2 The first law 2.4 The pressure 2.6 Isothermal and Adiabatic Expansion for Ideal Gasses and Engine Cycles

2.5 Energy

(September 1, 2023)

The total energy of the substance is $U(T,V,N)$ , and usually the $N$ is considered fixed and not notated $U(T,V)\equiv U(T,V,N)$ .

The energy is $U(T,V,N)$ is extensive. This means that if I consider twice as many particles at the same temperature and density, the energy is twice as large. The energy per particle is notated $e_{N}$ :

e_{N}\equiv\frac{U}{N}

and is intensive. Like with the pressure, the energy per particle $U/N$ is a function of temperature and the volume per particle, $U/N=e_{N}(T,v_{N})$ . If you prefer you may parametrize the energy per particle by the temperature and density $n\equiv N/V$ , that is $U/N=e_{N}(T,n)$ . Summarizing

U(T,V,N)=Ne_{N}(T,v_{N})=Ne_{N}(T,n)

(2.39)

At low densities or large volume we can make a Taylor series expansion in powers of the density $N/V$ , leading to the following series expansion for $e_{N}$ at low densities:

U(T,V,N)=Ne_{0}(T)\left[1+C_{1}(T)\frac{N}{V}+\ldots\right]

(2.40)

The first term, $Ne_{0}(T)$ , is finite in the limit of infinite volume. This is the ideal gas limit. $Ne_{0}(T)$ represents the energy of the individual atoms, and hence is proportional to $N$ . The next term in the series represents the interactions between the particles and is therefore proportional to $N^{2}$ .

For an ideal gas, we neglect the interactions and have:

U(T,V,N)=Ne_{0}(T)

any ideal gas (2.41)

This implies that for an ideal gas⁷⁷ 7 As is common, we are suppressing the $N$ . More precisely the equation is written: $\left(\frac{\partial U}{\partial V}\right)_{T,N}=0\,.$ (2.42)

\left(\frac{\partial U}{\partial V}\right)_{T}=0

any ideal gas (2.43)

As discussed above the function $e_{0}(T)$ determines the specific heat $C_{V}$ for an ideal gas. For a classical mono-atomic or classical diatomic gas the function $e_{0}(T)$ is just proportional to $T$ . For instance for a diatomic gas, where $U=\tfrac{5}{2}N{k_{\scriptscriptstyle B}T}$ , then

e_{0}=\frac{5}{2}{k_{\scriptscriptstyle B}T}

diatomic ideal gas (2.44)

However, if the gas is not entirely classical, e.g. the quantum mechanical vibrations of dilute ${\rm H}_{2}{\rm O}$ vapor, then $e_{0}(T)$ will have a non-trivial dependence on $T$ . We will calculate $e_{0}(T)$ for some cases as the course progresses.

Response: As with the pressure we need to characterize the response

dU=\left(\frac{\partial U}{\partial T}\right)_{V}\,dT+\left(\frac{\partial U}{% \partial V}\right)_{T}\,dV

(2.45)

The first law allows one to relate these derivatives to the measured specific heats and the response coefficients $\beta_{p}$ that we have already defined.

Fixed volume: If the volume is held fixed $dV=0$ and ${\mathchar 22\mskip-12.0mu d}W=0$ . The change in energy at fixed volume is

dU=\left(\frac{\partial U}{\partial T}\right)_{V}dT

(2.46)

and so from the first law, $dU={\mathchar 22\mskip-12.0mu d}Q$ , and so

\boxed{\left(\frac{\partial U}{\partial T}\right)_{V}=C_{V}}

(2.47)

Fixed pressure: Consider a change in temperature and volume at fixed pressure. From the first law

dU=C_{V}\,dT_{p}+\left(\frac{\partial U}{\partial V}\right)_{T}\,dV_{p}={% \mathchar 22\mskip-12.0mu d}Q_{p}-pdV_{p}

(2.48)

where we have put subscript “ $p$ ” to remind ourselves that the path taken is at fixed pressure. Dividing by $dT_{p}$ we see that:

C_{V}+\left(\frac{\partial U}{\partial V}\right)_{T}V\beta_{p}=C_{p}-pV\beta_{p}

(2.49)

where we have used the definition of $C_{p}$ ( )and $\beta_{p}$ Thus the second response coefficient is given by $C_{V}$ and $C_{p}$

\boxed{\left(\frac{\partial U}{\partial V}\right)_{T}=\frac{C_{p}-C_{V}}{V% \beta_{p}}-p}

(2.50)

This relation gives an experimental way to determine $(\partial U/\partial V)_{T}$ , from the measured specific heats.

For an ideal gas you should be able to show that $V\beta_{p}=Nk_{B}/p$ , and recognize $(\partial U/\partial V)_{T}=0$ , to produce the relation between $C_{p}$ and $C_{V}$ given in Eq. (any ideal gas (2.17)).