# Thermodynamical Equilibrium

• Gabriel Maia
In summary, a cylinder with rigid, adiabatic and impermeable external walls is divided into two volumes (A and B) filled with ideal monoatomic gas at identical initial conditions (P0, V0, T0). The heated chamber (A) undergoes a quasi-statical heat introduction until its pressure becomes 32 times the initial pressure (PA=32P0). The final volumes VA and VB can be obtained in terms of V0. Additionally, the final temperatures of A and B (TA and TB) can be calculated using the ideal gas law. The unheated chamber (B) experiences an adiabatic compression, and its final volume and temperature can be determined based on its initial conditions and the
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.

Gabriel Maia said:
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:
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?

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?

Gabriel Maia said:
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

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.

## 1. What is thermodynamical equilibrium?

Thermodynamical equilibrium is a state in which the temperature, pressure, and chemical potential of a system are uniform and there is no net flow of energy or matter within the system. It is a state of balance where all forces and energy within the system are evenly distributed.

## 2. How is thermodynamical equilibrium achieved?

Thermodynamical equilibrium can be achieved through various processes such as heat transfer, diffusion, and chemical reactions. It can also occur naturally over time as a system reaches a state of maximum entropy.

## 3. What are the laws of thermodynamics and how do they relate to thermodynamical equilibrium?

The laws of thermodynamics describe the fundamental principles governing energy and its transformation within a system. The first law states that energy cannot be created or destroyed, only transferred or converted. The second law states that entropy (disorder) of a system will always increase over time. The third law states that absolute zero temperature cannot be reached. These laws help to explain the conditions necessary for a system to reach thermodynamical equilibrium.

## 4. How does thermodynamical equilibrium differ from thermal equilibrium?

Thermodynamical equilibrium refers to a state of balance in all physical properties of a system, while thermal equilibrium specifically refers to a state of balance in temperature. In thermal equilibrium, two systems in contact with each other will have the same temperature, while in thermodynamical equilibrium, all properties of the systems will be equal.

## 5. What are some real-world examples of thermodynamical equilibrium?

Examples of thermodynamical equilibrium can be seen in various natural and man-made systems. For example, a cup of coffee left on a table will eventually reach thermodynamical equilibrium with the surrounding room temperature. The Earth's atmosphere is also an example of a system in thermodynamical equilibrium, as the temperature, pressure, and chemical composition are relatively stable. Additionally, a battery-powered device will reach thermodynamical equilibrium when the battery is fully drained and no more energy can be extracted from it.

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