What is the force escaping gas exerts on its container? There is a simple problem that I met today, one that I couldn't find a solution for that I was satisfied with.
Assuming we have a tank filled with gas at a certain pressure $p$ and temperature $T$, with the molecules of the gas having the mass $m$. There is a hole on the side of the tank where the pressurized gas escapes into vacuum.
I believe that the force would then be calculated using $$F=\dot m \cdot v_{th}$$ with $v_{th}$ being the thermal velocity $$v_{th}=\sqrt{\frac {8k_BT}{m\pi}}.$$
I would further assume that $\dot m$ is directly proportional to the pressure of the tank, however I don't know any equation I'd be able to substitute it for.
Is this right so far, or am I making completely false assumptions here? What would the final equation for $F$ look like?
 A: Chester Miller points out that my simplistic answer is incorrect. Please see his derivation in the comment section below. 
A: The force would be simply the product between the pressure and the surface of the hole in the tank, say $F = PS_h$, where $F$ is the force, $P$ is the pressure and $S_h$ the surface of the hole. But the force is changing in time, than it's interesting to see how the force is varying with time.
A solution might be this one:
Gas molecules are hitting the tank surface right? What is the chance of escaping the tank assuming a single molecule is hitting the surface? I'd say
$P_e = S_h/S$
where S is the total surface of the tank.
Then, what is the average time $\Delta t$ in which a collision with the surface occurs? Given the thermal velocity (which is constant! it is a function of the mass of the molecule which does not change in time, and the temperature that is fixed as well), we can compute this time from the second principle of dynamics.
$F_p = \frac{\Delta p}{\Delta t}$, where $F_p$ is the force that a single particle applies to the surface of the tank, $\Delta p$ is the variation of momentum, and $\Delta t$ the time interval.
$F_p = PS/N$, where N is the number of particle.$\mid\Delta p\mid = 2mv_{th}$ because the particle is bouncing back at the same velocity. 
So we get
$\Delta t = \frac{2Nmv_{th}}{PS}$.
Now, I'm sorry if it is messy, the 'getting stuck' probability when the particle hits the surface several times, say $N_h$, is
$P_g(N_h) = (1-P_e)^{N_h}$.
How many collision happen during the time interval t? $N_h = t/{\Delta t}$!
Then the mass in the tank should be proportional to this probability, that decreases with time.
The mass in the tank is also proportional to the pressure, so, lastly, we can write the pressure as a function of time as
$P(t) = P(0)(1-S_h/S)^{\frac{t}{\Delta t}}$,
where $P(0)$ is the initial pressure.
Let me know what you think about this solution.
