentally accessible quantities. The compressibility of a substance is the change in its volume that results from a change in the applied pressure, at a constant temperature. The thermal expansion of a substance is the change in its volume that results from a change in its temperature, at a constant applied pressure. It is convenient to convert measurements of these properties into intensive functions of the state of the substance by expressing the volume change as a fraction of the original volume. That is, we define the coefficient of thermal expansion:

\[\alpha = \frac{\mathrm{1}}{V}{\left(\frac{\partial V}{\partial T}\right)}_P \nonumber \]

and the coefficient of isothermal compressibility:

\[\beta \mathrm{=-}\frac{\mathrm{1}}{V}{\left(\frac{\partial V}{\partial P}\right)}_T \nonumber \]

Coefficients of thermal expansion and isothermal compressibility are available in compilations of thermodynamic data for many liquids and solids. In general, both coefficients are weak functions of temperature. We have

\[\left(\frac{\partial P}{\partial V}\right)_T=-\frac{1}{\beta V} \nonumber \]

and

\[\left(\frac{\partial P}{\partial T}\right)_V=-\left(\frac{\partial V}{\partial T}\right)_P/ \left(\frac{\partial V}{\partial P}\right)_T=\frac{\alpha }{\beta } \nonumber \]

Using these coefficients, we can estimate a pressure change, for example, as a line integral of

\[dP=\left(\frac{\alpha }{\beta }\right)dT-\left(\frac{1}{\beta V}\right)dV \nonumber \]