2.3 Thermodynamics of dilute gases

The pressure of a gas in a container is produced by the incessant drumming of the gas molecules upon the walls. At each such wall collision - say, against the right wall of a cubic box - the respective momentum of the molecule $p_{x}\equiv m v_{x}$ is reversed. The momentum transferred to the wall is thus $\Delta p_{x}=2mv_{x}$. The force acting on the unit area of the wall is then just the time average of the momentum transfer:

$$P \equiv \frac{\langle K \rangle_{t}}{F} \equiv \langle \fra... ...\langle \frac{1}{Ft} \sum_{k=1}^{M(t)} 2 m v_{x}^{(k)} \rangle$$ (2.35)

where $M(t)$ is the number of wall impacts within the time $t$.

To obtain a theoretical prediction for the value of the pressure we argue as follows:

The number of particles having $x$-velocity $v_{x}$ and impinging on the right wall per unit time is obviously proportional to $v_{x}$. The momentum transfer from such a particle to the wall is $2mv_{x}$. Thus we find

$$P = \frac{N}{V} \int_{v_{x}> 0} d\vec{v} 2mv_{x}^{2} f(\vec{v})$$ (2.36)

Inserting the Boltzmann density for $f(\vec{v})$ and performing the integrations we have

$$P = \frac{2}{3} \frac{N \langle E \rangle}{V}$$ (2.37)

Applet Hspheres: Start

$\parbox{360pt}{ \mbox{}\\ {\bf Simula... ...m equation \ref{EQPTH} ($P_{th}$). \\ \vspace{2ex} {\small [Code: Hspheres]} }$

Where do we stand now? Just by statistical reasoning we have actually arrived at a prediction for the pressure - a thermodynamic quantity!

Having thus traversed the gap to macroscopic physics we take a few steps further. From thermodynamics we know that the pressure of a dilute gas at density $\rho \equiv N/V$ and temperature $T$ is $P = \rho k T$. Comparing this to the above formula we find for the mean energy of a molecule

$$\langle E \rangle = \frac{3 k T}{2} ,$$ (2.38)

and the parameter $\beta$ (which was introduced in connection with the equilibrium density in velocity space) turns out to be just the inverse of $kT$.

The further explication of the dilute gas thermodynamics is easy. Taking the formula for the internal energy, $E \equiv N \langle E\rangle$ $= 3 NkT/2$, and using the First Law

$$dQ = dE + P dV$$ (2.39)

we immediately find that amount of energy that is needed to raise the temperature of a gas by $1$ degree - a.k.a. the heat capacity $C_{V}$:

$$C_{V} \equiv \left. \frac{dQ}{dT} \right\vert _{V} = \frac{dE}{dT} = \frac{3}{2} Nk$$ (2.40)