# How do we choose the path when deriving the ideal gas law?

So, my textbook details the derivation of ideal gas law in this way:

$$m_0v -(-m_0v)=2m_0v \tag{1}$$ $$F=\frac{2m_0v} {\Delta t} \tag{2}$$ $$\Delta t = \frac {a} v \tag{3}$$ $$F=\frac{2m_0v^2} a \tag{4}$$ $$\bar F = \frac{2m_0\bar v^2} a \tag{5}$$ $$\bar F = \frac{N} 6 \frac {2m_0\bar v^2} a \tag{6}$$ $$p = \frac {1} 3 \frac {N} {a^3} m_0\bar v^2 \tag{7}$$

However, I can't seem to grasp what happened in step 3. So, the force is equal to the change in momentum over a time, right, but the time was expressed over the distance of the furthest molecule from that cube wall (equal to a, or the length of cube side). This perfectly makes sense if there were one molecule bouncing left and right, (I'm sure I'm missing something obvious here) but couldn't have I just as well thought of that molecule moving by the diagonal with the speed $v$? The force would still be the same, but the path would be longer ($a\sqrt{2}$) and therefore end up in the 7th step as $a^3\sqrt{2}$.

• Hint: instead of thinking about the distance the particle travelled, think about the different components of the velocity of the particle. In particular the components parallel and perpendicular to the surface it hits. – By Symmetry Jan 10 '18 at 2:20
• How do you understand 5 --> 6 ? It might (or might not) help to understand that one. – garyp Jan 10 '18 at 2:23
• N/6 pops out of nowhere – QuIcKmAtHs Jan 10 '18 at 6:03
• Even if a molecule is not bouncing left-and-right, but rather hits the walls at an angle, then the momentum change is still only perpendicular to the wall it hits. The parallel momentum component is unchanged. – Steeven Jan 10 '18 at 6:23
• Note that you have written $\bar v^2$ where it should be $\bar {v^2}$ – Farcher Jan 10 '18 at 14:10

## 2 Answers

A slight variation on your derivation.

You start off considering a molecule moving at $v_{\rm x}$ in the x-direction.
The magnitude of the change in momentum of the molecule on hitting a wall and rebounding is $2mv_{\rm x}$.

The time it takes for the molecule to start at some position in the box and return to the same position after rebounding off both walls is $\Delta t = \dfrac {2a}{v_{\rm x}}$

So the magnitude of the force on the molecule is $\dfrac{m v_{\rm x}^2}{a}$

So the average force for all the molecules moving in the x-direction is $\dfrac{m \left <v_{\rm x}^2 \right >}{a}$

Assuming that the actual magnitude of the velocity of the molecule is $v$ then $v^2 = v_{\rm x}^2+ v_{\rm y}^2 + v_{\rm z}^2$

So $\left <v^2 \right> = \left< v_{\rm x}^2 \right>+ \left< v_{\rm y}^2 \right>+ \left< v_{\rm z}^2 \right>$

Assume random motion so $\left< v_{\rm x}^2 \right>= \left< v_{\rm y}^2 \right>= \left< v_{\rm z}^2 \right> \Rightarrow \left< v_{\rm x}^2 \right> = \dfrac{\left< v^2 \right>}{3}$

And with $N$ molecules in the box magnitude of the force on the molcules due to the wall is $Nm\dfrac{\left< v^2 \right>}{3}$ and this is equal to the force on the wall due to the molecules by Newton's third law.

The time quantity is totally wrong. The molecule undergoes a change in momentum only while it is in contact with the container wall. When it(the molecule) rebounds it is no longers being acted upon by a force. I don't understand what [a] represents in your expressions.