Thermally isolated cylinder - work done and temperature changes in each chamber

  • Thread starter davidray
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Homework Statement



There is a closed, thermally insulated cylinder, separated into two chambers by a moveable, frictionless, gas-tight, thermally insulating piston.

Initial, the volume of each chamber is 3 liters, and each contains an identical monoatomic ideal gas at pressure of 1 atmosphere and a temperature of 20C. You may assume that all relevant properties of the gas are temperature independent, and that classical equipartition theory is valid.

A heating coil inside chamber A is used to slowly supply heat to the gas in this chamber. This gas expands until equilibrium is reached and the pressure in chamber B is 3.5 atmospheres.

Calculate:
1. The work done on the gas in chamber B.
2. The final temperature of the gas in chamber B.
3. The final temperature of the gas in chamber A.
4. The ammount of heat supplied by the heating coil.

Any help with this would be greatly appreciated!
I really have no idea where to start! I've been reading over all of my notes and books, but I can't work out which equation or bit of the question to start with. Thank you so much if you can help with this!
 

Answers and Replies

  • #2
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Ok, what can you say about the pressures in the two cylinders once the heating is done?
 
  • #3
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It's until equilibrium is reached, so they'll be the same?
I wasn't sure because the piston is thermally insulating
 
  • #4
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Yes the pressures will be the same. If you calculate the change in volume needed to create a pressure of 3.5 atmospheres in chamber B, a bunch of the answers should fall out. But be careful because both pressure and temperature are dependent on volume.
 
  • #5
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P1V1/P2=V2=8.79 litres

I can then use W=-(P2xV1)-(P2vV2)
=2227.55J?
 
  • #6
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Following this, how do I work out the change in temperature in chamber B?
 
  • #7
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P1V1/P2=V2=8.79 litres

I can then use W=-(P2xV1)-(P2vV2)
=2227.55J?
I'm pretty sure you can't do that. You will need a differential equation.

EDIT: The problem is that nothing is constant. Chamber B experiences a change in pressure volume and temperature. The one thing that you have on your side is that it is an adiabatic process. You will need to use the fact that there is no heat added to the system to relate all of these variables. The key things are that you have a final pressure, and its adiabatic. with these two parameters you should be able to draw a connection.
 
Last edited:
  • #8
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It's from w=-integral between V1 and V2 (p dV)
That then leads to -[pv]between v1 and v2
which then leads to -[pv1]-[pv2]
?
 
  • #9
1,033
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see here is the problem. You want to integrate across all volumes. But the pressure is changing with the volume. in other terms, as you change V, P changes as well. Your equation is one for a constant pressure.
 
  • #10
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You may assume that all relevant properties of the gas are temperature independent
I don't know what this means.
 
  • #11
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"You may assume that all relevant properties of the gas are temperature independent."



Teacher is talking about specific heats.
 

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