Thermodynamical Equilibrium

[Gabriel Maia]

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.

 

[Chestermiller]

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?

 

[Gabriel Maia]

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?

 

[Chestermiller]

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

 

[paranormal]

I can provide some insights into the concept of thermodynamic equilibrium and how it applies to this scenario. Thermodynamic equilibrium refers to a state where there is no net transfer of energy or matter between different parts of a system. In this case, the cylinder with rigid, adiabatic, and impermeable walls is a closed system, meaning that no energy or matter can enter or leave it. This allows the system to reach a state of equilibrium, where the pressure, volume, and temperature are uniform throughout.

Now, in order to understand the concept of "quasi-statical heat introduction," we need to first understand that heat is a form of energy that can be transferred from one object to another. In this scenario, heat is being introduced into volume A in a slow and controlled manner, allowing the system to reach equilibrium at each stage. This is known as a quasi-static process, where the system is always close to equilibrium.

Moving on to the calculations, we can use the ideal gas law, PV = nRT, to determine the final volume and temperature of each volume. Since the initial conditions for both volumes A and B are the same (P0, V0, T0), we can set up the following equations:

For volume A: (PA)(VA) = (nR)(TA) and for volume B: (PB)(VB) = (nR)(TB)

Since we know that PA = 32P0 and that the total number of moles (n) remains constant, we can solve for VA and VB in terms of V0:

VA = (PA/P0)V0 = (32)V0 and VB = (PB/P0)V0 = (1)V0

This means that the final volume of volume A will be 32 times the initial volume, while the final volume of volume B will remain unchanged.

Similarly, we can use the ideal gas law to solve for the final temperatures:

TA = (PA/P0)(T0) = (32)(T0) and TB = (PB/P0)(T0) = (1)(T0)

This means that the final temperature of volume A will be 32 times the initial temperature, while the final temperature of volume B will remain unchanged.

In terms of the amount of heat introduced, we can use the first law of thermodynamics (U = Q - W) to calculate the change in internal energy (U) for volume A.