The diagram shows two systems that may exchange both thermal and mechanical energy via a movable, heat-conducting partition. Because both Eand Vmay change. We consider the entropy of each system to be a function of both:$S(E,V)$. Considering the exchange of thermal energy only, we argued in Section 9.2 that was reasonable to define$\frac{1}{T}$as$\frac{\delta S}{\delta E}$. In the more general case, $\frac{P}{T}$is also defined as something.

a) Why should pressure come into play, and to what might$\frac{P}{T}$be equated.

b) Given this relationship, show that$dS=\frac{dQ}{T}$(Remember the first law of thermodynamics.)

The first law of thermodynamics, if$\Delta U=0$for an ideal gas, then:

$0=TdS-PdV$

a)

Both systems can exchange heat because the partition is thermally conductive, and both can interchange mechanical energy because the partition may be moved, based on the configuration of the two systems. Let's pretend that each system is filled with a perfect gas. Because one of the two systems is constantly supplied with heat in return for releasing mechanical energy by applying constant pressure, the internal energy of either system filled with an ideal gas must be zero:

$\Delta E=0$

From the first law of thermodynamics, if$\Delta U=0$ for an ideal gas, then:

$0=TdS-PdV$

As can be observed, heat transmission diminishes the thermal energy of the system. It is compensated, however, by being subjected to continual pressure, which reduces its volume and so increases its mechanical energy. As a result, the system's temperature remains constant. This is known as an isothermal process, and the change in internal energy for an ideal gas undergoing such a process must be zero. As a result, it is evident that:

$$\begin{gathered} TdS=PdV \\ \frac{P}{T}=\frac{\partial S}{\partial V}. \end{gathered}$$

Express the units in their base form.

For $P/T$we have:

$\frac{P}{T}:\frac{\text{kg}}{\text{m}·{\text{s}}^2·\text{K}}$

For the partial derivative we have:

$$\begin{gathered} \frac{\partial S}{\partial V}:\frac{\text{J}/\text{K}}{{\text{m}}^3} \\ :\frac{\text{kg}·{\text{m}}^2}{{\text{m}}^3·{\text{s}}^2·\text{K}} \\ :\frac{\text{kg}}{\text{m}·{\text{s}}^2·\text{K}} \end{gathered}$$

Hence the expression for$\frac{P}{T}$ is $\frac{P}{T}=\frac{\partial S}{\partial V}.$and unit is$\frac{P}{T}:\frac{\text{kg}}{\text{m}·{\text{s}}^2·\text{K}}$

b)

First, let us recall that S is a function of E and V :

$S=f(E,V)$

Using the chain rule for derivatives, we obtain:

$dS=\frac{\partial S}{\partial E}dE+\frac{\partial S}{\partial V}dV$

$\frac{\partial S}{\partial E}=\frac{1}{T};\frac{\partial S}{\partial V}=\frac{P}{T}$

Put these expressions into dS we have:

$\begin{array}{rcl}S& =& \frac{1}{T}dE+\frac{P}{T}dV\\ & =& \frac{1}{T}(dE+PdV)\\ & & \end{array}$

According to the first law of thermodynamics:

$dE=dQ-dW$

Where, $dW=PdV$we have:

$dQ=dE+PdV$

$\begin{array}{rcl}S& =& \frac{1}{T}(dE+PdV)\\ & =& \frac{1}{T}dQ\\ & =& \frac{dQ}{T}·\\ & & \end{array}$

We have thus proved that $dS=\frac{dQ}{T}$using the first law of thermodynamics.