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## Homework Statement

Suppose we have a container divided into equal halves. Right half is fixed at temperature ##T##, volume ##V_2##.

Initially it has pressure ##P_0##, a hole of area ##A## is opened between them.

Part (a): Show that the pressure in the right half is:

[tex]p = \frac{p_0}{2} \left[ 1 + exp \left(-\langle v \rangle\frac{A}{V}t \right) \right][/tex]

Part (b): Vessel has total volume ##10^{-4} m^3##, area of ##7.9 \times 10^{-9} m^2##, ##T = 290 K##. Find the time taken for the pressure on the left half to become half that in the right part.

## Homework Equations

## The Attempt at a Solution

__Part (a)__[tex]dN = -A\phi_{(t)} dt[/tex]

Using flux as ##\phi = \frac{1}{4} n\langle v \rangle##:

[tex]dN = - \frac{A}{4} n \langle v \rangle dt = -\frac{A}{4}\left( \frac{N}{V/2} \right) \langle v \rangle dt = -\frac{NA}{2V} \langle v \rangle dt[/tex]

The graph looks like:

We solve for the dotted graph, then translate upwards by ##\frac{N_0}{2}##. Integrating, and translating upwards i get: ##p = \frac{p_0}{2} \left[ 1 + exp \left(-\langle v \rangle\frac{A}{2V}t \right) \right]##. Not sure why I'm missing a factor of ##2##..

__Part (b)__I know the way to solve this question is to find the pressure on the left half as a function of time, using a process similar to part (a). The graph we want to solve for is (L):

But I'm not sure how to begin..