# Total pressure of a gas?

I was just obtaining the total pressure of a gas enclosed in a cubical container of side $L$.



My attempt at solving the problem:

$N$ = Total number of gas molecules

$m$ = Mass of one molecule

$V$ = Volume of container

$V_{x1}$ = Velocity of a single particle along x-axis

$V_r$ = Root Mean Square velocity of the gas

$d$ = Density of gas

 Now let $A_1$ and $A_2$ be two opposite faces of the cube,

The time taken to move from $A_1$ to $A_2$ and back to $A_1 = \frac{2L}{V_{x1}}$

Change in momentum $= 2mV_{x1}$

Force applied on $A_1$ by single particle = $\frac{dp}{dt}$ = $\frac{m(V_{x1})^2}{L}$

Forced applied on $A_1$ by n particles = $\frac{Nm(V_r)^2}{3V}$

Pressure on wall $A_1$= $d(V_r)^{2/3}$

There are six walls, total pressure = $6PA_1$

 But my book refers to $PA_1$ as the total pressure, why?

## 1 Answer

Just because there are 6 walls, you shouldn't multiply the pressure by 6. The pressure on each wall is the same as the total pressure.

Look at it this way. Pressure is force applied by the gas particles per unit area. In your problem, you have found out this force. This is the same for each of the six walls of the cube. Each wall has the same area as well. So, for total pressure, if you divide total force by total area , you get back the same pressure as that on one wall since the factor 6 gets cancelled out. Hence, you get the correct answer as written in your book.

• You might want to add a reference to Pascal's law. – Yashas Feb 9 '17 at 6:58
• That would indeed be a good way to look at it. But considering the approach in the question, I figured that this would be the simplest way to explain it. Nevertheless, these very comments should draw the user's attention to Pascal's law. – TheFool Feb 9 '17 at 7:15