Why Is Time Interval Between Gas Particle Collisions Calculated as Δt=2l/v?

In summary: So the average force would be too small if we included the time when there is no force.In summary, the discussion was about finding the RMS speed of an ideal gas using classical mechanics. There was a question about why the time interval between collisions is used instead of the contact-time between particle and wall. The reason for this is that we want to find the average pressure over a long period of time, not just during the particle-wall collisions. The use of V rms also helps with this approximation. There was some discussion about the derivation and the use of sigma (σ) as the root-mean-square axial speed in statistical mechanics. Overall, there are some uncertainties but the group agrees that this is a good derivation to avoid misrepresentation of Δt terms
  • #1
gema
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Hello everyone,

When I was studying about the way to find the RMS speed of an ideal gas by using classical mechanics, I wondered that why the time interval of the collisions can be approached as the time between two collisions, instead of contact-time between particle and wall.

remember? Δt=2l/v,

Thankyou
 
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  • #2
gema said:
Hello everyone,

When I was studying about the way to find the RMS speed of an ideal gas by using classical mechanics, I wondered that why the time interval of the collisions can be approached as the time between two collisions, instead of contact-time between particle and wall.

remember? Δt=2l/v,

Thankyou

Off hand guess. It might depend on the size of the container. Wall collisions would be relatively rare compared to collisions between particles.
 
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  • #3
mathman said:
Wall collisions would be relatively rare compared to collisions between particles.

For a cubic meter container of humid air at a temperature of 25°C, a total pressure of 1,000 hectopascals, and a vapor pressure of 100 pascals, I get a collision rate with a wall of some 1.79 X 1028 impacts per square meter per second.

The number of intermolecular collisions would approximate some 6.21 X 1034 collisions per cubic meter per second. This makes intermolecular collisions some 112 thousand times more common than collisions with a wall [there are six walls].

By the by, this is only a rough approximation because intermolecular collisions depend upon estimates of effective molecular "size"--a subject for which precise answers are lacking.
 
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  • #4
Thanks for your helps,
I know that collisions between particles and walls is relatively rare compare to collisions between particles themselves. But, when we're trying to find the pressure at the wall using classical mechanics law, we're beginning with F=Δp/Δt right?

where Δp= -2mvx (If x is perpendicular to the wall)
and why should Δt=2l/vx , instead of contact-time between particle and wall?
This breaks Newton's 2nd law isn't it?
or do you have some explanations for that?

Thankyou
 
  • #5
In Feynman's lecture, I found his derivation different from common derivation.[itex]P=\frac{N}{A_{collision}}[/itex]

[itex]N=F_{net,acted\:by\:wall\:eek:n\:particles}=\frac{dp_{colliding\:particles}}{Δt_{collision}}[/itex]

[itex]dp_{colliding\:particles}=N_{colliding\:particles}\:\times\:dp_{one\:particle}[/itex]

[itex]N_{colliding\:particles}=\frac{1}{2}ρA_{collision}V_{rms,x}\:Δt_{collision}[/itex]

[itex]ρ=density=\frac{N_{total}}{V_{total}}[/itex]

The reason why there is [itex]\frac{1}{2}[/itex] in front of ρ is that the center velocity is zero so that there is only half particles colliding with wall in that space.

[itex]dp_{one\:particle}=m\times(2V_{rms,x})[/itex]

[itex]∴P=\frac{dp_{colliding\:particles}}{A_{collision}Δt_{collision}}[/itex]

[itex]∴P=N_{colliding\:particles}\times\frac{dp_{one\:particle}}{A_{collision}Δt_{collision}}[/itex]

[itex]∴P=\frac{1}{2}ρA_{collision}V_{rms,x}\:Δt_{collision}\frac{2mV_{rms,x}}{A_{collision}Δt_{collision}}---------------(*)[/itex]

[itex]∴P=ρmV^{2}_{rms,x}=\frac{N_{total}}{V_{total}}mV^{2}_{rms,x}[/itex]

And...

[itex]V^{2}_{rms,x}=\frac{V^{2}_{rms}}{3}[/itex]

[itex]∴P=\frac{N_{total}}{3V_{total}}mV^{2}_{rms}[/itex]

[itex]∴N_{total}\frac{1}{2}mV^{2}_{rms}=E_{k,total}=\frac{3}{2}PV_{total}[/itex]In my opinion, the only problem is... why can we use [itex]V_{rms,x}[/itex] in (*)?...
 
  • #6
I don't know if this will be of any help, but in statistical mechanics the root-mean-square axial speed is easily defined as:

σ=(kT/m)1/2

Here, sigma (σ) is the root-mean-square axial speed in meters per second, k is Boltzmann's Constant in joules per molecule per Kelvin, T is the temperature in Kelvins, and m is the unique molecular mass in kilograms. σ is used instead of vrms, because this quantity is also the standard deviation of the axial velocity distribution.
 
  • #7
I think this derivation is good, in terms to avoid misrepresentation (so far) of Δt terms in common derivation.

In my opinion, the only problem is... why can we use Vrms,x in (*)?...

I guess we should use V rms because it contains the average and variance. It means rms is arbitrary value, and it's good for approximation.

by definition, rms quantity is

<[itex]x^{2}[/itex]>=[itex]σ^{2}[/itex]+[itex]<x>^{2}[/itex]
 
  • #8
gema said:
and why should Δt=2l/vx , instead of contact-time between particle and wall?

It's because we want the average pressure over a long period of time (many round-trips between the walls of the container), not the pressure during (only) the particle-wall collisions.

During a particle-wall collision, the force is (relatively) large, but during the flight between walls, the force is zero.
 
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Related to Why Is Time Interval Between Gas Particle Collisions Calculated as Δt=2l/v?

What is the RMS speed of an ideal gas?

The RMS speed of an ideal gas is the root mean square speed of the gas particles, which is the square root of the average of the squared speeds of all the particles in the gas. It is a measure of the average speed of the gas particles.

How is the RMS speed of an ideal gas calculated?

The RMS speed of an ideal gas can be calculated using the formula: RMS speed = √(3RT/M), where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas.

What factors affect the RMS speed of an ideal gas?

The RMS speed of an ideal gas is affected by temperature, molar mass, and the gas constant. As temperature increases, the RMS speed also increases. A gas with a lower molar mass will have a higher RMS speed compared to a gas with a higher molar mass. And the gas constant determines the units of the RMS speed.

Why is the RMS speed of an ideal gas important?

The RMS speed of an ideal gas is important because it helps us understand the behavior of gas particles and how they move. It is also used in various gas laws and equations to calculate properties of gases, such as pressure and volume.

How does the RMS speed of an ideal gas compare to the average speed?

The RMS speed of an ideal gas is higher than the average speed of the gas particles. This is because the RMS speed takes into account the squared speeds of all the particles, while the average speed only considers the absolute values of the speeds. The difference between the two values increases as the temperature of the gas increases.

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