Writing physics notes on Ideal Gases?
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Explain how molecular movement causes the pressure exerted by a gas and hence
deduce the relationship,
(N = number of molecules)
**Can u please deduce this??**
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err... isn't this in the notes?
Anyway for deduction, you have to imagine a cubic volume of sides d. And we consider only 1 side of the wall.
for a molecule of mass m, it's change in momentum. So, that's mc - (-mc) = 2mc
time taken by a molecule between 2 sides of a wall = 2d / c (distance / speed)
hence, num of collisions per unit time = (N/3)/t = Nc/6d
We take N/3 because there are 3 directions, and we are only considering one direction. It's an average value.multiply above, you get Nmc²/3d for total change in momentum per unit time.
Remember Newton's 2nd law, force is d(mv)/dt, which means change in momentum per unit time. This means Nmc²/3d is the force on one side of the wall. Thus, pressure = force/area
p = â…“(Nmc²/d) / A
p = â…“(Nmc²/Ad) ============> A*d = V (volume)
p = â…“(Nmc²/V)(deduced)
btw, c is the root-mean-squared speed... You should use <c> in your workings. I was lazy to type out...
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I'm copying these from a study guide
The pressure exerted by an ideal gas could be explained by the motion of the gas molecules as follows:
- A gas is made up of many molecules that are constantly in random motion, colliding with one another and with the walls of the container.
- Each time a molecule strikes and rebounds from a wall, it undergoes a change of momentum. According to Newton's 2nd Law, a force is exerted by the wall on the molecule. According the Newton's 3rd Law, a the molecule would have exerted an equal but opposite force on the wall.
- Pressure is force exerted per unit area of contact. The pressure on the walls of the container is therefore the result of the reaction on the walls as the many molecules strike and rebound.