Why pressure stays the same when doubling both volume and temperature?

[lost captain]

TL;DR Summary

Viewing pressure as force per area, not the frequency of collisions. Can someone explain the reason why pressure stays the same when doubling both the volume and the temperature.

In thermodynamics we tend to think of pressure as the frequency of collisions with the walls of the container. And we say that the more the collisions the higher the pressure, the less the collisions the lower the pressure.

So lets say we have an ideal monoatomic gas enclosed in a cube container. According to the ideal gas law: PV = nRT, if we double the volume and the temperature then the pressure will remain the same
P = nRT / V
P= nR2T / 2V
And the explanation for that is that the number of collisions per second stays the same. Now the atoms have double the kinetic energy but they also have the double space to move into. So, that makes sense when viewing pressure as the frequency of collisions with the container walls.

But what if i were to think about pressure as the total force exerted over the total (inside) area of the cube?
When doubling the temperature the kinetic energy also doubles, so what happens to the forces of the collisions, do they also double? If they do double then the area should also double in order for the pressure to remain the same, But that doesn't happen, doubling the volume of a cube doesn't double the cubes total area.

 

[DaveC426913]

But what if i were to think about pressure as the total force exerted over the total (inside) area of the cube?

This is not my wheelhouse so I'm open to being corrected (as I often am, here) but pressure isn't the total force over the total area. It's force per unit area.

A one litre bottle at 2 atmos is under the same pressure (29.4 PSI) as a 100,000 litre shipping container at 2 atmos. Note the units: per square inch.

More generically P=f/a, where
P = pressure
f = force
a = area

 

[Ibix]

Let the initial box have size $L\times L\times L$ and contain $N$ atoms which have RMS velocity $v$. Each atom collides with average frequency $v/L$ and transfers momentum $2mv$, for a total momentum transfer per unit time (i.e., total force) to the wall of $F=2Nmv^2/L$ and hence a pressure over the wall of $P=2Nmv^2/L^3$ (edit: corrected missing $N$ - see next post).

Now let the temperature and volume double. The box's side length is now $L'=2^{1/3}L$, and the kinetic energy per atom doubles so the new RMS atom velocity is $v'=2^{1/2}v$. The number of atoms is unchanged. Now each atom collides with average frequency $v'/L'$ and transfers momentum $2mv'$, for a total momentum transfer per unit time (i.e., total force) to the wall of $F'=2Nmv'^2/L'$ and hence a pressure over the wall of $P'=2Nmv'^2/L'^3$ (edit: corrected missing $N$ here too). But $v'^2=2v^2$ and $L'^3=2L^3$, so $P'=P$ as expected.

 

[lost captain]

to the wall of F=2Nmu²/L and hence a pressure over the wall of P=2mu²/L³.

First of all, thank you very much for taking the time to reply
As i understand the formula for the force is only for one of the six walls of the container, so shouldn't it be F= 1/6N 2mu² , since the total number of atoms N will be divided by 6 for each wall ?
And then when we calculate the pressure why does the number of atoms dissappear, shouldn't it be P= N2mu²/L³ , according to your formula. If I'm right about the 1/6 of atoms, then pressure should be P= 1/6N2mu²/L³ ?

Also, and im sorry if this a stupid question, force is the change in momentum over the time of the collision, or the duration of the collision, right?
So we found the frequency to be u/L and hence the period will be L/u, meaning that every L/u seconds a collision happens, but this time is not the duration of the collision, it's just how much time it takes for a collision to happen.

 

[Ibix]

As i understand the formula for the force is only for one of the six walls of the container, so shouldn't it be F= 1/6N 2mu² , since the total number of atoms N will be divided by 6 for each wall ?

Well, you should really do the analysis separately for all three vector components. If you do that you will probably get different numeric constants, yes. It doesn't change the argument since the same numeric constant will appear in the primed and unprimed maths.

And then when we calculate the pressure why does the number of atoms dissappear, shouldn't it be P= N2mu²/L³ , according to your formula.

Yes, the missing $N$ is a typo. I've corrected it above - thanks.

Also, and im sorry if this a stupid question, force is the change in momentum over the time of the collision, or the duration of the collision, right?
So we found the frequency to be u/L and hence the period will be L/u, meaning that every L/u seconds a collision happens, but this time is not the duration of the collision, it's just how much time it takes for a collision to happen.

Force is the rate of change of momentum, and is typically not constant during a collision. The details of this are extremely messy.

Fortunately, in this case we can ignore that. To see why, first imagine a single ball bouncing between two walls - the force it exerts on either wall is zero most of the time with peaks. The shape of the peak is difficult to determine, depending on the shape and material of the ball. But the total momentum transfer is trivial to state: $2mv$ each collision, assuming no losses. And that means you can write down the average force easily: $2mv/T$ where $T$ is the time between collisions.

In the case of a single ball the average force isn't particularly meaningful - if you're personally holding the wall up the peak force and the impact duration is what you need to know. You can wander off and have a coffee between impacts. But in a gas there are always many atoms colliding with a wall. Here the average force is meaningful, because "how hard it is to hold the wall in place" is a result of the collective random collisions of many, many, atoms. Even though each atom is generating discrete peaks in force, there are so many all at different times and places that you can model that perfectly well as a continuous constant force, rather than a stacatto series of impacts.

Note that this averaging does break down on small scales. Brownian motion is a result of the failure of pressure to be perfectly uniform over small regions like the surface of a smoke particle. At that point you need a more sophisticated model than "just take the average momentum change".

 

[lost captain]

the force it exerts on either wall is zero most of the time with peaks.

Thank you once again and sorry it took me this long to reply, i felt that i needed to review my physics on momentum and collisions to understand this better.
So if i understand this correctly the force is zero most of the time, because most of the time the ball doesn't colide with the walls, rather most of the time the ball is traveling between the walls ?

But the total momentum transfer is trivial to state: 2mv each collision, assuming no losses. And that means you can write down the average force easily: 2mv/T where T is the time between collisions.

So this T is not the actuall duration of the collision.... we dont calculate the force as the change in momentum over the time of the collision, rather the change in momentum over the total time?

 

[Ibix]

So if i understand this correctly the force is zero most of the time, because most of the time the ball doesn't colide with the walls, rather most of the time the ball is traveling between the walls ?

For a single ball (or any single atom in the gas), yes, most of the time it's flying through the volume and only occasionally bouncing off a wall.

So this T is not the actuall duration of the collision.... we dont calculate the force as the change in momentum over the time of the collision, rather the change in momentum over the total time?

It depends what you're calculating. The force from a single ball or single atom varies from nothing (most of the time) to high (when the collision occurs). If we were dealing with one ball this is what would matter - most of the time there is no force, then there's a thump as the ball bounces.

But in the case of many atoms, you find that there's always some atoms in the process of bouncing. If (for the sake of argument) an atom spends 1% of its time in the process of bouncing, that means that all the time 1% of atoms are in the process of bouncing. So the average force is what we're interested in because we can average across all the atoms.

 

[lost captain]

If (for the sake of argument) an atom spends 1% of its time in the process of bouncing, that means that all the time 1% of atoms are in the process of bouncing.

So the pressure at one wall of the container is actually created by 1% of the atoms(for example, not the actual percentage)...? And why do we calculate the average force from all 99% plus 1%? Since only this 1% of atoms is responsible for the pressure and the change in momentum and the force?
I'm arguing that the pressure should be the average force from all the 1%atoms over the duration of the collision

 

[Ibix]

So the pressure at one wall of the container is actually created by 1% of the atoms(for example, not the actual percentage)...?

Worth stressing that 1% is a number I made up, and hugely over-estimates the amount of time atoms spend interacting with the wall. But let's stick with it, because the actual value doesn't matter to the argument.

So let's say an atom hits the wall every time $T$, and it spends $0.01T$ interacting with the wall. Of the $N$ atoms in the box, $0.01N$ are currently interacting with the wall, so the total momentum transfer in that short time is $2mv\times 0.01N$. Thus the force in that short time is $2mv\times 0.01N/(0.01T)$ and the 1% cancels out. The same argument can be made an instant later about a different 1% of the atoms. So the force is constant at $2mvN/T$. Noting that $L=vT$ you recover the expression for force from post #3. (Returning to your comment #4, laying it out like this suggests there probably is a factor of two missing at least, but since it's missing in both the before and after expansion cases it cancels out when comparing $P$ and $P'$.)

The point is that the momentum transfer is cumulative. One atom transfers $2mv$, two atoms transfer $4mv$, one million atoms transfer $2000000mv$, and so on. But the rate of momentum transfer is that $2mv$ per atom multiplied by the number of collisions divided by the time taken for those collisions to occur. But whatever time period you consider ($0.01T$, $17T$, $fT$, etc) the number of collisions will always be the same fraction times the number of atoms in the box ($0.01N$, $17N$, $fN$, etc). So you always get a force of $2mvN/T$.

Also note that atoms are nowhere near so regimented as I'm describing. The actual number of collisions in a small time can and does vary randomly, but over reasonable large box sizes and human time scales the fluctuations average out.

 

[lost captain]

So let's say an atom hits the wall every time T, and it spends 0.01T interacting with the wall. Of the N atoms in the box, 0.01N are currently interacting with the wall

Why does one atom spend 0.01T interacting with the wall? We assumed for the sake of the argument that 0.01N atoms hit the wall, we didn't assume anything about the duration of collision.
Am i missing something obvious here?
Isn't the number of atoms hitting the wall independent of the collision's duration?

 

[russ_watters]

Duration of the collision only matters if you're trying to show why the force is constant instead of chaotically varying when looking at/modeling a small number of particles. You don't need it to calculate the average force (only the momentum and frequency of collissions).

 

[Ibix]

Why does one atom spend 0.01T interacting with the wall? We assumed for the sake of the argument that 0.01N atoms hit the wall,

They're the same assumption. If an atom spends 1% of its time interacting with the wall then 1% of them are interacting at any given time.

 

[lost captain]

They're the same assumption. If an atom spends 1% of its time interacting with the wall then 1% of them are interacting at any given time.

Thank you for your patience, i still can't fully understand your explanation so let me give you a way too simplified numerical example, so i can be sure i get what you are saying.

Once again: ideal gas enclosed in cubic container(10meter side). Each atom has velocity 5m/s and when it hits the wall, it stays in contact for 0.5 seconds (Our only assumption is the duration of collision)
So for an atom to travel straight from one wall to the other and then bounce back takes 2.5 seconds, two seconds to travel 10m and 0.5 seconds during the collision.
And obsiously this holds true for all 20 atoms.
So assuming the duration of collision to be 0.5 seconds how can you find out how many atoms hit the red wall?

According to your logic: 0.5s is the one fifth of 2.5s , so one atom spends one fifth of its time in contact with the wall, while the other four fifths the atom travels be tween the walls.

So for a half a second atoms will be in contact with the wall, then half a second passes these atoms have left and new atoms are now in contact with the wall.

Now, the question is how many atoms hit-stay in contact for 0.5 s with the red wall, every 0.5s ?
Your answer: 0,5N = 0.5 * 20 = 4 atoms
....and i don't understand. Why should we multiply 0.5 times N? All we know is that for every 0.5 seconds new atoms will be in contact with the wall. Why should it be 4 atoms? Why not 2 atoms every 0,5 sec? Or why not 5?
Thank you, and I'm sorry for troubling you

 

[A.T.]

So for a half a second atoms will be in contact with the wall, ...

The whole wall contact duration is completely irrelevant. All you care about is the average momentum transfer rate over many collisions (force).

Consider this minimal symmetrical particle arrangement, with center of mass staying at rest:
- 6 particles in total in a cube
- all moving at the same speed
- 2 particles per each XYZ-dimension, moving in opposite directions, repeatedly striking the opposite walls simultaneously, rebounding perfectly elastically

Now compare what happens under those assumptions for a cube with double the volume and double particle kinetic energy (primed case):

Speed of particles:
$v' = 2^{1/2} \cdot v$

Travel distance between collisions:
$L' = 2^{1/3} \cdot L$

Time elapsed between collisions:
$t' = L'/v' = 2^{-1/6} \cdot t$

Momentum transferred per collision:
$M' = 2^{1/2} \cdot M$

Area:
$A' = 2^{2/3} \cdot A$

Pressure:
$P' = F'/A' = M'/(t' \cdot A') = 2^{1/2}/(2^{-1/6} \cdot 2^{2/3}) \cdot M/(t\cdot A) = P$

 

[lost captain]

average momentum transfer rate

And that is the change in momentum(dP) over the change in time in which that momentum has changed (dt) ? Then that change in time is the duration of the collision

 

[A.T.]

And that is the change in momentum(dP) over the change in time in which that momentum has changed (dt) ? Then that change in time is the duration of the collision

No. If I give you an apple every hour, then the average rate of apple transfer is 1 apple/hour.

The duration of me stretching out my hand and passing you the apple is irrelevant for this average rate of apple transfer. So you can also assume it is instantaneous, just like for the collisions.

 

[lost captain]

Oh okay so the thing i was describing was just the momentum? Not the average momentum that we need in this situation

 

[A.T.]

Oh okay so the thing i was describing was just the momentum? Not the average momentum that we need in this situation

You were describing the average momentum transfer rate during a single collision.

But what is relevant for pressure is the average momentum transfer rate over the duration of many collisions, which is based on the entire time interval, including the times between the collisions.

You won't get more apples per day from me, if I move my hand quicker during passing you the apple, but still do it only once an hour.

 

[lost captain]

You were describing the average momentum transfer rate during a single collision.

But what is relevant for pressure is the average momentum transfer rate over the duration of many collisions, which is based on the entire time interval, including the times between the collisions.

You won't get more apples per day from me, if I move my hand quicker during passing you the apple, but still do it only once an hour.

Thank you very much i think I'm really close to understand this completely.
So one more thing i want to ask

In one period time: L/U, do the collisions of the atoms with the wall happen all at the same moment or at different moments?
To use your analogy, if you had sliced the apple to 10 pieces and over the period of one hour you give me pieces of the apple , you dont give me the pieces all together but at the end of this one hour you have given me all ten pieces and i have in my hands a whole aple. So i end up with an apple per hour
Is this whats going on with the atoms, or is it that you give me all the ten pieces together each time one hour ends, like you give me 10 pieces when the clock shows 8am and then another 10 pieces when its 9am ?

 

[A.T.]

do the collisions of the atoms with the wall happen all at the same moment?

In my 6 particle scenario yes. Because you want to keep the center of mass static, so all the particles' KE is associated with temperature, and not with bulk motion of the gas.

In reality no. But with many particles, each having just a tiny fraction of the total mass, and many random collisions per second, there is no relevant jiggle of the center of mass, even if not every collisions has an simultaneous equal but opposite collision.

 

[lost captain]

In my 6 particle scenario yes

That was one particle for each wall.
If for example the particles were 12 then 2 whould hit every wall simultaneously ?
What im trying to understand is the reason we take dT to be the time between the collisions, that makes sense if the collisions don't happen all together in the same time, rather they all happen in the period of L/U

We have an average momentum change every dt right? So in that period dt, the total amount of collisions results in a momentum change: 2mu
(where m is the total mass of these atoms hitting the wall and u is their mean velocity)
So why the collisions happen all at once, or not?

In your example, but with 12 particles, the total momentum change will still be 2mu even if the 2 particles wont hit the wall together?

 

[A.T.]

What im trying to understand is the reason we take dT to be the time between the collisions,

Same reason that we use the average time between two apples to get the average apple transfer rate over long time.

 

[lost captain]

Same reason that we use the average time between two apples to get the average apple transfer rate over long time.

If you were giving me 2 apples an hour but the first one you give it to me in the first half hour and the second one you give it to me the second half of the hour, whould this senario be the same as if you were giving me both apples at the same time each hour?

 

[lost captain]

You were describing the average momentum transfer rate during a single collision.

If as you said i was describing the momentum transfer rate for a single collision then for all the collisions we should use the mean duration time of these collisions. Still we use a time interval where in most of it no collisions happen

 

[A.T.]

Still we use a time interval where in most of it no collisions happen

For the average rate during a time interval you use that interval, regardless of how much of that interval was actually used to do something.

 

[lost captain]

It's like.. for example a car that doesn't move for 59 minutes and then at the 60nth minute it moves 3 meters.
So in this 1 minute it has a velocity: 3meters/ 60 seconds = 0.05 m/s But! Instead we say in this whole hour, in the entire 60minutes (where in the first 59 minutes the velocity was 0 and in the last 60nth minute the velocity was 0.05m/s) that
the car mean velocity is :
3meters/ 3600 seconds = 0.00083 m/s

 

[A.T.]

It's like.. for example a car that doesn't move for 59 minutes and then at the 60nth minute it moves 3 meters.
So in this 1 minute it has a velocity: 3meters/ 60 seconds = 0.05 m/s But! Instead we say in this whole hour, in the entire 60minutes (where in the first 59 minutes the velocity was 0 and in the last 60nth minute the velocity was 0.05m/s) that
the car mean velocity is :
3meters/ 3600 seconds = 0.00083 m/s

Yes, because if the car keeps repeating that process, then over many such 1h cycles its effective mean speed, which determines how far it has traveled in total, will be 0.00083 m/s.

 

[lost captain]

Okay thank you very much, again.
Another question that rises from this:
The pressure exerted on the wall i thought it as beeing constant all the time and i guess mAcroscopically is constant but actually the pressure is being applied every period T every time the particles collide with the wall so mIcroscopically it isnt constant?

 

[A.T.]

The pressure exerted on the wall i thought it as beeing constant all the time and i guess mAcroscopically is constant but actually the pressure is being applied every period T every time the particles collide with the wall so mIcroscopically it isnt constant?

Yes, on small scales there is fluctuation. Not just for gas pressure, but pretty much for everything in physics.

 

[russ_watters]

Another question that rises from this:
The pressure exerted on the wall i thought it as beeing constant all the time and i guess mAcroscopically is constant but actually the pressure is being applied every period T every time the particles collide with the wall so mIcroscopically it isnt constant?

From post #11:

Duration of the collision only matters if you're trying to show why the force is constant instead of chaotically varying when looking at/modeling a small number of particles. You don't need it to calculate the average force (only the momentum and frequency of collissions).

You've been creating scenarios designed to imply chaotically varying force by looking at a tiny number of particles/collissions. This is far, far from the reality. The reality is that the number of collisions is huge, so the variation in force is tiny. If you're still doubting this, it might be instructive to calculate how many particles and how often there are collissions in real life.

 

[lost captain]

From post #11:

You've been creating scenarios designed to imply chaotically varying force by looking at a tiny number of particles/collissions. This is far, far from the reality. The reality is that the number of collisions is huge, so the variation in force is tiny. If you're still doubting this, it might be instructive to calculate how many particles and how often there are collissions in real life.

Yeah i guess my initial point of view and how i imagined the particles and their collisions with the wall was wrong from the start