# Adiabatic Free Expansion and Reversible Isothermal Path

• says
In summary, when considering an ideal monatomic gas undergoing an adiabatic free expansion, a reversible isothermal path can be constructed to join the initial and final states. The total entropy change between the two states can be calculated using the equation ΔS = nRln(Vf/Vi) and is expected to be the same as the entropy change for the irreversible free expansion, with the only difference being the sign due to the increase in entropy in the irreversible process.
says

## Homework Statement

Consider an ideal monatomic gas that undergoes an adiabatic free expansion starting from equilibrium state A with volume 500 cm3, pressure 40 kPa and temperature 300K to state B, which has a final equilibrium volume of 1000 cm3.
1. Construct an reversible isothermal path that joins the initial and final states and show it on a schematic p-V diagram.
2. Calculate the total entropy change between the two states using the reversible path.
3. Would you expect this to be the same as the entropy change for the irreversible free expansion?

## Homework Equations

1) P1*V1 = P2*V2

2) ΔS = nRln(Vf/Vi)

3) Ideal Gas Law pV=nRt

## The Attempt at a Solution

1) 40*500 = P2 * 1000
P2 = 20 kPa

I've attached a P-V diagram. What I'm failing to understand is how i draw a reversible isothermal path that joins the states. If ΔT in an adiabatic free expansion = 0, and then I'm asked to draw a reversible isothermal path that joins the initial and final states together then wouldn't this be the exact same path because ΔT for isothermal path = 0?

2) P1 = 40,000Pa
V1 = 0.0005m3
T = 300 K
R (gas constant) = 8.3144
∴ = PV / RT = n = 0.008018217

ΔS = nRln(Vf/Vi)

ΔS = (0.008018217)*(8.3144)* ln(0.0005/0.001)
= - 0.046209810 J/K

3) The entropy change in the adiabatic free expansion is the same as in question two, only it is positive because entropy is increasing.

#### Attachments

• photo (8).JPG
49.1 KB · Views: 739
says said:

## Homework Statement

Consider an ideal monatomic gas that undergoes an adiabatic free expansion starting from equilibrium state A with volume 500 cm3, pressure 40 kPa and temperature 300K to state B, which has a final equilibrium volume of 1000 cm3.
1. Construct an reversible isothermal path that joins the initial and final states and show it on a schematic p-V diagram.
2. Calculate the total entropy change between the two states using the reversible path.
3. Would you expect this to be the same as the entropy change for the irreversible free expansion?

## Homework Equations

1) P1*V1 = P2*V2

2) ΔS = nRln(Vf/Vi)

3) Ideal Gas Law pV=nRt

## The Attempt at a Solution

ΔS = nRln(Vf/Vi)

ΔS = (0.008018217)*(8.3144)* ln(0.0005/0.001)
Vi and Vf are interchanged.[/QUOTE]

A reversible process is exactly what it says, that is you can at any stage stop and reverse the process
and the system will be exactly like before. This means that the process need to be well-behaved or as it
is called - in equilibrium - at all stages, no funny stuff going on like friction or tubulence is allowed. Free
expansion is an example of an irreversible process, since the gas expands in vacuum. This means that the
pressure is probably varying through the gas while it is expanding and it definitely is not in equilibrium during the expansion.

Also, for the free expansion, if P represents the force per unit area applied by the surroundings to the gas, then the P-V diagram for the free expansion is incorrect.

Chet

Chestermiller said:
Also, for the free expansion, if P represents the force per unit area applied by the surroundings to the gas, then the P-V diagram for the free expansion is incorrect.

Chet
Thanks! I understand that the adiabatic p-v diagram is incorrect now. I only have to provide the isothermal p-v diagram.

says said:
Thanks! I understand that the adiabatic p-v diagram is incorrect now. I only have to provide the isothermal p-v diagram.
Aside from the correction that ehild indicated, very nice analysis.

Chet

I think I've misread question 1. I don't think it's asking for isothermal compression, but isothermal expansion. In this case:

ΔS = nRln(Vf/Vi)

ΔS = (0.008018217)*(8.3144)* ln(0.001/0.0005)
= 0.046209810 J/K (positive entropy and not negative)

## 1. What is adiabatic free expansion?

Adiabatic free expansion is a process in thermodynamics where a gas expands rapidly into a vacuum without any heat exchange with its surroundings. This means that there is no transfer of energy as heat during the expansion.

## 2. What is a reversible isothermal path?

A reversible isothermal path is a thermodynamic process where the temperature of a system remains constant throughout the process. This is achieved by maintaining a balance between the heat added to the system and the work done by the system.

## 3. How are adiabatic free expansion and reversible isothermal path related?

These two processes are related because they represent two extreme cases of thermodynamic processes. Adiabatic free expansion occurs when no heat is exchanged with the surroundings, while a reversible isothermal path occurs when heat is constantly exchanged to maintain a constant temperature.

## 4. What are some real-life examples of adiabatic free expansion and reversible isothermal path?

Adiabatic free expansion can be seen in the expansion of compressed air in a tire or the release of a compressed gas from a tank. Reversible isothermal paths can be observed in refrigeration and air conditioning systems, where a refrigerant undergoes a constant temperature change while absorbing or releasing heat.

## 5. What are the applications of adiabatic free expansion and reversible isothermal path?

Adiabatic free expansion is used in industrial processes to quickly and efficiently expand gases. Reversible isothermal paths have applications in various industries, including refrigeration, chemical processing, and power generation, where maintaining a constant temperature is crucial for efficient operation.

• Introductory Physics Homework Help
Replies
8
Views
926
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Thermodynamics
Replies
11
Views
622
• Introductory Physics Homework Help
Replies
3
Views
4K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Thermodynamics
Replies
22
Views
2K
• Introductory Physics Homework Help
Replies
12
Views
1K