11.3.4 Pressure Relationship with Mean-Square-Speed

In this section, we will discover the quantitative link between a gas’s pressure (a macroscopic property) and the gas particles’ mean-square-speed (a microscopic property). Buckle up. It is going to be a long but exhilarating ride.

To start off, let’s imagine a space enclosed by a cube of sides L. To keep things simple, let’s imagine that there is only one solitary gas particle of mass m travelling at constant speed v in the x-direction in this cube.

Every time it hits the right wall (shaded in green), it rebounds with speed v since an elastic collision is assumed.

During each collision, the particle undergoes momentum change of $\Delta p=2mv$ .

The time taken for the particle to return to make the next collision is $\displaystyle \Delta t=2L\div v$ .

So the average force exerted by one gas particle $\displaystyle F=\frac{{\Delta p}}{{\Delta t}}=\frac{{2mv}}{{2L\div v}}=\frac{{m{{v}^{2}}}}{L}$ .

Now let’s upgrade our model to having $\displaystyle i=1,2,...N$ number of gas particles. For the time being, let’s imagine that all the N particles are travelling along the x-axis.

If we denote the speed of the i^th particle by v_i, then the i^th particle exerts a force of $\displaystyle \frac{{m{{v}_{i}}^{2}}}{L}$ on the wall.

This means that the total force exerted on the wall is

$\displaystyle \begin{aligned}F&=\frac{{m{{v}_{1}}^{2}}}{L}+\frac{{m{{v}_{2}}^{2}}}{L}+\frac{{m{{v}_{3}}^{2}}}{L}+...\frac{{m{{v}_{N}}^{2}}}{L}\\&=\frac{m}{L}({{v}_{1}}^{2}+{{v}_{2}}^{2}+{{v}_{3}}^{2}+...{{v}_{N}}^{2})\end{aligned}$

Instead of working with individual v_i² values, it is a lot smarter to work with the mean-square value <v²>. So

$\displaystyle \begin{aligned}F&=\frac{m}{L}(N<{{v}^{2}}>)\\&=\frac{{Nm}}{L}<{{v}^{2}}>\end{aligned}$

The pressure on the wall is thus

$\displaystyle \begin{aligned}p&=\frac{F}{A}\\&=\frac{{Nm}}{L}<{{v}^{2}}>\div {{L}^{2}}\\&=\frac{{Nm}}{{{{L}^{3}}}}<{{v}^{2}}>\\&=\frac{{Nm}}{V}<{{v}^{2}}>\end{aligned}$

where V is the volume of the gas.

Now let’s address the elephant in the room: the gas particles are not constrained to move only long the x-axis. Each particle, instead of having a single-directional velocity v_i, actually has a three-dimensional velocity $\displaystyle {{\mathbf{c}}_{i}}=[{{v}_{{x,i}}},{{v}_{{y,i}}},{{v}_{{z,i}}}]$ .

Upon closer inspection, we realize that the pressure exerted on the green wall has nothing to do with v_y and v_z. As there is no change to the momentum in the y and z directions when the particles collide with the green wall, the pressure exerted on the green wall only depends on v_x.

So let’s fine tune our formula replacing $\displaystyle \left\langle {{{v}^{2}}} \right\rangle$ with $\displaystyle \left\langle {{{v}_{x}}^{2}} \right\rangle$

$\displaystyle p=\frac{{Nm}}{V}<{{v}_{x}}^{2}>$

Now, the final push. Let’s replace the single-directional component v_x with the 3-deminsional velocity c to. But what’s the link between c and v_x?

From Pythagoras theorem, we know the magnitudes are related by

$\displaystyle {{c}^{2}}={{v}_{x}}^{2}+{{v}_{y}}^{2}+{{v}_{z}}^{2}$

Working with statistical averages, we can write

$\displaystyle \begin{aligned}\left\langle {{{c}^{2}}} \right\rangle &=\left\langle {{{v}_{x}}^{2}+{{v}_{y}}^{2}+{{v}_{z}}^{2}} \right\rangle \\&=\left\langle {{{v}_{x}}^{2}} \right\rangle +\left\langle {{{v}_{y}}^{2}} \right\rangle +\left\langle {{{v}_{z}}^{2}} \right\rangle \end{aligned}$

Since the motion of the gas particles is random, we expect the speed distribution to be uniform in all three directions, meaning $\displaystyle \left\langle {{{v}_{x}}^{2}} \right\rangle =\left\langle {{{v}_{y}}^{2}} \right\rangle =\left\langle {{{v}_{z}}^{2}} \right\rangle$ . So

$\displaystyle \begin{aligned}\left\langle {{{c}^{2}}} \right\rangle &=\left\langle {{{v}_{x}}^{2}} \right\rangle +\left\langle {{{v}_{y}}^{2}} \right\rangle +\left\langle {{{v}_{z}}^{2}} \right\rangle \\&=3\left\langle {{{v}_{x}}^{2}} \right\rangle \end{aligned}$

Replacing $\displaystyle \left\langle {{{v}_{x}}^{2}} \right\rangle$ with $\displaystyle \frac{1}{3}\left\langle {{{c}^{2}}} \right\rangle$ , we obtain

$\displaystyle p=\frac{1}{3}\frac{{Nm}}{V}\left\langle {{{c}^{2}}} \right\rangle$

Since Nm is actually the mass of the gas M, and density is $\displaystyle \rho =\frac{M}{V}$ , this equation is also often presented as

$\displaystyle p=\frac{1}{3}\rho \left\langle {{{c}^{2}}} \right\rangle$

We now know exactly how the macroscopic properties of pressure (and density) is related to the microscopic property of the molecular motion (of godzillion number of gas particles)! Note that gas pressure is directly proportional to the mean-square speed of the gas particles (and not the average speed).

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Video Explanation

Derivation of p=1/3ρ<c²>

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xmPhysics

11.3.4 Pressure Relationship with Mean-Square-Speed

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