Why ds=0 only in a steady rev. process?

[Cosmossos]

Hello
Why ds=0 only in a steady rev. process? Can't it be zore for a rev. and not steady process?

And if the process isn't rev. but still circular. since the entropy is still a state function it should still be zero if we come back to the starting point ,right?
if it isn't true then dU also shouldn't be zero in non rev. circular process.

In a rev. process , we don't have to "complete the circle" in order to call the process a rev. process, right?
please help me understand it.

 

[Andrew Mason]

Hello
Why ds=0 only in a steady rev. process? Can't it be true for a rev. process alone?

And if the process isn't rev. but still circular. since the entropy is still a state function it should still be zero ,right? because we came back to the starting point.
if it isn't true then dU also shouldn't be zero in non rev. circular process.

please help me understand it.

The change in entropy of the system + surroundings is 0 in a reversible process, not just the entropy change of the system.

There is no change in entropy of the system itself over complete reversible or non-reversible cycles, since it returns to its same thermodynamic state. But the entropy of the system + surroundings will change in a non-reversible process.

AM

 

[Cosmossos]

Hello again,
I asked why ds=0 in a steady and rev. process? Why is the steady condition neccesery ?
The entropy is zero if the system comes back to its starting state.

One more question:
why in a isothermic process ds=dq/T? We know that q=integral(c*dt) and since there is no chance in themprature q should be zero.

 

[Cosmossos]

Can ds=0 in rev. but not steady process?
is ds=0 in rev. process because in that kind of process every infi. step is said to be in equilibrium and therefore there is no change in the systems energetic parameters?
Is rev. state also a steady state?

 

[Cosmossos]

The change in entropy of the system + surroundings is 0 in a reversible process, not just the entropy change of the system.

AM

How come? in a rev. process there is no change in the energy of the system (no heat losses for ex.) therefore ds=0

There is no change in entropy of the system itself over complete reversible or non-reversible cycles, since it returns to its same thermodynamic state. But the entropy of the system + surroundings will change in a non-reversible process

We have the cyclic rule (with the integral and q/t>=0) and ds=0 only when we are talking about rev. process

 

[Andrew Mason]

How come? in a rev. process there is no change in the energy of the system (no heat losses for ex.) therefore ds=0

There is a change in energy within the system + surroundings. Heat flows and work is done so heat is converted to work. The total energy does not change. This is not entropy. Conservation of energy has little to do with change in entropy. They are very different: dS = dQ/T; dQ = dU + dW

We have the cyclic rule (with the integral and q/t>=0) and ds=0 only when we are talking about rev. process

If you are speaking about the system and surroundings, dS = 0 only in a reversible process. But not if you are looking only at the system. There is no change in entropy of the system (alone) when it returns to the same thermodynamic state, which it does every cycle even if the process is non-reversible. The surroundings will have an increase in entropy if the process is not reversible but there will be no increase in the entropy of the system itself.

AM

 

[kntsy]

There is no change in entropy of the system itself over complete reversible or non-reversible cycles, since it returns to its same thermodynamic state. But the entropy of the system + surroundings will change in a non-reversible process.

AM

Thanks AM!
But Why But the entropy of the system + surroundings will change in a non-reversible process. ?
Change of Entropy is Not computed from irreversible process, it is calculated by reversible process parameter changes
$$dS\geq\left(\frac{dq}{T}\right)_{irreversible}$$
Whether the process is reversible or not change of entropy is always the same.
If the irreversible process is adiabatic, entropy of the system + surroundings will NOT change in a non-reversible process as there exists a adiatic reversible process linking the 2 states.

 

[Andrew Mason]

Whether the process is reversible or not change of entropy is always the same.
If the irreversible process is adiabatic, entropy of the system + surroundings will NOT change in a non-reversible process as there exists a adiatic reversible process linking the 2 states.

If there exists and adiabatic reversible process from state A to state B, then a non-reversible adiabatic process beginning at state A cannot reach state B. To put it another way, for a non-reversible adiabatic process from state A to state B, the reversible path from state A to state B is not adiabatic.

Example: consider an adiabatic free expansion of a gas from state A (P1,V1,T1) to state B (P2, V2, T1) (temperature does not change since it does no work and there is no heat flow). The reversible path from A to B results in work being done. In that case, since there is no heat flow into the gas, the internal energy (temperature) will decrease. So in the reversible process, heat must flow into the gas to maintain the original temperature.

AM

 

[kntsy]

If there exists and adiabatic reversible process from state A to state B, then a non-reversible adiabatic process beginning at state A cannot reach state B. To put it another way, for a non-reversible adiabatic process from state A to state B, the reversible path from state A to state B is not adiabatic.

Example: consider an adiabatic free expansion of a gas from state A (P1,V1,T1) to state B (P2, V2, T1) (temperature does not change since it does no work and there is no heat flow). The reversible path from A to B results in work being done. In that case, since there is no heat flow into the gas, the internal energy (temperature) will decrease. So in the reversible process, heat must flow into the gas to maintain the original temperature.

AM

I regret saying that for a non-reversible adiabatic process from state A to state B, the reversible path from state A to state B is also adiabatic.
The example of free expansion you given is brilliant.

But further question:
[1]
By analogy can i say for a "non-reversible isotermal process from state A to state B, the reversible path from state A to state B is not isothermal"?
[2]
WHY free expansion is always irreversible and adiabatic? Is it due to friction? Free expansion in "vacuum and no insulator protection" also irreversible and adiabatic?Reference:wiki:Free expansion is an irreversible process in which a gas expands into an insulated evacuated chamber.

 

[kntsy]

I regret saying that for a non-reversible adiabatic process from state A to state B, the reversible path from state A to state B is also adiabatic.
The example of free expansion you given is brilliant.

But further question:
[1]
By analogy can i say for a "non-reversible isotermal process from state A to state B, the reversible path from state A to state B is not isothermal"?
[2]
WHY free expansion is always irreversible and adiabatic? Is it due to friction? Free expansion in "vacuum and no insulator protection" also irreversible and adiabatic?


Reference:wiki:Free expansion is an irreversible process in which a gas expands into an insulated evacuated chamber.

Can anyone answer this question about free expansion? I just realize my concept is very weak. Help!

 

[Andrew Mason]

[1]
By analogy can i say for a "non-reversible isotermal process from state A to state B, the reversible path from state A to state B is not isothermal"?

This is not true. Consider n moles of an ideal gas in these states:
State A = (P1, V1, T1); State B = (P2, V2, T1).

By definition the process is isothermal. But there can be any number of paths that achieve this. You just have to end up with P2V2 = P1V1 = nRT1.

A reversible isothermal path is one in which the system's temperature is constant and the system is in thermal equilibrium at all times with its surroundings -- that means that the system and surroundings are at the same temperature (ie there is an arbitrarily small temperature difference between the system and surroundings in order for heat to flow into or out of the system). In a non-reversible isothermal process, the temperature of the system is constant but it is not in thermal equilibrium with its surroundings (there is a material temperature difference between the system and surroundings).

[2]
WHY free expansion is always irreversible and adiabatic? Is it due to friction? Free expansion in "vacuum and no insulator protection" also irreversible and adiabatic?

Free expansion is always irreversible because the system is not in equilibrium with its surroundings. The surroundings are initially at 0 pressure whereas the gas is at P = nRT/V. There is not an arbitrarily small difference in pressure. So the direction of change cannot be reversed by an infinitessimal change in conditions. Free expansion cannot be reversed without doing work. But no work is done during the free expansion. So you cannot reverse the process using the work generated in the forward process.

AM