Themes > Science > Physics > Statistical Mechanics > Statistical Mechanics Topics > Kinetic Theory of Gases > Temperature

We have so far been able to define both the pressure $P$ and energy $U$ of the gas from the atomic point of view. We now need to define temperature. Consider two gases in a cylinder separated by a frictionless piston. When the piston reaches equilibrium there is no further change in the system. We hence conclude that the temperature of both of the gases must be the same, since the very definition of temperature is that there will be heat flows, and consequently, no equilibrium, unless and until the temperatures of the two gases becomes equal.
Figure 10.6: Piston Separating out Two Gases
\begin{figure} \begin{center} \epsfig{file=core/figure26.eps, height=5cm} \end{center} \end{figure}

We now examine the conditions under which there will be equilibrium. Let us label the gas on the left of the cylinder as $1$ and that on the right as $2$. For equilibrium, the pressure exerted by both the gases on the piston must be equal. Hence, from (10.16) we have
\begin{displaymath} \frac{N_1}{3V_1}<m_1{\bf v_1}^2>=P= \frac{N_2}{3V_2}<m_2{\bf v_2}^2> \end{displaymath} (10.19)

Are the densities of the two gases the same on two sides of the piston, that is $\displaystyle \frac{N_1}{V_1}$ equal to $\displaystyle \frac{N_2}{V_2}$? The answer is yes, although to prove this is quite difficult. The intuitive proof that the two densities are equal is that if there was a difference in the densities, there would be a net ``osmotic'' pressure on the piston forcing it to move, and consequently the system would not be in equilibrium. Hence, in equilibrium, we have
\begin{displaymath} \frac{N_1}{V_1}=\frac{N_2}{V_2} \end{displaymath} (10.20)

and from (10.19) we obtain
\begin{displaymath} <\frac{1}{2}m_1{\bf v_1}^2>=<\frac{1}{2}m_2{\bf v_2}^2> \end{displaymath} (10.21)

We see that the equation above is simply a statement that the average kinetic of the atoms in two gases which are in equilibrium is the same. Hence temperature is defined to be proportional to the average kinetic energy of a single atom of the gas. Fixing the constant of proportionality to be the Boltzmann constant we finally arrive at the following definition of temperature $T$.
\begin{displaymath} \frac{3}{2}k_BT\equiv <\frac{1}{2}m{\bf v}^2> \end{displaymath} (10.22)

Temperature is a measure of how fast, on the average, that the atoms of a gas are moving. At room temperature $k_BT \simeq \frac{1}{40}$ eV. The faster the atoms move, the hotter the temperature. The sensation of burning that we have on putting our hands into, say a fire, is because fast moving atoms from the fire impart high amounts of kinetic energy to our hands, causing atoms in our hand to move very fast and result in the sensation of burning. Hence, from eqns.(10.2) and (10.22)we have
$\displaystyle U=<E_G>$ $\textstyle =$ $\displaystyle \frac{1}{2}m\sum_{n=1}^{N}<{\bf v}_n^2>$ (10.23)
  $\textstyle =$ $\displaystyle Nk_BT$ (10.24)

 

Combining our results, from (10.16) and (10.22) we finally obtain the ideal gas law
\begin{displaymath} PV=Nk_BT \end{displaymath} (10.25)

The result above has the remarkable implication that no matter what the gas is composed of, for example, be it nitrogen, helium and so on, equal volumes of the various gases at the same pressure and temperature have the same number N of atoms. Note this result follows directly from Newton's Laws as is seen by the derivation given. This remarkable property of the ideal gas led Avoagardo to postulate that one molar volume of any gas will have the same number of atoms, given by Avagardo's number $N_{\mathrm{Avogardo}}$.


Information provided by: http://www.srikant.org