How Does Pressure in a Sealed Container Change Over Multiple Cycles?

In summary: Pn=-Px and the equation becomes( (Pn ) -1 ) = (\frac {10}{11})^n ## (\frac {P0 - Px}{P1 + P2})^n ##
  • #1
jstefanop
1
0
Im tryng to solve an equation for how pressure changes over time in a container. It goes something like this

There is a 10 liter sealed container(V1) with x amount of initial pressure (P1), and a 1 liter container attached to it (V2) with one valve that opens between them, and then another valve that opens to atmosphere. The initial pressure in the 1 liter attachment is atmospheric (P2).

For each cycle the valve opens between the pressureised 10 liter container and the one liter at 1 atm, the pressure then equalizes in both containers. Then the inner valve is closed and the outer valve is opened on the 1 liter container and pressure is vented to atmosphere (so the 1 liter container returns to 1atm) then the outside valve is closed and the cycle continues.

I need to find out after how many cycles does the large 10 liter container reach a certain pressure above 1 atm...

I can figure out what the pressure is in the container after they equalize each time which is
P= (P1*V1 +P2*V2)/(V1+V2)

So let's say the initial pressure is of P1 is 2atm after the first cycle the pressure would be P = (2*10l +1l*1atm)/(1l+10l) = 1.91 atm

then the second cycle would be P = (1.91*10l +1l*1atm)/(1l+10l) = 1.83 atm

etc etc

so what would be the equation t describe this pressure drop over a certain number of cycles? i.e. P(c) = ?
 
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  • #2
jstefanop said:
P= (P1*V1 +P2*V2)/(V1+V2)
Okay.
jstefanop said:
say the initial pressure is of P1 is 2 x atm
Now try it.
 
  • #3
I'll have to accept your physics, because I don't know how to do that, but juggling your equation round and using V1, V2, and P2 as constants, I get a formula for Pn after n cycles, as

( (Pn ) -1 ) = ## (\frac {10}{11})^n ## ((P0) -1 )

Edit - added excess brackets just to make clear
 
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  • #4
Or in a more general form

( Pn - Px ) = ( P0 - Px ) ##( \frac{V1}{V1 + V2})^n##

Where Px is the sink pressure, P0 is the container starting pressure and Pn is the container pressure after n cycles.

Edit - BTW this looks so nice to me, that I think your physics must be correct!
Edit2 - and it looks as if you could start with container pressure below atmospheric as well.
 
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1. What is basic pressure change over time?

Basic pressure change over time, also known as pressure gradient, refers to the rate at which pressure changes in a given amount of time. It is often measured in units of Pascals per second (Pa/s) or millibars per hour (mb/h).

2. How is basic pressure change over time calculated?

Basic pressure change over time is calculated by taking the difference in pressure between two points over a certain time interval. This can be done using a pressure gauge or by analyzing data from a weather station.

3. What factors can affect basic pressure change over time?

There are several factors that can affect basic pressure change over time, including temperature, altitude, and weather patterns. Changes in these factors can cause shifts in air pressure, resulting in fluctuations in basic pressure change over time.

4. Why is basic pressure change over time important to study?

Basic pressure change over time is important to study because it can provide valuable information about weather patterns and atmospheric conditions. It can also help in predicting changes in air pressure, which can have implications for aviation, agriculture, and other industries.

5. How does basic pressure change over time relate to other scientific concepts?

Basic pressure change over time is related to other scientific concepts such as air pressure, air density, and atmospheric pressure. It is also an important factor in understanding weather systems and climate change.

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