Daraan on 28/2/2012 at 21:06
I haven't found a better place for this off-topic question.
I got an exercise but I can't find the correct solution for it (and I don't know the correct solution).
So I hope that maybe some of you can help me.
I hope this sketch can help you.
Code:
___________________________________________________
| a ‡ b |
| ‡ |
'''''''''''''''''''''''''''''''''''''''''''''''''''
We have a closed system with the two parts a and b. The plunger ‡ is fixed at the beginning and we got the following data
volume a = 1.5m^3 ; volume b = 5m^3
pressure a=3.0MPa ; pressure b = 0.1MPa
the temperature is constant at 500°K
Now the plunger ‡ is released.
Task: calculate the new pressure and the new volume for the parts a and b.
I first thought that at the end both parts have the same pressure - (3+0.1)/2 = 1.55MPa - but that attempt is wrong.
How to do it correct?
Koki on 28/2/2012 at 21:13
They will for sure have same pressure, but their volumes are different so you can't just take an average of both I think
st.patrick on 28/2/2012 at 21:37
Let me see how much of my high school knowledge can I refresh with a healthy dose of Wikipedia.
A closed isothermal process, ergo Boyle's Law applies. The product of multiplying pressure and volume is constant. We're however missing one element here, the molar concentration of the gas, which would let us find out the rate at which the increased pressure diminishes the volume of the gas.
Hypothetically, if this rate were linear (it isn't), initially we have a container with the total volume of 6.5 m^3, with:
1.5 m^3 @ 3.0 MPa ~ 4.5 m^3 @ 1.0 MPa, and
5.0 m^3 @ 0.1 MPa ~ 0.5 m^3 @ 1.0 MPa, which gives us the total of
--------------------- 5.0 m^3 @ 1.0 MPa.
However, the total volume of the container is 6.5 m^3, so the gas expands to this volume and the pressure diminishes to 5/6.5 = 0.77 MPa. Given that the dimensions of the container do not change, the proportions between parts a and b should stay equal, i.e. 1.5 m^3 and 5.0 m^3.
Oh, and for the record, "hypothetically" means "I have no clue what I'm talking about so take this with a substantial amount of NaCl."
Al_B on 28/2/2012 at 21:38
It's been years since I did chemistry or physics but isn't it just Boyle's law? i.e. the product of the pressure and volume for both compartments must be the same before and after the plunger is released (assuming constant temperature). If the pressure is the same you should be able to find the relationship between the volumes in the two compartments. You know the overall volume so you should be able to work out the volumes in each compartment and derive the absolute pressure.
Of course - the above may be completely way off base...
jay pettitt on 28/2/2012 at 21:42
I can do the volume bit. And I'm guessing the pressure is 0.8 Mpa
--edit--
being able to solve units in equations is a super awesome skill.
SubJeff on 28/2/2012 at 22:19
Yep, looks like a Boyles law question to me. It can't be a universal gas law question because you don't have any info about the number of moles present.
The substances (gases, I'm assuming) will alter volume until the pressure across the plunger is zero, so you need to work out when the pressure in each sub-container is equal. Just get the Boyles constant for each section and do a bunch of calculations to get some ballpark figures for pressure and extrapolate from there.
Martin Karne on 29/2/2012 at 00:35
And what is the plunger function, speed and volume?
Or just its passing by the landscape to amuse the scene as a whole?
Brian The Dog on 29/2/2012 at 01:11
Also note that the total Volume of regions A and B is a constant, which provides the other "limiting condition" - you should now have 2 unknowns and 2 conditions (the other being that the pressure of regions A and B must be identical if the system is not moving any more) which have to be true, so it should just be a case of solving simultaneous equations:
I get:
Volume of A = 5.85 m^3
Volume of B = 0.65 m^3
Pressure = 769,230 Pa
ZylonBane on 29/2/2012 at 01:22
Oh boy, we haven't had a "Do my homework for me" thread in a while.
Briareos H on 29/2/2012 at 10:39
Wait 'til he comes back and asks for the oscillations equation of the piston.
