Free expansion process and temperature change

In summary, for the case of Joule expansion, the final temperature of the air after adiabatic expansion will be very close to the initial temperature, due to the fact that the process is not reversible and no work is being done. Therefore, the final temperature can be approximated as equal to the initial temperature.
  • #1
Amaelle
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Good day all!
I have the following exercise
One kilo-mole of air expands adiabatically from an initial state p1=1 bar, T1=340 K to a final state V2= 2V1. Compute the temperature of the final state, for the following cases: Joule expansion (gas free expansion with W12=0)

My problem is with the joule expansion : in the joule expansion W12=0 and because we have an adiabatic process Q=0
so according to the first law of thermodynamics dU12=0 which means T2=T1 which I totally agree with

On the other hand we have an adiabatic process which means T1V1(k-1)=T2V2(k-1) so T2=T1*(V1/V2)(k-1) so accordingly T2 ≠T1?!

THIS IS THE POINT I'M STRUGGLING WITH AND ANY HELP WOULD BE HIGHLY APPRECIATED THANKS!
 
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  • #2
Amaelle said:
On the other hand we have an adiabatic process which means T1V1(k-1)=T2V2(k-1)
This equation does not apply to every adiabatic process of an ideal gas. Do you know what other condition must be met? Does a free expansion meet this additional requirement?
 
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  • #3
thanks for your prompt reply!
I really don't know, this is why I'm asking
I would be very grateful if you could elaborate more
 
  • #4
Amaelle said:
Good day all!
I have the following exercise
One kilo-mole of air expands adiabatically from an initial state p1=1 bar, T1=340 K to a final state V2= 2V1. Compute the temperature of the final state, for the following cases: Joule expansion (gas free expansion with W12=0)

My problem is with the joule expansion : in the joule expansion W12=0 and because we have an adiabatic process Q=0
so according to the first law of thermodynamics dU12=0 which means T2=T1 which I totally agree with

On the other hand we have an adiabatic process which means T1V1(k-1)=T2V2(k-1) so T2=T1*(V1/V2)(k-1) so accordingly T2 ≠T1?!
The equation
T1V1(k-1)=T2V2(k-1)
is valid for a quasistatic, , "equilibrium" process, and free expansion is not that. The intensive parameters, temperature, pressure are defined for the static case. During free expansion, the pressure outside is zero, the piston moves fast, neither pressure nor temperature is defined inside the gas .
 
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  • #5
Awesome! thanks a million! to both of you ehild and Tsny
 
  • #6
Amaelle said:
thanks for your prompt reply!
I really don't know, this is why I'm asking
I would be very grateful if you could elaborate more
It applies to an adiabatic reversible process. Joule expansion is not a reversible expansion.
 
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  • #7
Chestermiller said:
It applies to an adiabatic reversible process. Joule expansion is not a reversible expansion.
Thanks Chestermiller! you nail it as usual!
 
  • #8
Amaelle said:
One kilo-mole of air expands adiabatically from an initial state p1=1 bar, T1=340 K to a final state V2= 2V1. Compute the temperature of the final state, for the following cases: Joule expansion (gas free expansion with W12=0)
For free expansion the change in the internal energy is 0 and as the process is adiabatic dQ is 0 and from that we conclude that dW (work done) is also 0.
now dU(T,v)=CvdT+(∂U/∂v)Tdv ,
Now the final temp. is Cv(Tf-Ti)=-∫(∂U/∂v)Tdv
 
  • #9
Apashanka said:
For free expansion the change in the internal energy is 0 and as the process is adiabatic dQ is 0 and from that we conclude that dW (work done) is also 0.
now dU(T,v)=CvdT+(∂U/∂v)Tdv ,
Now the final temp. is Cv(Tf-Ti)=-∫(∂U/∂v)Tdv
This is basically correct, except that, at 340 K and 1 bar, air behaves essentially as an ideal gas, in which case the temperature change would be very close to zero.
 

1. What is the free expansion process?

The free expansion process is a thermodynamic process in which a gas expands into a vacuum without any external work being done on the gas. This means that the gas expands freely and no energy is exchanged between the gas and its surroundings.

2. How does free expansion affect temperature?

During free expansion, the gas molecules have more space to move around and therefore, their average kinetic energy decreases. This leads to a decrease in temperature of the gas.

3. Is the temperature change in free expansion always the same?

No, the temperature change in free expansion can vary depending on the initial conditions of the gas. If the gas is initially at a high pressure and low temperature, the temperature change may be more significant compared to a gas at a lower pressure and higher temperature.

4. Can free expansion cause a change in internal energy?

No, since no external work is done on the gas during free expansion, there is no change in the internal energy of the gas. This is because internal energy is a measure of the total energy of a system, including both the kinetic and potential energies of its molecules.

5. How is free expansion different from other thermodynamic processes?

In free expansion, no external work is done on the gas, whereas in other thermodynamic processes, external work is involved. Additionally, free expansion is an irreversible process, meaning that it cannot be reversed to its original state, while other processes, such as isothermal and adiabatic, can be reversed.

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