What is the Ideal Gas Force on a Container at Temperature T?

[Raghav Gupta]

Homework Statement

In an ideal gas at temperature T, the average force that a molecule applies on the walls of a closed container depends on T as T^q. A good estimate for q is:
A. 2
B. 1
C. 1/2
D. 1/4

Homework Equations

PV= nRT
I think, Volume of container and moles are constant.
So P ∝ T

The Attempt at a Solution

Answer should be B. 1, as force = pressure* area.
Here all would be ideal case.
But it is C. 1/2. Why?

 

[DrClaude]

Answer should be B. 1, as force = pressure/ area.

First, that should be force = pressure × area. Second, the pressure depends not only on the force of each collision of a molecule with a wall, but also on the frequency of those collisions. Both change with temperature. You need another equation to find the force of each molecular collision.

 

[Raghav Gupta]

Okay edited that force= pressure * area. That was a typo.
I don't know that second equation, can you give a hint?

 

[DrClaude]

Newton's second law.

 

[Raghav Gupta]

F ∝ d(mv)/dt
Now mass and velocity remains constant in ideal gas, so zero force?

 

[DrClaude]

velocity remains constant in ideal gas

?

 

[Raghav Gupta]

Okay on colliding it changes direction.
momentum change = 2mv ?

 

[DrClaude]

Okay on colliding it changes direction.
momentum change = 2mv ?

Yes, but only if v is the component of the velocity perpendicular to the wall.

Now, what is the relation between v and T?

 

[Raghav Gupta]

$v ∝ \sqrt{T}$
Now f ∝ dv/dt
So, f ∝ √T
Got it q= 1/2.

 

[Raghav Gupta]

First, that should be force = pressure × area. Second, the pressure depends not only on the force of each collision of a molecule with a wall, but also on the frequency of those collisions. Both change with temperature. You need another equation to find the force of each molecular collision.

But still pressure changes linearly with temperature,
According to ideal gas law, as p is proportional to T as V and moles are constant.
If you are saying it depends on frequency, where is the frequency part in other equation?

 

[DrClaude]

But still pressure changes linearly with temperature,
According to ideal gas law, as p is proportional to T as V and moles are constant.
If you are saying it depends on frequency, where is the frequency part in other equation?

Lets consider what happens in 1D, in a direction perpendicular to the wall. The average pressure is
$$
\langle P \rangle = \frac{\langle F \rangle}{A} = - \frac{m \langle \frac{ \Delta v }{\Delta t}\rangle}{A}
$$
When calculating the pressure, you are using the average force, which includes an average over time: you need to consider the average force per collision times number of collisions per unit time. When $T$ increases, two things change in that equation: the average force per collision increases and the average rate of collision increases. Overall, you get $P \propto T$, but the problem only asks for the force per collision, which is $\propto T^{1/2}$.

 

[Raghav Gupta]

I am not getting it correctly.
Here force ∝ pressure , as area of wall is constant.
So how if we are getting overall P∝ T, then force is not proportional to T ?
Does that word average makes a difference?

 

[DrClaude]

So how if we are getting overall P∝ T, then force is not proportional to T ?

The average total force applied to the wall is proportional to T, not the average force per collision.

 

[Raghav Gupta]

Okay, got it. Thanks.

 

[Raghav Gupta]

@DrClaude you must see this. I think you are incorrect.

 

[collinsmark]

@DrClaude you must see this. I think you are incorrect.

But the video is considering the "time-averaged" force on the walls (over many collisions). Note that the instructor in the video, around 1:12 or so, begins calculating the time it takes to go from once side of the wall to the other, and relates that to molecule's (average) velocity. Essentially what he's doing there bringing the frequency of collisions into the overall picture.

If the problem in this thread is asking for the time-averaged force on the wall of the container, time-averaged over many collisions (however many collisions that happen within a unit amount of time [for a fixed volume, fixed number of molecules, etc.]), then yes, what's said in the video applies to this thread.

On the other hand if the problem statement is asking for the average force for per individual collision, then that's a whole different story and the video doesn't fully apply to this thread.

[Edit: In my opinion, the way the problem statement is worded it's a little ambiguous.]

[Another edit: I should point out that if the answer C. is correct, the question really is asking about the force per collision (not time-averaged over all collisions within a unit period of time), and DrClaude's interpretation is correct. The problem statement could have been worded more clearly to avoid ambiguities, whatever the case.]

 

[Raghav Gupta]

Thanks collinsmark. Yeah, I think the problem statement is ambiguous.