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

From the atomic point of view, how does pressure arise? Even before going into the detailed mechanism, we expect that pressure should be a macroscopic manifestation of microscopic motion. Consider a frictionless piston, which has an area $A$, contains a gas in some volume $V$ and with total number of atoms given by $N$. The piston is in equilibrium with the gas at some temperature $T$. Outside the piston is a perfect vacuum.

 

Figure 10.3: Gas Inside the Piston and Vacuum Outside
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The atoms of the gas are constantly bombarding the piston and this will cause a force to be exerted on the piston, and to keep the piston in place (stationary) we need to counter this force. Consider for an atom traveling straight towards the piston with velocity ${\bf v}=(v_x,v_y,v_z)\equiv (v,u,w)$; we will show later the case of the atom moving with arbitrary velocity gives the same result. The particle hits the piston and bounces back. We assume that the collision is elastic in that the energy of the atom is the same before and after bouncing off the piston. The assumption of the collision being elastic means that the atom does not lose any of its energy to the piston. This is reasonable, since if it lost energy to the piston, the piston would heat up; and once the piston reached equilibrium with the gas, the assumption of collisions being elastic would be correct. As shown in Figure 10.4, for a wall placed along the $y$-axis, an elastic collision leads to a velocity ${\bf v'}=(-v_x,v_y,v_z)\equiv (-v,u,w)$. Hence, for simplicity we consider only the special case of ${\bf v}=(v,0,0)$. Let the velocity of the atom after the collision be ${\bf v'}=(v',0,0)$. Since the atom possesses only kinetic energy, we have from energy conservation
$\displaystyle \frac{1}{2}mv^2$ $\textstyle =$ $\displaystyle \frac{1}{2}mv'^2$ (10.3)
$\displaystyle \Rightarrow v'$ $\textstyle =$ $\displaystyle -v$ (10.4)

Figure 10.4: Diagram v to -v
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Figure 10.5: Diagram v to -v
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In other words, in colliding off the piston, the particle's velocity changed from $v$ to $-v$, and hence the momentum imparted to the piston is
\begin{displaymath} \mbox{\rm {Momentum imparted to piston}}=mv-m(-v)=2mv \end{displaymath} (10.5)

In time $t$, how many atoms will bounce off the piston? All the atoms with velocity $v$ can reach the piston in time $t$ if they are at a distance less than $x=vt$ from the piston. Consequently, all the atoms in a volume of size $xA=vtA$ will bounce off the piston. Hence, since the density (particles per unit volume) is $\displaystyle n=\frac{N}{V}$, in time $t$ the momentum imparted to the piston is
$\displaystyle \mbox{\rm {Momentum imparted to piston in time t}}$ $\textstyle =$ $\displaystyle 2mvxA\times n$ (10.6)
  $\textstyle =$ $\displaystyle 2\frac{Nmv^2A}{V}t$ (10.7)

Since the force on the piston is nothing but the rate at which momentum changes on the piston due to collisions of the gas atoms, we have
$\displaystyle \mbox{\rm {Force}}$ $\textstyle =$ $\displaystyle \frac{\mbox{\rm {Momentum imparted to piston in time t}}}{t}$ (10.8)
$\displaystyle \Rightarrow F$ $\textstyle =$ $\displaystyle 2\frac{Nmv^2A}{V}$ (10.9)

Recall pressure $P$ is defined to be force per unit area, and hence the pressure on the piston due to the gas is
$\displaystyle \mbox{\rm {Pressure}}$ $\textstyle =$ $\displaystyle \mbox{\rm {Force per unit area}}$ (10.10)
  $\textstyle =$ $\displaystyle \frac{F}{A}$ (10.11)
$\displaystyle \Rightarrow P$ $\textstyle =$ $\displaystyle 2\frac{Nmv^2}{V}$ (10.12)

From the ensemble point of view, the velocity $v$ of the atom is a random variable, and what the piston really experiences is the average value over all possible velocities that the atoms has as it bounces off the piston. Hence, denoting as usual average values by $<>$, we have

\begin{displaymath} PV=N<mv^2> \end{displaymath} (10.13)

The reason we have dropped the factor of $2$ in going from (10.12) to (10.13) is that we need to perform the average over only those particles which are heading towards the piston and not away from it. Since we are taking the average value of $v^2$, we are over-counting by a factor of $2$ since we are also including the particles moving away from the piston. Recall we had considered a very special set of velocities, namely, those heading straight for the piston, and hence with ${\bf v}=(v,0,0)$. In general, the velocity of an arbitrary atom has the form ${\bf v}=(v,u,w)$. Since all directions for the gas are equivalent, we have
$\displaystyle <v^2>=<u^2>=<w^2>$     (10.14)
$\displaystyle \Rightarrow <{\bf v}^2>=<v^2>+<u^2>+<w^2>=3<v^2>$     (10.15)

We finally have, from eq.(10.13), the following

\begin{displaymath} PV=\frac{1}{3}N<m{\bf v}^2> \end{displaymath} (10.16)

The total energy, $U$ of the gas is solely composed of kinetic energy. Hence we have
\begin{displaymath} U= N\frac{1}{2}m<{\bf v}^2> \end{displaymath} (10.17)

From (10.13) and (10.15) and (10.17) we have
$\displaystyle PV=\frac{2}{3}U$     (10.18)


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