Two cylinders A and C of equal volume V contain the same ideal gas at temperature T and at pressures 2p and p respectively. A valve connecting the two cylinders is opened slightly and as the gas leaks from A to B, the pressure in A is maintained at 2p by pushing in a piston. The process is continued until the gas in cylinder B is also at 2p. If there is good thermal contact between the cylinders but the are thermally insulated from their surroundings, find:
(i) The final temperature in terms of T
(ii) the final volume of gas in cylinder A in terms of V.
Before I attempt the question, may I ask if the final temperature will be the same for both sides of the cylinder? What about the pressure? I'm still visualising the problem. What other tips can you give me? Thank you!
Cambridge Physics Problem 20: Thermal System Problem
|Study Help needs new mods!||14-04-2014|
|Post on TSR and win a prize! Find out more...||10-04-2014|
- 0 followers
Yes good "thermal contact" should mean that the cylinders reach thermal equilibrium rapidly, the pressure in A is kept constant while the pressure in B increases as the moles of gas there increases.
Tips: this is kinda dependent on my answer being right but it seems reasonable :P , anyway...
Set up an equation of state for cylinder B, then one for the system at equilibrium when the pressure is the same in both cylinders I treated the system as one cylinder : if you were to just weld the cylinders together there would be no movement of gas because the pressure is the same in both)
Then consider the work done on the system : You have compressed the system by a certain volume against a constant pressure of 2P, the work done is the increase in internal energy of the gas (which for a monoatomic gas should = 3/2 nRT), from the surroundings as the cylinders are thermally insulated
Any other comments from anyone?
I'm having trouble eliminating the VA' and VB' . Can anyone help me on this?
Is molar heat capacity necessary to solve the question?
RegisterThanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following: