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

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Discussion Overview

The discussion centers around the behavior of pressure in an ideal gas when both volume and temperature are doubled. Participants explore the implications of the ideal gas law and the relationship between pressure, volume, and temperature, considering both kinetic theory and force dynamics within a container.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants explain pressure as related to the frequency of collisions with container walls, suggesting that doubling both volume and temperature keeps pressure constant due to unchanged collision frequency.
  • Others argue that pressure should be viewed as force per unit area, emphasizing that pressure remains the same regardless of the size of the container, as long as the conditions of temperature and volume are met.
  • A participant presents a mathematical analysis using kinetic theory, showing how pressure is derived from momentum transfer during collisions, leading to the conclusion that pressure remains constant when both volume and temperature are doubled.
  • Concerns are raised about the assumptions made in the calculations, particularly regarding the distribution of atoms across the walls of the container and the implications for pressure calculations.
  • Questions arise about the nature of force during collisions, with discussions on whether the force should be considered over the duration of the collision or the total time between collisions.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of pressure and the implications of doubling volume and temperature. There is no consensus on the best approach to understand the relationship between these variables, and the discussion remains unresolved.

Contextual Notes

Participants note the complexity of force dynamics during collisions and the averaging of forces in a gas, which may not hold true at small scales. There are also discussions about the assumptions made in the mathematical models used to describe the system.

  • #31
russ_watters said:
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
 
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  • #32
lost captain said:
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
Key is that you don't have to know the details of how the particles collide with the wall, in order to answer the original question. All you need to know is how the average particle momentum, average rate of collisions and wall area change.

But the hand wavy explanation you mention in your OP is indeed not quite correct:
lost captain said:
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.
As you see in post #14 the rate of collisions is not the same. And it is not that obvious that the various powers of 2 involved in the pressure ratio cancel. But they do.
 
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  • #33
A.T. said:
As you see in post #14 the rate of collisions is not the same.
Sorry, but where in post #14 ? Isn't the rate of collisions, the pressure? And pressure stays the same. Probably i didn't understand what you are referring to.
edit: wait you mean the time between collisions right?
 
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  • #34
lost captain said:
Sorry, but where in post #14 ? Isn't the rate of collisions, the pressure? And pressure stays the same. Probably i didn't understand what you are referring to.
The rate of collisions is not the pressure. It is just a rate of collisions. It is not even a force.

The [average] rate of collisions multiplied by the [average] momentum transferred per collision is not the pressure. It is a rate of momentum transfer. A force. [A force is a momentum transfer without an associated mass transfer].

The [average] rate of collisions multiplied by the [average] momentum transferred per collision and divided by the surface area over which the collisions are distributed is a force divided by an area. That is to say, a pressure.
lost captain said:
edit: wait you mean the time between collisions right?
It does not really matter whether we describe the rate of collisions as how many collisions per unit time or how many units of time between collisions. Either way will work. One number is the reciprocal of the other.

One may need to be careful about how one computes an "average" rate in order to get the averages to match. But that is subject matter for a different thread.
 
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  • #35
jbriggs444 said:
The [average] rate of collisions
Just a clarification: Average rate of collisions and average frequency u/L in post#3 is the same thing?
 
  • #36
lost captain said:
Just a clarification: Average rate of collisions and average frequency u/L in post#3 is the same thing?
Yes, it's also the reciprocal of the ##t## in post #14, which is different in the primed case, and so is its reciprocal.
 
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