Free expansion of an ideal gas

[Apashanka]

If considering the free expansion of an ideal gas adibatically such that the final volume is double of the initial volume.
Since dU=0 for free expansion ,which implies T_i=T_f for ideal gas...(1)
From adiabatic relation ,
T_iV_i^γ-1=T_fV_f^γ-1, to satisfy (1) ,γ-1=0
and for ideal gas C_p-C_v=nR
or, γ-1=nR/C_v
And hence nR/C_v→0,
Or ,C_v→∞,?

 

[DrClaude]

From adiabatic relation ,
T_iV_i^γ-1=T_fV_f^γ-1

That equation is only valid for a reversible adiabatic process. Free expansion is not reversible.

 

[Apashanka]

That equation is only valid for a reversible adiabatic process. Free expansion is not reversible.

Yah sir,
Is reversible process are those which follow the same path in the p-v diagram , when taken from final to initial state as of initial to final state.??

 

[DrClaude]

Is reversible process are those which follow the same path in the p-v diagram , when taken from final to initial state as of initial to final state.??

In a PV diagram, there is always a reversible path between any two points. This is actually the usual method used for calculating things when a non-reversible path is followed: find a reversible path between the initial and final states, and used that to find, e.g., the amount of additional entropy created along the non-reversible path.

See for instance https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node51.html

 

[Apashanka]

This is actually the usual method used for calculating things when a non-reversible path is followed: find a reversible path between the initial and final states, and used that to find, e.g., the amount of additional entropy created along the non-reversible path.

The entropy change only depends on the initial and final temp.,volume.for ideal gas.
So sir how can we say by entropy change that the system has evolved through reversible or irreversible path??

 

[Chestermiller]

For a process to be reversible, you need to be able to return both the system and its surroundings to their initial states without causing a significant change in anything else. Please describe, for your free expansion situation, a path for returning both the system and its surroundings to their initial states such that nothing else changes.

 

[Chestermiller]

The entropy change only depends on the initial and final temp.,volume.for ideal gas.
So sir how can we say by entropy change that the system has evolved through reversible or irreversible path??

In a reversible process, the entropy change for the combination of system and surroundings is zero. In an irreversible process, the entropy change for the combination of system and surroundings is positive. Also, for a reversible process on a system alone, the equality sign applies in the Clausius inequality; for an irreversible process on a system alone, the inequality sign applied in the Clausius inequality.

 

[Apashanka]

For a process to be reversible, you need to be able to return both the system and its surroundings to their initial states without causing a significant change in anything else. Please describe, for your free expansion situation, a path for returning both the system and its surroundings to their initial states such that nothing else changes.

So for free expansion of an ideal gas if the volume is reduced from double to initial volume ,the initial state is reached as temp. remains same throughout ,so why is it reversible process??

 

[Chestermiller]

So for free expansion of an ideal gas if the volume is reduced from double to initial volume ,the initial state is reached as temp. remains same throughout ,so why is it reversible process??

Please describe the process you would use to get the system back to its original state (including how the surroundings would be handled for this change). In other words, I'm asking you to devise and describe the (reversible) process you would use.

 

[Apashanka]

Please describe the process you would use to get the system back to its original state (including how the surroundings would be handled for this change). In other words, I'm asking you to devise and describe the (reversible) process you would use.

Ohh yes I got it.
It can't be since for the first case the gas is allowed to expand against vacuum to twice it's original volume(2V) by removing a partition in which no work is done by the system

But to restore the system to it's initial state the volume is to be reduced to V by doing work on the system and in that case the change of internal energy will not be zero,otherwise it would be dU=dW and for that the temp.will not be the same as of initial temp.
So one of the parameters T change and hence the free expansion is not a reversible process .
Am I right??

 

[Chestermiller]

Ohh yes I got it.
It can't be since for the first case the gas is allowed to expand against vacuum to twice it's original volume(2V) by removing a partition in which no work is done by the system

But to restore the system to it's initial state the volume is to be reduced to V by doing work on the system and in that case the change of internal energy will not be zero,otherwise it would be dU=dW and for that the temp.will not be the same as of initial temp.
So one of the parameters T change and hence the free expansion is not a reversible process .
Am I right??

Yes. But, in addition to this, you have had to make a change in the surroundings. To recompress the gas, the surroundings have to do work on the system, say by sliding small weights at a sequence of elevations onto the top of the piston. The elevation of these weights would change from beginning to end of the compression. So the ability of the surroundings to do work (its gravitational potential energy) has decreased from beginning to end of the process. So, not only have you returned the system to a different state than initially, you have also changed the surroundings from its initial state.

One way of getting the system back to its initial state would be to do the compression isothermally and reversibly by putting it in contact with a constant-temperature reservoir. However, in this case, in addition to the surroundings doing work on the system, the reservoir (part of the surroundings) would also have to remove heat from the system. So, in this case, the system would not change, but the surroundings would change in two ways.