Question

A key feature of thermodynamics is the fact that the internal energy, \(E_{\text {int }}\) of a system and its entropy, \(S\), are state variables; that is, they depend only on the thermodynamic state of the system and not on the processes by which it reached that state (unlike, for example, the heat content, \(Q\) ). This means that the differentials \(d E_{\text {int }}=T d S-p d V\) and \(d S=T^{-1} d E_{\text {int }}+p T^{-1} d V\) where \(T\) is temperature (in kelvins), \(p\) is pressure, and \(V\) is volume, are exact differentials as defined in calculus. What relationships follow from this fact?

Short answer

In this problem, the relationships between the differentials are given by: 1. The change in internal energy with respect to the entropy is equal to the temperature: \(\frac{\partial E_{int}}{\partial S}=T\). 2. The change in entropy with respect to the internal energy is the inverse of the temperature: \(\frac{\partial S}{\partial E_{int}}=T^{-1}\). Also, the relationships involving volume and pressure changes are given by the original equations: 1. \(dE_{int} = TdS - pdV\) 2. \(dS = T^{-1}dE_{int} + pT^{-1}dV\) These relationships provide information on the system's internal energy and entropy, considering the changes in volume and pressure.

Step-by-step solution

Identify the given equations and variables

We are provided with the following relationships: 1. \(d E_{int}=T d S-p d V\) 2. \(d S=T^{-1} d E_{int}+p T^{-1} d V\) We need to find the relationships between these differentials. We are dealing with the following variables: - Internal energy (\(E_{int}\)) - Entropy (\(S\)) - Temperature (\(T\)) - Pressure (\(p\)) - Volume (\(V\))

Differentiate the given equations

First, we need to find the differential of these equations. It can help us identify the connection between these exact differentials. 1. Differentiating the first equation with respect to \(S\): \(\frac{\partial}{\partial S}(dE_{int})=\frac{\partial}{\partial S}(TdS - pdV)\) 2. Differentiating the second equation with respect to \(E_{int}\): \(\frac{\partial}{\partial E_{int}}(dS)=\frac{\partial}{\partial E_{int}}(T^{-1}dE_{int}+pT^{-1}dV)\)

Find the relationship between the variables

Now that we have the differentials, we can express the relationships between the variables in the problem. For the first equation: \(\frac{\partial E_{int}}{\partial S}=T\) For the second equation: \(\frac{\partial S}{\partial E_{int}}=T^{-1}\) These relationships state that the change in internal energy with respect to the entropy is equal to the temperature, and the change in entropy with respect to the internal energy is the inverse of the temperature. The change between these variables depends solely on the temperature, regardless of the method by which the system achieves its state. We can also express the relationships in terms of volume and pressure, as given in the original equations involving differentials: 1. \(dE_{int} = TdS - pdV\) 2. \(dS = T^{-1}dE_{int} + pT^{-1}dV\) These relationships provide information on the system's internal energy and entropy when the volume and pressure change as well.