Supplement 2.7: Differentials in thermodynamics

State quantities and other quantities

In Supplement 2.3 the change of the entropy $S$ of a medium at a temperature $T$ is explained through the transport of heat $q$ :

d S = \frac{δ q}{T}

The first law is used to visualise heat transfer.

δ q = d E + p d V

in which $E$ is the internal energy of the medium, $p$ is the pressure, $d V$ is a change in the volume, and $p d V$ is the volume work transferred from the medium to the environment. It is noticeable here that the relationships only contain differential quantities and there are two symbols for the notation: $d$ and $δ$ . The question also arises as to why the equations are not formulated with finite quantities instead of differentials, for example the first law in the form $q = E + p V$ .

Heat is one of the variables whose change in a medium does not always lead to the same result. In fact, the result is also dependent on other physical variables. A gas that moves a piston in a cylinder - i.e. performs pressure-volume work during expansion or absorbs heat during compression - and in which heat is released through the combustion of fuel, reaches a different final state after the piston has completed a full cycle, depending on the time of heat release: the heat content of the gas has changed. The integral of the converted heat over the complete cycle of the piston movement therefore applies:

\oint δ q \neq 0

the closed path integral is therefore not equal to zero. Now let's look at the pressure in the cylinder. If the piston is moved without burning fuel and its movement is frictionless, if the cylinder is gas-tight and insulated from the environment against heat loss (i.e. adiabatically sealed), the pressure in the cylinder after a complete cycle is the same as before:

\oint d p = 0

If we consider a movement of the piston between two positions 1 and 2 with the volumes $V_{1}$ and $V_{2}$ of the gas enclosed in the cylinder, the following applies to the associated pressures, irrespective of the details of the path travelled by the piston:

\int_{V_{1}}^{V_{2}} d p = p_{2} - p_{1} = Δ p

As the pressure can be clearly assigned to the state of the medium, it is referred to as a thermodynamic state variable. The corresponding integral of heat, on the other hand,

\int_{V_{1}}^{V_{2}} δ q = q

provides a result that depends on the path and cannot be represented as the difference between the values of the initial and final state of the medium. It is one of the "other thermodynamic quantities", for which a separate designation has not yet been found. Formulated with finite instead of differential values, it therefore applies to the first law:

q = Δ E + p Δ V

The fact that the heat cannot be clearly integrated independently of the displacement means that an integral of the differential of the heat $δ q$ does not exist in a mathematical sense. To emphasise this, the differential is written with the $δ$ symbol instead of $d$ . However, the integral of a differential state variable does exist.

In order to specify this in a general way, we consider the differential already discussed in the right-hand column of Supplement 2.6

d z = \frac{\partial z}{\partial x} d x + \frac{\partial z}{\partial y} d y

Whether an antiderivative function $z$ exists depends on the partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ . This is the case if the function $z$ is continuous with its derivatives. Then the following applies:

\frac{\partial}{\partial y} (\frac{\partial z}{\partial x}) = \frac{\partial}{\partial x} (\frac{\partial z}{\partial y})

resp.

\frac{\partial^{2} z}{\partial y \partial x} = \frac{\partial^{2} z}{\partial x \partial y}

For a test of integrability, the partial derivatives must therefore be differentiated according to the other variables and the results checked for equality. If they are equal, the integral exists, otherwise it does not exist. Applied to thermodynamic variables: If the corresponding expressions are equal, a state variable exists, otherwise it is a "different variable".

This will be explained using three very simple examples.

Example 1:

For the differential

d z = \frac{\partial z}{\partial x} d x + \frac{\partial z}{\partial y} d y = y d x + x d y

the crossed derivative results in

\frac{\partial}{\partial y} (\frac{\partial z}{\partial x}) = 1

and

\frac{\partial}{\partial x} (\frac{\partial z}{\partial y}) = 1

The differential is integrable and therefore it is: $z = x y$ .

Example 2:

For the differential

d z = \frac{\partial z}{\partial x} d x + \frac{\partial z}{\partial y} d y = y d x - x d y

the crossed derivative results in the terms

\frac{\partial}{\partial y} (\frac{\partial z}{\partial x}) = 1

and

\frac{\partial}{\partial x} (\frac{\partial z}{\partial y}) = - 1

so that no antiderivative is existent.

Example 3:

It is quite possible to transform a differential which is not integrable by division by a suitable expression into one which is integrable. The differential from example 2 divided by $y^{2}$ ,

d z = \frac{\partial z}{\partial x} d x + \frac{\partial z}{\partial y} d y = \frac{y d x - x d y}{y^{2}}

becomes integrable. Please calculate it yourself!

With respect to thermodynamics, the third example shows that the differential of the heat $δ q$ divided by the temperature may become a state quantity, the entropy

d S = \frac{δ q}{T}

This is explained in detail in the specialist literature, e.g.,
    Theories and Problems of Thermodynamics
    Authors: Michael M. Abbott & Hendrik C. van Ness
    Publisher: Schaum's Outline, McGraw-Hill, New York, 1972
from which the examples mentioned are copied.

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Supplement 2.7: Differentials in thermodynamics

State quantities and other quantities