# Explaining Avogadro's Law using kinetic theory

1. Sep 3, 2012

### Bipolarity

So far I have found a kinetic-theory explanation of the Boyle's Law, Charle's Law, and Pressure-Temperature Law. For example, for the pressure-temperature law: increasing the temperature of a gas while holding the volume constant causes the gas molecules to collide more frequently with the container of the gas, resulting in increased pressure.

Is there an explanation of the Avogadro's Law that uses the kinetic theory?

BiP

2. Sep 3, 2012

### bigerst

i think avogadro's law relies on the fact that microscopically gas particles are like tiny billiard balls, aside from the mass difference they have essentially the same properties and are ideal (ideal collisions, no interaction between molecules etc)

3. Sep 3, 2012

### Bipolarity

I don't see how the analogy is relevant here.

BiP

4. Sep 4, 2012

### Staff: Mentor

Avogadro's law is true for ideal gases (and those behave like a collection of point-like billard balls), and an approximation for gases which are nearly ideal.
What do you want to explain? Why some gases are nearly ideal?

5. Sep 4, 2012

### Philip Wood

In the case of the pressure law, it's not just that the more frequent hits increase the pressure, but that, on average each hit imparts a greater impulse. That's why $\overline{c^2}$, with the 'squared', appears in the kinetic theory formula for pressure:
$$pV = \frac{1}{3} Nm\overline{c^2}$$.

Now let's look at Avogadro. We need the additional input that molecules of all ideal gases at the same temperature have the same mean KE of translational motion, $\frac{1}{2}m \overline{c^2}$. [Jeans, in The kinetic Theory of Gases has a nice justification for this, using the fact that on average there must be no energy exchange in collisions between gas molecules and wall molecules if the gas is in equilibrium with its container walls.]

So for any two gases at the same temperature $$m_1\overline{c_1^2} = m_2\overline{c_2^2}$$.

So, using the pressure formula above:
$$\frac{p_1V_1}{N_1} = \frac{p_2V_2}{N_2}$$

This formula applies for equal temperatures, but if we also impose the conditions that $p_1=p_2$ and $V_1=V_2$, then $N_1=N_2$.

So at the same temperature and pressure, equal volumes of gases contain the same number of molecules!

[You can reach the same conclusion using $\frac{1}{2}m \overline{c^2}=\frac{3}{2}kT$, but this isn't quite as economical because the argument above does not require a specific relationship between temperature and mean KE, merely a knowledge that if two gases have the same mean molecular KE, their temperatures are the same, and the converse.]

Last edited: Sep 4, 2012