Work done in a chamber with gas undergoing adiabatic free expansion?

  • #1
zenterix
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Homework Statement
A container with rigid, well-insulated walls is divided into two parts by a partition. One part contains a gas, and the other is evacuated. If the partition suddenly breaks, show that th einitial and final internal energies of the gas are equal. (Note: this process is called an adiabatic free expansion).
Relevant Equations
##\Delta U = Q+W##
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Let's consider the system as both internal chambers together, ie everything inside the adiabatic walls.

We have ##Q_{sys}=Q_{gas}+Q_{evac}=0## because we have adiabatic walls and ##W_{gas}+W_{evac}=0## because of the rigid walls.

##\Delta U = U_f-U_i=Q_{gas}+Q_{evac}+W_{gas}+W_{evac}=0##.

How do we reason about ##W_{gas}## and ##W_{evac}## individually? Their sum is zero because it represents work done on the system as a whole. But when the gas expands suddenly into the chamber with no gas, what is the work done on the evacuated chamber for example (of course this is just the negative of the work done by the chamber with the gas).
 

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  • #2
In your 1st law equation, ##W_{gas}## represents work done on the gas by something external to the gas. In order for something external to the gas to do work on the gas, that something must be in contact with some of the gas. What objects are in contact with the gas as the gas expands into the vacuum? Do any of these objects do work on the gas? Why or why not?
 
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  • #3
zenterix said:
Homework Statement: A container with rigid, well-insulated walls is divided into two parts by a partition. One part contains a gas, and the other is evacuated. If the partition suddenly breaks, show that th einitial and final internal energies of the gas are equal. (Note: this process is called an adiabatic free expansion).
Relevant Equations: ##\Delta U = Q+W##

Their sum is zero because it represents work done on the system as a whole
Not quite sure about the reasoning behind that statement.

You seem to have switched systems.
ΔU = Q + W, is for the system boundary around the whole container.

zenterix said:
the system as both internal chambers together, ie everything inside the adiabatic walls.
has two systems, each with a separate system boundary.
There is a moving boundary connecting one system to the other ( ie think if the gas-evac boundary as a massless partition that moves as the gas expands )
One is the gas within the end and side walls of the adiabatic rigid container, and with a moving boundary being the moving partition.
Similarly, with the evac volume being the second system.

Thus you can evaluate the work done on or by each system, whether the second is another gas, a spring, your finger pushing/ pulling the partition, or an evacuated space, or something else.
 
  • #4
TSny said:
In your 1st law equation, ##W_{gas}## represents work done on the gas by something external to the gas. In order for something external to the gas to do work on the gas, that something must be in contact with some of the gas. What objects are in contact with the gas as the gas expands into the vacuum? Do any of these objects do work on the gas? Why or why not?
Other than the adiabatic rigid walls, the only thing in contact with the gas when the partition disappears is the evacuated chamber. In other words, nothing (no force acting on the gas as it expands).

The gas had some pressure when the partition existed. When it suddenly disappears, what happens next is a very sudden change in volume such that we don't have a uniform pressure to speak of.

The final state is with the volume being that of the entire two chambers, It is not clear to me how temperature and pressure change to satisfy the ideal gas law.
 
  • #5
256bits said:
Not quite sure about the reasoning behind that statement.

You seem to have switched systems.
ΔU = Q + W, is for the system boundary around the whole container.has two systems, each with a separate system boundary.
There is a moving boundary connecting one system to the other ( ie think if the gas-evac boundary as a massless partition that moves as the gas expands )
One is the gas within the end and side walls of the adiabatic rigid container, and with a moving boundary being the moving partition.
Similarly, with the evac volume being the second system.

Thus you can evaluate the work done on or by each system, whether the second is another gas, a spring, your finger pushing/ pulling the partition, or an evacuated space, or something else.
The total work on the system (the gas + the evacuated nothingness) is the sum of work on the gas and work on the nothingness which is the lack of gas.

If energy is conserved, there is no heat exchange between the system and the surroundings, and the surroundings do no work on the chambers, then isn't the total energy of the system constant?

Is it not the case then that whatever work is done on the gas is work done by the evacuated chamber, or work done by the gas is work done on the evacuated chamber?
 
  • #6
Do you think that the ideal gas law is satisfied by a gas experiencing a rapid expansion like this, even if it behaves like an ideal gas under static conditions?
 
  • #7
Chestermiller said:
Do you think that the ideal gas law is satisfied by a gas experiencing a rapid expansion like this, even if it behaves like an ideal gas under static conditions?
I think that there is an initial equilibrium state and a final equilibrium state and we have a state equation for those two states (ideal gas law)

In fact, now that I think about it, since we have an adiabatic process, no matter how we get from initial to final states, the amount of work is the same and is equal to the change in internal energy.

For the intermediate states, no, we can't use a state equation because the gas has all kinds of non-uniform characteristics.
 
  • #8
zenterix said:
I think that there is an initial equilibrium state and a final equilibrium state and we have a state equation for those two states (ideal gas law)

In fact, now that I think about it, since we have an adiabatic process, no matter how we get from initial to final states, the amount of work is the same and is equal to the change in internal energy.

For the intermediate states, no, we can't use a state equation because the gas has all kinds of non-uniform characteristics.
This is all a correct assessment. Very nice.

Next question. Do you think that there is a named physical law that describes the relationship between the state of stress and the kinematics of the deformation for an ideal gas experiencing a rapid irreversible deformation like this (at least, say, locally)?
 
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  • #9
Chestermiller said:
This is all a correct assessment. Very nice.

Next question. Do you think that there is a named physical law that describes the relationship between the state of stress and the kinematics of the deformation for an ideal gas experiencing a rapid irreversible deformation like this (at least, say, locally)?
You used a few expressions there

1) State of stress

2) Kinematics of the deformation of an ideal gas

3) Rapid irreversible deformation

I can't think of a physical law describing a relationship between 1) and 2). Maybe I haven't learned about it yet. My wild guess would be something from statistical mechanics like brownian motion.
 
  • #10
zenterix said:
You used a few expressions there

1) State of stress

2) Kinematics of the deformation of an ideal gas

3) Rapid irreversible deformation

I can't think of a physical law describing a relationship between 1) and 2). Maybe I haven't learned about it yet. My wild guess would be something from statistical mechanics like brownian motion.
The state of stress of a material (solid, liquid, or gas) is described by the stress tensor. That is used to determine the force per unit area acting on a surface of arbitrary orientation within the material.

The kinematics of a deformation of a fluid (liquid or gas) are determined by the specific volume and velocity gradients within the fluid.

The relationship between the kinematics of the deformation and stress tensor is determined by Newton's Law of Viscosity, expanded to 3D to be tensorially (mathematically) correct and independent of the frame of reference of the observer.

Are you familiar with the concept of a Newtonian fluid?
 
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  • #11
Chestermiller said:
The state of stress of a material (solid, liquid, or gas) is described by the stress tensor. That is used to determine the force per unit area acting on a surface of arbitrary orientation within the material.

The kinematics of a deformation of a fluid (liquid or gas) are determined by the specific volume and velocity gradients within the fluid.

The relationship between the kinematics of the deformation and stress tensor is determined by Newton's Law of Viscosity, expanded to 3D to be tensorially (mathematically) correct and independent of the frame of reference of the observer.

Are you familiar with the concept of a Newtonian fluid?
I am not. Haven't ever heard of it before.
 
  • #12
zenterix said:
I am not. Haven't ever heard of it before.
Are you familiar with the concept of viscosity?? Any experience with stress and strain in solid mechanics?
 
  • #13
Chestermiller said:
Are you familiar with the concept of viscosity?? Any experience with stress and strain in solid mechanics?
Nope, haven't studied those things yet.
 
  • #14
Well, when you study fluid mechanics, you will learn about these concepts, and you will then understand why the ideal gas law does not describe the forces during an irreversible expansion, even for a gas that is considered ideal under thermodynamic equilibrium conditions. In other words, the ideal gas law applies only at thermodynamic equilibrium.
 
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FAQ: Work done in a chamber with gas undergoing adiabatic free expansion?

What is adiabatic free expansion?

Adiabatic free expansion is a thermodynamic process in which a gas expands into a vacuum without exchanging heat with its surroundings. This means that the process is both adiabatic (no heat transfer) and free (no external work is done).

Does the temperature of the gas change during adiabatic free expansion?

In an ideal gas undergoing adiabatic free expansion, the temperature remains constant. This is because no work is done on or by the gas, and there is no heat exchange, so the internal energy, and hence the temperature, does not change.

Is any work done during adiabatic free expansion?

No, during adiabatic free expansion, no work is done. The gas expands into a vacuum without any external pressure to do work against, so the work done is zero.

How does the internal energy of the gas change during adiabatic free expansion?

For an ideal gas, the internal energy remains constant during adiabatic free expansion. This is because the internal energy of an ideal gas is a function of temperature, and the temperature does not change during this process.

What happens to the pressure of the gas during adiabatic free expansion?

The pressure of the gas decreases during adiabatic free expansion because the gas molecules spread out into a larger volume. Since the number of molecules remains the same but the volume increases, the pressure drops according to the ideal gas law.

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