Ideal Gas in Piston (Constant Pressure vs Rate of collision)

In summary, the rate of collision between molecules and piston decreases as the temperature increases.
  • #1
Stunner
1
0
A syringe contains ideal gas. The piston is frictionless and no gas escapes. Once heated slowly, the piston moves outwards. The piston stops moving when the temperature is steady. The pressure of the gas after the piston stops moving remains the same (no change in pressure). What can we say about the rate of collision between the gas molecules and piston?
1) Ok we know that the average speed of the molecules increases (hence increased KE).
2) It is said that the rate of collision “decreases”.
Reason: KE increases. If frequency of collision remains the same (or increases), the pressure would increase. But pressure remains constant, so the frequency of collision has to be lower.

This is a very logical approach to answering the problem. But is there a mathematical derivation to show how this is so? The closest I can find is P=(Nmv^2)/(3V) where v is the average speed per molecule and V is the volume. N is the number of molecules and m is the mass.
Is there a formula showing how rate of collision is linked to an increased temperature (given increased volume and constant pressure).

Any help would be greatly appreciated. Also let me know if there’s anyone I can ask. Thank you so much in advance!
 
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  • #2
We can show that the number, f, of collisions per unit wall area, per unit time is given by [tex]f\:=\:\frac{1}{4}\frac{N}{V}\overline{v}[/tex]
in which [itex]\overline{v}[/itex] is the mean speed of the molecules.

It can be shown from the Maxwell distribution of molecular speeds that [itex]\overline{v}=0.921v_{rms}[/itex].

So [itex]f\:=\:\frac{1}{4}0.921\frac{N}{V}v_{rms}[/itex].

But [itex]pV=\frac{1}{3}Nmv_{rms}^2[/itex]

Eliminating [itex]v_{rms}[/itex] we obtain
[tex]f\:=\:\frac{1}{4}0.921\sqrt{\frac{3Np}{mV}}[/tex]
and, finally, eliminating V using pV = NkT

We have [itex]f\:=\:\frac{1}{4}0.921\sqrt{\frac{3p^2}{mRT}}[/itex]

So we have established that for constant p, f is proportional to [itex]\frac{1}{\sqrt T}[/itex]
 
  • #3
Stunner: Is this what you wanted?
 

Related to Ideal Gas in Piston (Constant Pressure vs Rate of collision)

1. What is an ideal gas in piston?

An ideal gas in piston is a theoretical concept in physics and thermodynamics that describes the behavior of a gas in a confined space, such as a piston. It is assumed to have no intermolecular forces and follow the gas laws, making it easier to study and calculate.

2. What is constant pressure in the context of an ideal gas in piston?

Constant pressure in an ideal gas in piston refers to a situation where the external pressure on the piston remains constant while the gas inside the piston expands or contracts. This can be achieved by adjusting the weight on top of the piston or using a pressure regulator.

3. How does constant pressure affect the behavior of an ideal gas in piston?

Constant pressure will result in a direct relationship between the volume and temperature of the gas, known as Charles' Law. As the gas expands, the temperature will increase, and vice versa. This is because the gas particles are colliding more frequently with the piston, increasing their kinetic energy and temperature.

4. What is the rate of collision in an ideal gas in piston?

The rate of collision refers to the number of collisions between gas particles and the walls of the piston per unit time. In an ideal gas in piston, the rate of collision is directly proportional to the pressure of the gas and the temperature of the gas.

5. How does the rate of collision change in a constant pressure situation compared to a changing pressure situation in an ideal gas in piston?

In a constant pressure situation, the rate of collision will increase as the temperature increases, keeping the pressure constant. However, in a changing pressure situation, the rate of collision will vary depending on the changing pressure, as well as the temperature and volume of the gas.

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