1.1: Review and Setting the Stage

Introduction

Using the mathematical properties of state functions we can extend our ability to use thermodynamics combined with physical measurements to make quantitative predictions about the behavior of real systems. This section reviews and defines the mathematics of total differentials and what it means for a total differential to be exact. Then we review the fundamental differentials of thermodynamics. These can be combined with their mathematical properties to generate important partial derivatives that provide rigorous thermodynamic definitions of many useful physical quantities and useful relations between more easily measured and hard to measure thermodynamic quantities.

Total derivatives

A partial derivative of  a function F(x₁, x₂ ...), often written as \((\partial{F}/\partial{x_i})_{x_{k_1}, x_{k_2}...k_{j}\ne i}\), can be thought of as the rate of change of F along the direction x_i with all the other coordinates held constant. Geometrically this can be thought of as the tangent to the surface defined by F along the coordinate direction x_i. The total derivative (sometimes referred to as the total differential) is defined as the sum of all the partial derivatives each multiplied by an infinitesimal change along the coordinate the partial is taken along. This sum is then the infinitesimal change in F for infinitesimal changes along all the coordinates. Assuming F is a function of x₁, x₂ and x₃ yields a total derivative:

\[dF = \left(\frac{\partial{F}}{\partial{x_1}}\right)_{x_2, x_3}dx_1 +\left(\frac{\partial{F}}{\partial{x_2}}\right)_{x_1, x_3}dx_2 +\left(\frac{\partial{F}}{\partial{x_3}}\right)_{x_1, x_2}dx_3\label{dF_generic}\]

For a single component closed thermodynamic system (fixed number of moles, n), state functions are always determined when two state variables are defined. In the case of internal energy, we might write \(U=f(V,T)\) or \(U(V,T)\). If there is a mathematical function that relates the internal energy to these two variables, it is easy to see how U changes when either (or both!) are changed. This can be written as a total differential:

\[ dU = \left( \dfrac{\partial U}{\partial V} \right)_T dV + \left( \dfrac{\partial U}{\partial T} \right)_V dT \label{total} \]

Even without knowing the actually mathematical function relating the variables to the property, we can imagine how to calculate changes in the property from this expression.

\[ \Delta U = \int _{V_1}^{V_2} \left( \dfrac{\partial U}{\partial V} \right)_T dV + \int _{T_1}^{T_2} \left( \dfrac{\partial U}{\partial T} \right)_V dT \nonumber \]

In words, this implies that we can think of a change in \(U\) occurring due to an isothermal change followed by an isochoric change. And all we need to know is the slope of the surface in each pathway direction. Clearly, finding ways to measure these derivatives (slopes) is key to being able to calculate thermodynamic changes.

Exact Differentials

Previously we have used the fact that changes in thermodynamic state functions are path independent and noted this is the result of them being exact differentials. Here we define carefully what is required for a differential to be exact.

In general, if a differential can be expressed as

\[ df(x,y) = X(x,y)dx + Y(x,y)dy \nonumber \]

the differential will be an exact differential if it follows the Euler relation

\[\left( \dfrac{\partial X}{\partial y} \right)_x = \left( \dfrac{\partial Y}{\partial x} \right)_y \label{euler} \]

In order to illustrate this concept, consider \(P(V, T)\) using the ideal gas law for a fixed n.

\[P= \frac{nRT}{V} \nonumber \]

The total differential of \(P\) can be written

\[ dP = \left( - \frac{nRT}{V^2} \right) dV + \left( \dfrac{nR}{V} \right) dT \label{dPIdeal} \]

Example \(\PageIndex{1}\): Euler Relation

Does Equation \ref{dPIdeal} follow the Euler relation (Equation \ref{euler})?

Solution

Let’s confirm!

\[ \begin{align*} \left[ \frac{\partial}{\partial T} \left( - \frac{nRT}{V^2} \right) \right]_V &\stackrel{?}{=} \left[ \frac{\partial }{\partial V} \left( \frac{nR}{V} \right) \right]_T \\[4pt] \left( - \frac{nR}{V^2} \right) &\stackrel{\checkmark }{=} \left( - \frac{nR}{V^2} \right) \end{align*} \nonumber \]

\(dP\) is, in fact, an exact differential.

The differentials of all of the thermodynamic functions that are state functions will be exact. Heat and work are not exact differential and \(dw\) and \(dq\) are called inexact differentials instead.

Four fundamental differentials of thermodynamics

The first law of thermodynamics in differential form is:

\[dU=\delta q+ \delta w \label{EQ:firstlaw}\]

For a reversible process we have defined the entropy as \(dS=\delta q_{rev}/T\) and the reversible work as \(\delta w=−P⋅dV\). Substituting these identities into equation \(\ref{EQ:firstlaw}\) gives the following differential form of the first law:

\[dU=TdS−PdV\label{EQ:firstlaw2}\]

Note that Equation \(\ref{EQ:firstlaw2}\) is valid for a reversible process in which the only work is due to compression/expansion.

The enthalpy is defined as:

\[H=U+PV\label{EQ:enthalpy}\]

From Equation \(\ref{EQ:enthalpy}\) we can write a differential form of the enthalpy as:

\[dH=dU+PdV+VdP\label{EQ:enthalpy2}\]

where we have again used the product rule from calculus on the PV term. Substituting equation \(\ref{EQ:firstlaw2}\) into equation \(\ref{EQ:enthalpy2}\) for the dU term gives another differential relation for dH:

\[dH = TdS - \cancel{PdV} + \cancel{PdV} + VdP = TdS + VdP \label{EQ:dHdiff}\]

Similar arguments, which you should be able to produce, can be made for the state functions G and A. This leads to the four total differentials in table \(\PageIndex{1}\).

Table \(\PageIndex{1}\): the four fundamental differential relations for dU, dH, dG, and dA.

Differential Relation	Equation
dU=TdS−PdV	\(\ref{EQ:firstlaw2}\)
dH = TdS + VdP	\(\ref{EQ:dHdiff}\)
dA=−PdV−SdT
dG=VdP−SdT

Some important partial derivatives

From the properties of total and exact differentials plus these fundamental equations we can develop thermodynamically rigorous definitions of thermodynamic quantities in terms of derivatives. Because derivatives reflect rates of change this gives us a handle on measuring these quantities as changes are often measurable even when absolute values of quantities are not.

T, V, P and S

Consider equation \(\ref{EQ:firstlaw2}\) from above:

\[dU=TdS−PdV\nonumber\]

The natural variables of internal energy are thus \(S\) and \(V\). So the total differential (\(dU\)) in equation \(\ref{EQ:firstlaw2}\) can also be expressed as:

\[dU = \left( \dfrac{\partial U}{\partial S} \right)_V dS + \left( \dfrac{\partial U}{\partial V} \right)_S dV \label{dU_expanded} \]

Comparing the two expressions for \(dU\) it is apparent that:

\[\left( \frac{\partial U}{\partial S} \right)_V = T \label{T_from_U} \]

and

\[\left( \frac{\partial U}{\partial V} \right)_S = -P \label{P_from_U} \]

The relations in equations \(\ref{T_from_U}\) and \(\ref{P_from_U}\) can be understood as ways to measure the partial derivatives, since T and P are values we can measure. Alternatively, they can be thought of as thermodynamic definitions of T under constant volume conditions and P under constant entropy conditions. For example we can rewrite equation \(\ref{T_from_U}\) as:

\[T = \left( \frac{\partial U}{\partial S} \right)_V\nonumber \]

Likewise the other fundamental differentials of thermodynamics lead to the additional relations between thermodynamic quantities and partial derivatives listed in table \(\PageIndex{2}\). You should be able to recreate the above logic for the other cases.

Table \(\PageIndex{2}\): All the differential relations that can be found by comparing the total differential of U, H, G and A with the four fundamental differentials of thermodynamics.

from dU = TdS - PdV	from dH = TdS + VdP	from dA = -PdV - SdT	from dG = VdP - SdT
\(\left( \frac{\partial U}{\partial S} \right)_V = T\)	\(\left( \frac{\partial H}{\partial S} \right)_P = T\)	\(\left( \frac{\partial A}{\partial V} \right)_T = -P\)	\(\left( \frac{\partial G}{\partial P} \right)_T = V\)
\(\left( \frac{\partial U}{\partial V} \right)_S = -P\)	\(\left( \frac{\partial H}{\partial P} \right)_S = -V\)	\(\left( \frac{\partial A}{\partial T} \right)_V = -S\)	\(\left( \frac{\partial G}{\partial T} \right)_P = -S\)

Notice that this means there are multiple thermodynamic definitions of T, P, V and S depending on what is being held constant.

Heat Capacities

Two other useful thermochemical quantities we have encountered that can be expressed in terms of partial derivatives are the heat capacities \(C_P\) and \(C_V\) which can be expressed as

\[ C_P = \left(\frac{\partial H}{\partial T}\right)_P \nonumber \]

and

\[ C_V = \left(\frac{\partial U}{\partial T}\right)_V \nonumber \]

These are properties that can be measured experimentally and tabulated for many substances. These quantities can be used to calculate changes in quantities since they represent the slope of a surface (\(H\) or \(U\)) in the direction of the specified path (constant \(P\) or \(V\)). This allows us to use the following kinds of relationships:

\[ dH = \left(\frac{\partial H}{\partial T}\right)_p dT \nonumber \]

and

\[ \Delta H = \int \left(\frac{\partial H}{\partial T}\right)_P dT \nonumber \]

Likewise ∆U can be calculated under constant volume conditions using C_V.

 

Maxwell Relations

The fact that total differentials of state functions are exact allows us to make some unexpected connections between partial derivatives, which are useful when determining what measurements can be used to determine a particular derivative.

As an example consider \(dU\) as written out in equation \(\ref{dU_expanded}\):

\[dU = \left( \dfrac{\partial U}{\partial S} \right)_V dS + \left( \dfrac{\partial U}{\partial V} \right)_S dV \nonumber \]

Being an exact differential, the Euler relation must hold. Thus:

\[ \left[ \dfrac{\partial}{\partial V} \left( \dfrac{\partial U}{\partial S} \right)_V \right]_S= \left[ \dfrac{\partial}{\partial S} \left( \dfrac{\partial U}{\partial V} \right)_S \right]_V \nonumber \]

By substituting using equations \ref{T_from_U} and \ref{P_from_U} (or from table \(\PageIndex{2}\), we see that

\[ \left[ \frac{\partial}{\partial V} \left( T \right)_V \right]_S= \left[ \frac{\partial}{\partial S} \left( -P \right)_S \right]_V \nonumber \]

or

\[ \left( \frac{\partial T}{\partial V} \right)_S = - \left( \frac{\partial P}{\partial S} \right)_V \nonumber \]

This is an example of a Maxwell Relation. These are very powerful relationships that allows one to a substitute partial derivatives with a more convenient one.

Similar results can be derived based on the definitions of H, G and A. The results are summarized in table \(\PageIndex{3}\).

Table \(\PageIndex{3}\): Maxwell Relations

Function	Differential	Natural Variables	Maxwell Relation
\(U\)	\(dU = TdS - PdV\)	\(S, \,V\)	\( \left( \dfrac{\partial T}{\partial V} \right)_S = - \left( \dfrac{\partial P}{\partial S} \right)_V \)
\(H\)	\(dH = TdS + VdP\)	\(S, \,P\)	\( \left( \dfrac{\partial T}{\partial P} \right)_S = \left( \dfrac{\partial V}{\partial S} \right)_P \)
\(A\)	\(dA = -PdV - SdT\)	\(V, \,T\)	\( \left( \dfrac{\partial P}{\partial T} \right)_V = \left( \dfrac{\partial S}{\partial V} \right)_T \)
\(G\)	\(dG = VdP - SdT\)	\(P, \,T\)	\( \left( \dfrac{\partial V}{\partial T} \right)_P = - \left( \dfrac{\partial S}{\partial P} \right)_T \)

The Maxwell relations are extraordinarily useful in deriving the dependence of thermodynamic values on the state variables of \(P\), \(T\), and \(V\).

Partial derivative reciprocal and cyclic relations

Two more mathematical manipulations involving partial derivatives are also very useful when working with thermodynamics. These relations are valid as long as each variable in the total derivative are differentiable functions of the other variables (for more discussion of this see "Partial differential relations" on Wikipedia.org and the links and references included there). This is essentially the requirement for being able to write a total differential, thus  the relations described below are valid for relationships based on the fundamental differentials of thermodynamics.

The reciprocal relation is

\[1= \left( \frac{\partial z}{\partial y} \right)_x \left( \frac{\partial y}{\partial z} \right)_x \quad \text{or} \quad  \left( \dfrac{\partial z}{\partial y} \right)_x = \dfrac{1}{\left( \dfrac{\partial y}{\partial z} \right)_x} \label{recip_rel}\]

As the choice of label for each variable is arbitrary this could equally well be stated as:

\[\left( \dfrac{\partial x}{\partial y} \right)_z = \dfrac{1}{\left( \dfrac{\partial y}{\partial x} \right)_z} \nonumber\]

Derivation of the reciprocal rule

Consider a system that is described by three variables, and for which one can write a mathematical constraint on the variables

\[F(x, y, z) = 0 \nonumber \]

Under these circumstances, one can specify the state of the system varying only two parameters independently because the third parameter will have a fixed value. As such one could define two functions: \(z(x, y)\) and \(y(x,z)\).

This allows one to write the total differentials for \(dz\) and \(dy\) as follows

\[dz = \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x dy \label{eq5} \]

and

\[dy= \left( \dfrac{\partial y}{\partial x} \right)_z dx + \left( \dfrac{\partial y}{\partial z} \right)_x dz \label{eq6} \]

Substituting the Equation \ref{eq6} expression into Equation \ref{eq5}:

\[ \begin{align} dz &= \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x \left[ \left( \dfrac{\partial y}{\partial x} \right)_z dx + \left( \dfrac{\partial y}{\partial z} \right)_x dz \right] \\[4pt] &= \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial x} \right)_z dx + \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial z} \right)_x dz \label{eq7} \end{align} \]

Collecting dz terms on one side and dy terms on the other yields:

\[\left[1 - \left( \frac{\partial z}{\partial y} \right)_x \left( \frac{\partial y}{\partial z} \right)_x \right]dz = \left[\left( \frac{\partial z}{\partial x} \right)_y + \left( \frac{\partial z}{\partial y} \right)_x \left( \frac{\partial y}{\partial x} \right)_z \right]dx\label{key_deriv_expr}\]

Because dz and dx can vary independently the the quantities in square brackets must equal zero. The reciprocal relation is derived from the left-hand-side:

\[1 - \left( \frac{\partial z}{\partial y} \right)_x \left( \frac{\partial y}{\partial z} \right)_x = 0 \implies \left( \frac{\partial z}{\partial y} \right)_x \left( \frac{\partial y}{\partial z} \right)_x = 1\nonumber\]

The cyclic relation (sometimes called the triple product rule) is: 

\[\left(\frac{\partial x}{\partial y}\right)_z\left(\frac{\partial y}{\partial z}\right)_x\left(\frac{\partial z}{\partial x}\right)_y = -1\label{cyclic_rel}\]

This is useful because it allows any one of the three partial derivatives to be expressed as a fraction involving the other two partial derivatives especially when combined with the reciprocal rule this can be very useful.

Example \(\PageIndex{1}\)

Imagine you are working with a real gas and have equipment to measure T and P and V and the ability to control T and V, but no way to maintain the sample at a constant pressure as T and V vary. What two measurements can be combined to yield \(\left(\frac{\partial V}{\partial T}\right)_P\)?

Solution

From the cyclic relation we can write:

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

Isolating the partial derivative of interest on the left-hand-side and using the reciprocal relation to invert the constant V derivative yields:

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

Notice that the two derivatives on the right-hand-side require measuring pressure while varying T (and holding V constant) or varying V (and holding T constant). These are conditions that can be met with the equipment at hand. We have made no assumptions about whether the gas is ideal or not.

 

Derivation of the cyclic relation

This right-hand-side of equation \(\ref{key_deriv_expr}\) implies:

\[\left( \frac{\partial z}{\partial x} \right)_y + \left( \frac{\partial z}{\partial y} \right)_x \left( \frac{\partial y}{\partial x} \right)_z = 0 \nonumber \]

This rearranges to:

\[\left( \frac{\partial z}{\partial x} \right)_y = - \left( \frac{\partial z}{\partial y} \right)_x \left( \frac{\partial y}{\partial x} \right)_z \nonumber \]

Using the reciprocal rule to rewrite the right-hand-side yields:

\[ \left(\frac{\partial z}{\partial x} \right)_y = - \frac{1}{\left( \frac{\partial y}{\partial z} \right)_x \left( \frac{\partial x}{\partial y} \right)_z } \nonumber\]

Multiplying through by the the denominator on the right-hand-side results in the cyclic relation:

\[ \left( \frac{\partial x}{\partial y} \right)_z\left( \frac{\partial y}{\partial z} \right)_x \left(\frac{\partial z}{\partial x} \right)_y = - 1 \nonumber \]

Exercises 

1. Write out the derivation of all the relations in table \(\PageIndex{1}\).
2. Write out the logic to develop the relations in table \(\PageIndex{2}\).
3. Write out the derivation of all the Maxwell relations in table \(\PageIndex{3}\).
4. Starting with the fundamental differential expression for dU divide through by \(dV\) to get an expression for \(\left(\frac{\partial U}{\partial V}\right)_T\). Then use a Maxwell relation to show that you can rewrite it to \(\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_V - P\), which contains easily measurable quantities.