Why Is Time Represented as 2d/v in Kinetic Gas Theory Calculations?

In summary: Suppose I have a single molecule bouncing between two parallel walls, normal to the walls. The distance between the walls is d, and the velocity of the molecule when it hits each wall is v. So the molecule bounces of one of the walls every 2d/v seconds. Each time it bounces off the wall its velocity reverses, so that its change in momentum is 2mv. Assume that the time in contact with the wall is negligible compared to the transit time between successive strikes of each wall. So the time average of the force that each wall exerts on the molecule is mv2/d. This is also equal to the time average of the force that each molecule exerts on
  • #1
sgstudent
739
3
To obtain force from momentum we use the formula change in momentum/time. The time in the equation refers to the amount of time the force is exerted on the gas molecules.

So when considering the kinetic theory of gases we are taught that F ∝ change in momentum/ time so F ∝ 2mv/(2d/v) and hence F ∝ mv2/d. The thing that I don't understand is why the time is 2d/v. 2d/v gives us the time taken for the molecule to travel from one wall to the other. But the "time" in the change in momentum/time equation shouldn't be the time taken for the molecule to travel from one wall to another and instead should be the time the molecule spends in collision with the wall instead.

So why is the 2d/v always used?
 
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  • #2
sgstudent said:
To obtain force from momentum we use the formula change in momentum/time. The time in the equation refers to the amount of time the force is exerted on the gas molecules.

So when considering the kinetic theory of gases we are taught that F ∝ change in momentum/ time so F ∝ 2mv/(2d/v) and hence F ∝ mv2/d. The thing that I don't understand is why the time is 2d/v. 2d/v gives us the time taken for the molecule to travel from one wall to the other. But the "time" in the change in momentum/time equation shouldn't be the time taken for the molecule to travel from one wall to another and instead should be the time the molecule spends in collision with the wall instead.

So why is the 2d/v always used?
This is the average rate of change of momentum for a single molecule. The instantaneous rate of change of momentum is zero during the time the molecule is traveling between the walls, and is very large for the brief time in which the molecule is in contact with the wall. If you add up all the average rates of change of momentum for all the molecules, you get the force exerted on a wall.

Chet
 
  • #3
Chestermiller said:
This is the average rate of change of momentum for a single molecule. The instantaneous rate of change of momentum is zero during the time the molecule is traveling between the walls, and is very large for the brief time in which the molecule is in contact with the wall. If you add up all the average rates of change of momentum for all the molecules, you get the force exerted on a wall.

Chet
Hmm but why would taking the time for the molecule to move from one wall to another give is the average force exerted? Shouldn't we had that very large value instead? I don't see how using 2d/v gives us the average force exerted on the wall by a single molecule..

Thanks
 
  • #4
sgstudent said:
Hmm but why would taking the time for the molecule to move from one wall to another give is the average force exerted? Shouldn't we had that very large value instead? I don't see how using 2d/v gives us the average force exerted on the wall by a single molecule..

Thanks
We want to add up the forces that all the molecules are exerting on the wall to get the total force, but they are not all hitting the wall as the same time. So this is the first step in beginning to take that into account.

Suppose I have a mass m, and I exert a force of 5 N on the mass for 10 seconds, and then I exert a force of 1 N on the mass for the next 50 seconds. What is the average force that I exerted on the mass over the entire 60 seconds? And, if I had exerted that same average force on the mass over the entire 60 seconds, would its momentum be the same as in the case of the two separate intervals?

Suppose I have a single molecule bouncing between two parallel walls, normal to the walls. The distance between the walls is d, and the velocity of the molecule when it hits each wall is v. So the molecule bounces of one of the walls every 2d/v seconds. Each time it bounces off the wall its velocity reverses, so that its change in momentum is 2mv. Assume that the time in contact with the wall is negligible compared to the transit time between successive strikes of each wall. So the time average of the force that each wall exerts on the molecule is mv2/d. This is also equal to the time average of the force that each molecule exerts on the wall. Now suppose you have N molecules bouncing back and forth, and they are all hitting the wall at different times. There are so many of them that, even though the contact time with the wall is very short, many of them are in contact with the wall at the same time (at different points during the contact interval). This has the effect of smoothing out the force at anyone time for those in contact with the wall. To get the overall force acting on the wall at anyone time, you just get Nnv2/d.

Chet
 

Related to Why Is Time Represented as 2d/v in Kinetic Gas Theory Calculations?

1. What is momentum in gases?

Momentum in gases refers to the quantity of motion that a gas particle possesses. It is a product of the mass and velocity of the particle and is a measure of how difficult it is to stop or change the direction of its motion.

2. How is momentum conserved in gases?

Momentum is conserved in gases according to Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. This means that in a closed system, the total momentum of all the particles before a collision is equal to the total momentum after the collision.

3. What is the relationship between force and momentum in gases?

Force and momentum in gases are directly proportional. This means that an increase in force will result in a proportional increase in momentum, and vice versa. This relationship is described by Newton's Second Law of Motion, which states that force is equal to the rate of change of momentum.

4. How do gases exert forces on each other?

Gases exert forces on each other through collisions between particles. When two gas particles collide, they exert equal and opposite forces on each other, resulting in a transfer of momentum.

5. How do temperature and pressure affect momentum in gases?

Temperature and pressure have a direct impact on the momentum of gases. An increase in temperature will result in an increase in the average velocity of gas particles, thus increasing their individual momentum. Similarly, an increase in pressure will result in more frequent collisions between particles, also increasing their momentum.

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