Force and pressure done by molecules

[songoku]

Homework Statement

Molecules of hydrogen, each has mass of 3.3 x 10^-27 kg, move with speed 1.6 kms^-1 hit a wall at angle π/3 rad to the normal. If there are 2.0 x 10²⁰ molecules s^-1 hit the area of 1.2 x 10^-4 m², find:
a. the average normal force on the wall if all the molecules are absorbed by the wall
b. the average normal force on the wall if all the molecules are reflected
c. pressure exerted by case (a) and (b)

Homework Equations

P = F/A
p = mv (momentum)
Δp = F.Δt

The Attempt at a Solution

Δp = F.Δt
Nmv = F cos (π/3) . t ; N = number of molecules
F = (N/t . m . v) / cos π/3

What is the difference when the molecules are absorbed and when reflected? Thanks

 

[ehild]

What is the difference when the molecules are absorbed and when reflected? Thanks

The absorbed molecule stays in rest after the collision, so it loses the initial normal component of momentum. The normal component of momentum of the reflected particle changes to the opposite.

ehild

 

[songoku]

The absorbed molecule stays in rest after the collision, so it loses the initial normal component of momentum. The normal component of momentum of the reflected particle changes to the opposite.

ehild

Oh I see. I think it will be like this:
a.
Δp = F.Δt
m(v₂ - v₁) = F . t

Because the molecules are absorbed, v₂ = 0 and v₁ = v cos π/3

b. Because the molecules are absorbed, v₂ = - v cos π/3 and v₁ = v cos π/3

Correct? Thanks

 

[ehild]

Because the molecules are absorbed, v₂ = 0 and v₁ = v cos π/3

b. Because the molecules are absorbed, v₂ = - v cos π/3 and v₁ = v cos π/3

Correct? Thanks

You meant reflected in the second case, did you not? Then correct.

ehild

 

[songoku]

You meant reflected in the second case, did you not? Then correct.

ehild

Ah yes. I meant reflected, not absorbed.

Thanks

 

[ehild]

You are welcome. Have you got the solution? ehild

 

[songoku]

You are welcome. Have you got the solution?


ehild

Haven't finished it, but this is my idea:

a. all the molecules are absorbed
Δp = F.Δt
m(v₂ - v₁) = F . t
-mv₁ = F.t
F = -N/t . m₀ . v cos π/3 ,where N is number of molecules and m₀ is the mass of one molecule

b. all the molecules are reflected
m(v₂ - v₁) = F . t
m(- v cos π/3 - v cos π/3) = F.t

The force for (b) is twice of (a)

Am I correct? Thanks

 

[ehild]

Haven't finished it, but this is my idea:

a. all the molecules are absorbed
Δp = F.Δt
m(v₂ - v₁) = F . t
-mv₁ = F.t
F = -N/t . m₀ . v cos π/3 ,where N is number of molecules and m₀ is the mass of one molecule

b. all the molecules are reflected
m(v₂ - v₁) = F . t
m(- v cos π/3 - v cos π/3) = F.t

The force for (b) is twice of (a)

Am I correct? Thanks

Yes, but the sign. You calculated the force the wall exerts on a molecule and multiplied it by N/t: You need to multiply the force a molecule exerts on the force, which is just of the opposite sign. And you need to give the numerical values.

ehild

 

[songoku]

Yes, but the sign. You calculated the force the wall exerts on a molecule and multiplied it by N/t: You need to multiply the force a molecule exerts on the force, which is just of the opposite sign. And you need to give the numerical values.

ehild

For (a), the force the molecules exert on the wall is:
F = N/t . m₀ . v cos π/3
= 2.0 x 10²⁰ x 3.3 x 10^-27 x 1.6 x 10³ cos π/3
= 5.28 x 10^-4 N

 

[ehild]

Right!

ehild

 

[songoku]

Right!

ehild

Thanks again