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Thermodynamical Equilibrium

  • #1
A cylinder has rigid, adiabatic and impermeable external walls. It is provided with a mobile internal ideal partition wall, dividing it in two volumes (A and B). Each volume is filled with an ideal monoatomic gas. Initially, the volume, pressure and temperature of A and B are identical (P0,V0,T0). A certain amount of heat is then introduced into A in a quasi-statical way until the pressure of A is PA=32P0.

Obtain the final volumes VA and VB in terms of V0.

Obtain the final temperatures TA and TB.



This is my problem. Well... first of all, what does it mean a "quasi-statical heat introduction?". I know the ideal gas internal energy is given by

U=3/2*nRT=3/2*PV and that U=Q-W but without knowing the amount of heat, how can I determine the internal energy and the work done, in order to calculate the new volumes?


Thank you.
 

Answers and Replies

  • #2
20,111
4,191
A cylinder has rigid, adiabatic and impermeable external walls. It is provided with a mobile internal ideal partition wall, dividing it in two volumes (A and B). Each volume is filled with an ideal monoatomic gas. Initially, the volume, pressure and temperature of A and B are identical (P0,V0,T0). A certain amount of heat is then introduced into A in a quasi-statical way until the pressure of A is PA=32P0.

Obtain the final volumes VA and VB in terms of V0.

Obtain the final temperatures TA and TB.



This is my problem. Well... first of all, what does it mean a "quasi-statical heat introduction?". I know the ideal gas internal energy is given by

U=3/2*nRT=3/2*PV and that U=Q-W but without knowing the amount of heat, how can I determine the internal energy and the work done, in order to calculate the new volumes?


Thank you.
quasi-statical heat introduction means reversibly. How are the pressures in the two chambers related during this process? How does the sum of the volumes of the two chambers change during the process?

Try starting out by focusing on the chamber that is not heated. What type of process does the gas within this chamber experience:
A) Adiabatic reversible
B) Adiabatic irreversible
C) Isothermal reversible
D) Isothermal irreversible
E) Something else

Based on the answers to these questions, what is the final volume of the unheated chamber?
What is the final temperature of the unheated chamber?
What is the final volume of the heated chamber?
Using the ideal gas law, what is the final temperature of the heated chamber?
 
  • #3
Well... the sum of the volumes is constant. The chamber that are not heated undergoes an adiabatic compression. Therefore, the change in its internal energy is

deltaU = P*deltaV

This pressure P... is it the 32P0 pressure of the heated chamber? Shouldn't it decrease as the wall moves?
 
  • #4
20,111
4,191
Well... the sum of the volumes is constant. The chamber that are not heated undergoes an adiabatic compression.
Yes. Very good.
Therefore, the change in its internal energy is

deltaU = P*deltaV
No. This is not the equation for the internal energy change in a reversible adiabatic compression. How are pressure and volume related in the reversible adiabatic compression of an ideal monotonic gas?
This pressure P... is it the 32P0 pressure of the heated chamber? Shouldn't it decrease as the wall moves?
No. The gas is being compressed, so the pressure in the unheated chamber should increase from P0 to 32P0. This should give you enough information to get the final volume of the unheated chamber in terms of V0.

Chet
 

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