Ratio of wall collisions when temperature quadruples

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 1K views
Any Help
Messages
79
Reaction score
2

Homework Statement


You have two samples of the same gas in the same size container, with the same pressure. The gas in the first container has a kelvin temperature four times that of the gas in the other container.

The ratio of number of collisions with the wall in the first container compared to that in the second is:
A) 1:1
B) 4:1
C) 1:4
D) 2:1
E) 1:2

Homework Equations


PV=nRT
Urms = sqrt(3.R.T/M)
collisions= m.Urms^2
Kinetic energy average= 3/2 RT

The Attempt at a Solution


I find that n2 = 4n1
but what I should do after that?
I tried relating the kinetic energy but it didn't work with me.
How we can approach to such problems? What rules should we use
 
Physics news on Phys.org
Any Help said:

Homework Statement


You have two samples of the same gas in the same size container, with the same pressure. The gas in the first container has a kelvin temperature four times that of the gas in the other container.

The ratio of number of collisions with the wall in the first container compared to that in the second is:
A) 1:1
B) 4:1
C) 1:4
D) 2:1
E) 1:2

Homework Equations


PV=nRT
Urms = sqrt(3.R.T/M)
collisions= m.Urms^2
Kinetic energy average= 3/2 RT

The Attempt at a Solution


I find that n2 = 4n1
but what I should do after that?
I tried relating the kinetic energy but it didn't work with me.
How we can approach to such problems? What rules should we use

It is a bit difficult to help since you don't define the meaning of any variables. Some of them are maybe obvious. But, what is ##m## and ##M##?
if ##m## is the mass/particle and ##M## is the total you would have ##M/m=n##. However, I would start by putting the expressions for ##U_{\mathrm{rms}}## in the expression for "collisions", leading to an expression for "collisions" depending on ##m/M##.