Gas Leak (Effusion): Solving Differential Equation

  • Thread starter Thread starter moonman
  • Start date Start date
  • Tags Tags
    Effusion Gas
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 7K views
moonman
Messages
21
Reaction score
0
I'm working on a question where gas is leaking out of a container from a small pinhole. I have a differential equation dN/dt = (Constants)*N. I solved this to get an exponential. The exponent is that group of constants, which are A/2V*(kT/m)^1/2. Shouldn't the exponent have something to do with time? Because if they don't, the dimensions don't work out. Can anyone clear this up for me?
 
Physics news on Phys.org
dN/dt = C * N
yields
N = N0 * exp (C * t)
,where t is time

BTW, this is not "Advanced Physics" ;-)
 
You're right, that's not advanced physics, its simple DE. But that's not what I asked.
For that equation to make sense, the exponent should be dimensionless. C*t is not. That's the part I don't understand
 
If A is area and V is volume (and yes, they are) then dimension of your constant is
m2/m3 * m/s = 1/s
and no problem.
I advise to use more precise formula
dN/dt = -1/4 n*<v>*A = -1/4 AN/V * sqrt[kT/(2*3.1416*m)]