# 10.4: Expressing Thermodynamic Functions with Independent Variables V and T

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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*}

$dA\mathrm{=-}SdT + PdV \nonumber$

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

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

This page titled 10.4: Expressing Thermodynamic Functions with Independent Variables V and T is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Paul Ellgen via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.