Entropy and reversibility
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I'm very confused right now, I need to make sure of the following if they're correct or not:
A reversible path is a path in which if if you return back to your initial state, there would be no change in the system and the surroundings
Let's say I now increase the temperature of something by adding heat and then remove heat again to bring it back to its initial temperature, along the adding heat path, I should have increased the entropy of the system, if I go back to the initial state, I should have decreased the entropy again to its initial state and so the total change of the entropy of the system is zero.
Now from the above, atleast from my understanding which I'm not 100 % convinced about, I just concluded that the change in entropy of the system is zero for a reversible process, even though the process was not adiabatic, but the equation for entropy states that $Delta S = int_i^f{frac{dQ}{T}}$ for a reversible path, so its not always equal to zero, it needs to be adiabatic, what is wrong in my statements? I hope you clear everything up for me please, thank you.
thermodynamics entropy reversibility
asked Nov 24 '18 at 14:47
khaled014zkhaled014z
1067
$endgroup$
|
show 1 more comment
$begingroup$
I'm very confused right now, I need to make sure of the following if they're correct or not:
A reversible path is a path in which if if you return back to your initial state, there would be no change in the system and the surroundings
Let's say I now increase the temperature of something by adding heat and then remove heat again to bring it back to its initial temperature, along the adding heat path, I should have increased the entropy of the system, if I go back to the initial state, I should have decreased the entropy again to its initial state and so the total change of the entropy of the system is zero.
Now from the above, atleast from my understanding which I'm not 100 % convinced about, I just concluded that the change in entropy of the system is zero for a reversible process, even though the process was not adiabatic, but the equation for entropy states that $Delta S = int_i^f{frac{dQ}{T}}$ for a reversible path, so its not always equal to zero, it needs to be adiabatic, what is wrong in my statements? I hope you clear everything up for me please, thank you.
thermodynamics entropy reversibility
asked Nov 24 '18 at 14:47
khaled014zkhaled014z
1067
$endgroup$
$begingroup$
How would you add heat? From some body at a higher temperature? Then how would you get it back to that object?
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– Pieter
Nov 24 '18 at 15:42
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I want to add heat from a higher temperature object then lose the heat back to a lower temperature object, I want to focus on only the state of the system
$endgroup$
– khaled014z
Nov 24 '18 at 15:56
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Then you changed the entropy of the surroundings.
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– Pieter
Nov 24 '18 at 16:12
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But reversibility implies no change in entropy, so the only way in which the entropy of the surroundings are changed is that the process itself I mentioned is not reversible, correct?
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– khaled014z
Nov 24 '18 at 16:14
1
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@khaled014z, reversibility does not imply no change in entropy, it implies no entropy generation. You can have a reversible entropy change, and it's given by $Delta S = int_i^f({frac{dQ}{T}})_{rev}$. Note the $rev$ subscript, as this is integrated along a reversible path. Entropy is a state function, so this same $Delta S$ will also apply for an irreversible interaction between 2 states. For an irreversible interaction, you have $Delta S = int_i^f{frac{dQ}{T}} + S_{gen}$, where we no longer integrate the entropy change term $int_i^f{frac{dQ}{T}}$ along a reversible path.
$endgroup$
– Drew
Nov 24 '18 at 18:52
|
show 1 more comment
$begingroup$
I'm very confused right now, I need to make sure of the following if they're correct or not:
A reversible path is a path in which if if you return back to your initial state, there would be no change in the system and the surroundings
Let's say I now increase the temperature of something by adding heat and then remove heat again to bring it back to its initial temperature, along the adding heat path, I should have increased the entropy of the system, if I go back to the initial state, I should have decreased the entropy again to its initial state and so the total change of the entropy of the system is zero.
Now from the above, atleast from my understanding which I'm not 100 % convinced about, I just concluded that the change in entropy of the system is zero for a reversible process, even though the process was not adiabatic, but the equation for entropy states that $Delta S = int_i^f{frac{dQ}{T}}$ for a reversible path, so its not always equal to zero, it needs to be adiabatic, what is wrong in my statements? I hope you clear everything up for me please, thank you.
thermodynamics entropy reversibility
asked Nov 24 '18 at 14:47
khaled014zkhaled014z
1067
$endgroup$
I'm very confused right now, I need to make sure of the following if they're correct or not:
A reversible path is a path in which if if you return back to your initial state, there would be no change in the system and the surroundings
Let's say I now increase the temperature of something by adding heat and then remove heat again to bring it back to its initial temperature, along the adding heat path, I should have increased the entropy of the system, if I go back to the initial state, I should have decreased the entropy again to its initial state and so the total change of the entropy of the system is zero.
Now from the above, atleast from my understanding which I'm not 100 % convinced about, I just concluded that the change in entropy of the system is zero for a reversible process, even though the process was not adiabatic, but the equation for entropy states that $Delta S = int_i^f{frac{dQ}{T}}$ for a reversible path, so its not always equal to zero, it needs to be adiabatic, what is wrong in my statements? I hope you clear everything up for me please, thank you.
thermodynamics entropy reversibility
thermodynamics entropy reversibility
asked Nov 24 '18 at 14:47
khaled014zkhaled014z
1067
asked Nov 24 '18 at 14:47
khaled014zkhaled014z
1067
asked Nov 24 '18 at 14:47
khaled014zkhaled014z
1067
asked Nov 24 '18 at 14:47
khaled014zkhaled014z
1067
asked Nov 24 '18 at 14:47
khaled014zkhaled014z
1067
1067
$begingroup$
How would you add heat? From some body at a higher temperature? Then how would you get it back to that object?
$endgroup$
– Pieter
Nov 24 '18 at 15:42
$begingroup$
I want to add heat from a higher temperature object then lose the heat back to a lower temperature object, I want to focus on only the state of the system
$endgroup$
– khaled014z
Nov 24 '18 at 15:56
$begingroup$
Then you changed the entropy of the surroundings.
$endgroup$
– Pieter
Nov 24 '18 at 16:12
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But reversibility implies no change in entropy, so the only way in which the entropy of the surroundings are changed is that the process itself I mentioned is not reversible, correct?
$endgroup$
– khaled014z
Nov 24 '18 at 16:14
1
$begingroup$
@khaled014z, reversibility does not imply no change in entropy, it implies no entropy generation. You can have a reversible entropy change, and it's given by $Delta S = int_i^f({frac{dQ}{T}})_{rev}$. Note the $rev$ subscript, as this is integrated along a reversible path. Entropy is a state function, so this same $Delta S$ will also apply for an irreversible interaction between 2 states. For an irreversible interaction, you have $Delta S = int_i^f{frac{dQ}{T}} + S_{gen}$, where we no longer integrate the entropy change term $int_i^f{frac{dQ}{T}}$ along a reversible path.
$endgroup$
– Drew
Nov 24 '18 at 18:52
|
show 1 more comment
$begingroup$
How would you add heat? From some body at a higher temperature? Then how would you get it back to that object?
$endgroup$
– Pieter
Nov 24 '18 at 15:42
$begingroup$
I want to add heat from a higher temperature object then lose the heat back to a lower temperature object, I want to focus on only the state of the system
$endgroup$
– khaled014z
Nov 24 '18 at 15:56
$begingroup$
Then you changed the entropy of the surroundings.
$endgroup$
– Pieter
Nov 24 '18 at 16:12
$begingroup$
But reversibility implies no change in entropy, so the only way in which the entropy of the surroundings are changed is that the process itself I mentioned is not reversible, correct?
$endgroup$
– khaled014z
Nov 24 '18 at 16:14
1
$begingroup$
@khaled014z, reversibility does not imply no change in entropy, it implies no entropy generation. You can have a reversible entropy change, and it's given by $Delta S = int_i^f({frac{dQ}{T}})_{rev}$. Note the $rev$ subscript, as this is integrated along a reversible path. Entropy is a state function, so this same $Delta S$ will also apply for an irreversible interaction between 2 states. For an irreversible interaction, you have $Delta S = int_i^f{frac{dQ}{T}} + S_{gen}$, where we no longer integrate the entropy change term $int_i^f{frac{dQ}{T}}$ along a reversible path.
$endgroup$
– Drew
Nov 24 '18 at 18:52
$begingroup$
How would you add heat? From some body at a higher temperature? Then how would you get it back to that object?
$endgroup$
– Pieter
Nov 24 '18 at 15:42
$begingroup$
How would you add heat? From some body at a higher temperature? Then how would you get it back to that object?
$endgroup$
– Pieter
Nov 24 '18 at 15:42
$begingroup$
I want to add heat from a higher temperature object then lose the heat back to a lower temperature object, I want to focus on only the state of the system
$endgroup$
– khaled014z
Nov 24 '18 at 15:56
$begingroup$
I want to add heat from a higher temperature object then lose the heat back to a lower temperature object, I want to focus on only the state of the system
$endgroup$
– khaled014z
Nov 24 '18 at 15:56
$begingroup$
Then you changed the entropy of the surroundings.
$endgroup$
– Pieter
Nov 24 '18 at 16:12
$begingroup$
Then you changed the entropy of the surroundings.
$endgroup$
– Pieter
Nov 24 '18 at 16:12
$begingroup$
But reversibility implies no change in entropy, so the only way in which the entropy of the surroundings are changed is that the process itself I mentioned is not reversible, correct?
$endgroup$
– khaled014z
Nov 24 '18 at 16:14
$begingroup$
But reversibility implies no change in entropy, so the only way in which the entropy of the surroundings are changed is that the process itself I mentioned is not reversible, correct?
$endgroup$
– khaled014z
Nov 24 '18 at 16:14
1
1
$begingroup$
@khaled014z, reversibility does not imply no change in entropy, it implies no entropy generation. You can have a reversible entropy change, and it's given by $Delta S = int_i^f({frac{dQ}{T}})_{rev}$. Note the $rev$ subscript, as this is integrated along a reversible path. Entropy is a state function, so this same $Delta S$ will also apply for an irreversible interaction between 2 states. For an irreversible interaction, you have $Delta S = int_i^f{frac{dQ}{T}} + S_{gen}$, where we no longer integrate the entropy change term $int_i^f{frac{dQ}{T}}$ along a reversible path.
$endgroup$
– Drew
Nov 24 '18 at 18:52
$begingroup$
@khaled014z, reversibility does not imply no change in entropy, it implies no entropy generation. You can have a reversible entropy change, and it's given by $Delta S = int_i^f({frac{dQ}{T}})_{rev}$. Note the $rev$ subscript, as this is integrated along a reversible path. Entropy is a state function, so this same $Delta S$ will also apply for an irreversible interaction between 2 states. For an irreversible interaction, you have $Delta S = int_i^f{frac{dQ}{T}} + S_{gen}$, where we no longer integrate the entropy change term $int_i^f{frac{dQ}{T}}$ along a reversible path.
$endgroup$
– Drew
Nov 24 '18 at 18:52
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
An example of a reversible path that is not adiabatic is a reversible isothermal process. During a reversible isothermal expansion of a gas, for example, heat is added. Since an isothermal process is constant temperature, the temperature comes out of the integral and the change in entropy is $frac {+Q}{T}$ for the system and $frac {-Q}{T}$ for the surroundings where Q is the heat transferred from the surroundings to the system undergoing expansion. The key for reversibility is the temperature difference between the system and the surroundings must be infinitesimally small and the expansion carried out very slowly (Quasi-statically).
Now reverse the process and compress the gas to its original state. The entropy change of the system is $frac{-Q}{T}$ and the surroundings is $frac{+Q}{T}$. The key again is an infinitesimally small difference between the system and surroundings and a slow (Quasi-static) process.
The total entropy change for the combination of the expansion and compression is zero for both the system and surroundings.
Hope this helps.
answered Nov 24 '18 at 16:52
Bob DBob D
2,9802214
$endgroup$
$begingroup$
It helped indeed, just one thing I need to be confirmed, if you have a reversible process then when you expand the gas then compress it back to the original state, the change of entropy of the system is zero and the change of entropy of the surroundings is also zero, but, if I don't reverse the path, say I only expand the gas so the change in entropy of the universe is zero, correct?
$endgroup$
– khaled014z
Nov 24 '18 at 16:59
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@khaled014z Yes, as long as it is reversible. Keep in mind that all real processes are irreversible and that the reversible process is an idealization. For the total entropy change to actually be zero, the temperature between the system and surroundings would have to be the same. But if there is no difference in temperature, you can not have heat transfer.
$endgroup$
– Bob D
Nov 24 '18 at 17:07
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@khaled014z You can prove it to yourself by letting the temperature of the surroundings be greater than the system. Then the temperature in the denominator for the surroundings will be greater than the temperature in the denominator of the system. When you add the two entropies (system + surroundings for the expansion) you will get a number greater that zero.
$endgroup$
– Bob D
Nov 24 '18 at 17:12
1
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@khaled014z The increase in entropy of the surroundings will be greater than the decrease in entropy of the coffee for a total entropy change greater than zero. The process is irreversible. Let Q be the heat transfer from the coffee to the surroundings and let the transfer be isothermal (no change in temp of the coffee and the surroundings). Let the temperature of the coffee be $T_H$ and the temperature of the surroundings be $T_L$. Then the total entropy change will be $frac{-Q}{T_H} + frac{+Q}{T_L}$. You can see that for all $T_{H}$ > $T_L$ the total entropy change will be >0.
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– Bob D
Nov 24 '18 at 18:05
1
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@khaled014z Another way to think about irreversibility and entropy increase. The transfer of heat from the coffee to the surroundings occurs spontaneously. Can you envision the transfer of the same heat from the surroundings to the coffee happening spontaneously?
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– Bob D
Nov 24 '18 at 18:40
|
show 2 more comments
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Reversible just like you said doesn't mean that there is no exchange, it just mean that if you go back there would be no change. In the formula you are using everything depends on your final state. If the final state is the same as the initial state (f = i) then you would find that $Delta S = 0$. But if you do not go back then it is expected that the entropy has changed.
If you separate the path in two part the first part will have $delta Q > 0$ for instance and when you go back you will have $delta Q < 0$ so the integral will be null.
answered Nov 24 '18 at 15:33
Q.ReindeersonQ.Reindeerson
3396
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So far so good, what if I now have an irreversible path from i to f, now I will not go back to i, how would I know that this process is irreversible without going back? Why is change in entropy restricted only on which it has to be calculated along a reversible path? Am I making sense?
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– khaled014z
Nov 24 '18 at 15:54
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Reversibility is just a property. You define that a path is reversible because $textbf{if}$ you chose to go back then there would not have any change. It is just the (kind of) the definition. It is an abstract situation that you don't encounter in everyday life. I'm not sure that i'm getting the "Why is change in entropy restricted only on which it has to be calculated along a reversible path?"
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– Q.Reindeerson
Nov 24 '18 at 16:07
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I mean that, entropy is a state function, which depends only on initial and final points, then why is entropy defined as $Delta S = int_i^f{frac{dQ}{T}}$ if the process is reversible, why can't we calculate the change in entropy on an irreversible path too?
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– khaled014z
Nov 24 '18 at 16:09
2
$begingroup$
@khaled014z you can calculate the change in entropy for an irreversible process, you just need to add the entropy generation term to your equation.
$endgroup$
– Drew
Nov 24 '18 at 17:26
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
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$begingroup$
An example of a reversible path that is not adiabatic is a reversible isothermal process. During a reversible isothermal expansion of a gas, for example, heat is added. Since an isothermal process is constant temperature, the temperature comes out of the integral and the change in entropy is $frac {+Q}{T}$ for the system and $frac {-Q}{T}$ for the surroundings where Q is the heat transferred from the surroundings to the system undergoing expansion. The key for reversibility is the temperature difference between the system and the surroundings must be infinitesimally small and the expansion carried out very slowly (Quasi-statically).
Now reverse the process and compress the gas to its original state. The entropy change of the system is $frac{-Q}{T}$ and the surroundings is $frac{+Q}{T}$. The key again is an infinitesimally small difference between the system and surroundings and a slow (Quasi-static) process.
The total entropy change for the combination of the expansion and compression is zero for both the system and surroundings.
Hope this helps.
answered Nov 24 '18 at 16:52
Bob DBob D
2,9802214
$endgroup$
$begingroup$
It helped indeed, just one thing I need to be confirmed, if you have a reversible process then when you expand the gas then compress it back to the original state, the change of entropy of the system is zero and the change of entropy of the surroundings is also zero, but, if I don't reverse the path, say I only expand the gas so the change in entropy of the universe is zero, correct?
$endgroup$
– khaled014z
Nov 24 '18 at 16:59
$begingroup$
@khaled014z Yes, as long as it is reversible. Keep in mind that all real processes are irreversible and that the reversible process is an idealization. For the total entropy change to actually be zero, the temperature between the system and surroundings would have to be the same. But if there is no difference in temperature, you can not have heat transfer.
$endgroup$
– Bob D
Nov 24 '18 at 17:07
$begingroup$
@khaled014z You can prove it to yourself by letting the temperature of the surroundings be greater than the system. Then the temperature in the denominator for the surroundings will be greater than the temperature in the denominator of the system. When you add the two entropies (system + surroundings for the expansion) you will get a number greater that zero.
$endgroup$
– Bob D
Nov 24 '18 at 17:12
1
$begingroup$
@khaled014z The increase in entropy of the surroundings will be greater than the decrease in entropy of the coffee for a total entropy change greater than zero. The process is irreversible. Let Q be the heat transfer from the coffee to the surroundings and let the transfer be isothermal (no change in temp of the coffee and the surroundings). Let the temperature of the coffee be $T_H$ and the temperature of the surroundings be $T_L$. Then the total entropy change will be $frac{-Q}{T_H} + frac{+Q}{T_L}$. You can see that for all $T_{H}$ > $T_L$ the total entropy change will be >0.
$endgroup$
– Bob D
Nov 24 '18 at 18:05
1
$begingroup$
@khaled014z Another way to think about irreversibility and entropy increase. The transfer of heat from the coffee to the surroundings occurs spontaneously. Can you envision the transfer of the same heat from the surroundings to the coffee happening spontaneously?
$endgroup$
– Bob D
Nov 24 '18 at 18:40
|
show 2 more comments
$begingroup$
An example of a reversible path that is not adiabatic is a reversible isothermal process. During a reversible isothermal expansion of a gas, for example, heat is added. Since an isothermal process is constant temperature, the temperature comes out of the integral and the change in entropy is $frac {+Q}{T}$ for the system and $frac {-Q}{T}$ for the surroundings where Q is the heat transferred from the surroundings to the system undergoing expansion. The key for reversibility is the temperature difference between the system and the surroundings must be infinitesimally small and the expansion carried out very slowly (Quasi-statically).
Now reverse the process and compress the gas to its original state. The entropy change of the system is $frac{-Q}{T}$ and the surroundings is $frac{+Q}{T}$. The key again is an infinitesimally small difference between the system and surroundings and a slow (Quasi-static) process.
The total entropy change for the combination of the expansion and compression is zero for both the system and surroundings.
Hope this helps.
answered Nov 24 '18 at 16:52
Bob DBob D
2,9802214
$endgroup$
$begingroup$
It helped indeed, just one thing I need to be confirmed, if you have a reversible process then when you expand the gas then compress it back to the original state, the change of entropy of the system is zero and the change of entropy of the surroundings is also zero, but, if I don't reverse the path, say I only expand the gas so the change in entropy of the universe is zero, correct?
$endgroup$
– khaled014z
Nov 24 '18 at 16:59
$begingroup$
@khaled014z Yes, as long as it is reversible. Keep in mind that all real processes are irreversible and that the reversible process is an idealization. For the total entropy change to actually be zero, the temperature between the system and surroundings would have to be the same. But if there is no difference in temperature, you can not have heat transfer.
$endgroup$
– Bob D
Nov 24 '18 at 17:07
$begingroup$
@khaled014z You can prove it to yourself by letting the temperature of the surroundings be greater than the system. Then the temperature in the denominator for the surroundings will be greater than the temperature in the denominator of the system. When you add the two entropies (system + surroundings for the expansion) you will get a number greater that zero.
$endgroup$
– Bob D
Nov 24 '18 at 17:12
1
$begingroup$
@khaled014z The increase in entropy of the surroundings will be greater than the decrease in entropy of the coffee for a total entropy change greater than zero. The process is irreversible. Let Q be the heat transfer from the coffee to the surroundings and let the transfer be isothermal (no change in temp of the coffee and the surroundings). Let the temperature of the coffee be $T_H$ and the temperature of the surroundings be $T_L$. Then the total entropy change will be $frac{-Q}{T_H} + frac{+Q}{T_L}$. You can see that for all $T_{H}$ > $T_L$ the total entropy change will be >0.
$endgroup$
– Bob D
Nov 24 '18 at 18:05
1
$begingroup$
@khaled014z Another way to think about irreversibility and entropy increase. The transfer of heat from the coffee to the surroundings occurs spontaneously. Can you envision the transfer of the same heat from the surroundings to the coffee happening spontaneously?
$endgroup$
– Bob D
Nov 24 '18 at 18:40
|
show 2 more comments
$begingroup$
An example of a reversible path that is not adiabatic is a reversible isothermal process. During a reversible isothermal expansion of a gas, for example, heat is added. Since an isothermal process is constant temperature, the temperature comes out of the integral and the change in entropy is $frac {+Q}{T}$ for the system and $frac {-Q}{T}$ for the surroundings where Q is the heat transferred from the surroundings to the system undergoing expansion. The key for reversibility is the temperature difference between the system and the surroundings must be infinitesimally small and the expansion carried out very slowly (Quasi-statically).
Now reverse the process and compress the gas to its original state. The entropy change of the system is $frac{-Q}{T}$ and the surroundings is $frac{+Q}{T}$. The key again is an infinitesimally small difference between the system and surroundings and a slow (Quasi-static) process.
The total entropy change for the combination of the expansion and compression is zero for both the system and surroundings.
Hope this helps.
answered Nov 24 '18 at 16:52
Bob DBob D
2,9802214
$endgroup$
An example of a reversible path that is not adiabatic is a reversible isothermal process. During a reversible isothermal expansion of a gas, for example, heat is added. Since an isothermal process is constant temperature, the temperature comes out of the integral and the change in entropy is $frac {+Q}{T}$ for the system and $frac {-Q}{T}$ for the surroundings where Q is the heat transferred from the surroundings to the system undergoing expansion. The key for reversibility is the temperature difference between the system and the surroundings must be infinitesimally small and the expansion carried out very slowly (Quasi-statically).
Now reverse the process and compress the gas to its original state. The entropy change of the system is $frac{-Q}{T}$ and the surroundings is $frac{+Q}{T}$. The key again is an infinitesimally small difference between the system and surroundings and a slow (Quasi-static) process.
The total entropy change for the combination of the expansion and compression is zero for both the system and surroundings.
Hope this helps.
answered Nov 24 '18 at 16:52
Bob DBob D
2,9802214
answered Nov 24 '18 at 16:52
Bob DBob D
2,9802214
answered Nov 24 '18 at 16:52
Bob DBob D
2,9802214
answered Nov 24 '18 at 16:52
Bob DBob D
2,9802214
2,9802214
$begingroup$
It helped indeed, just one thing I need to be confirmed, if you have a reversible process then when you expand the gas then compress it back to the original state, the change of entropy of the system is zero and the change of entropy of the surroundings is also zero, but, if I don't reverse the path, say I only expand the gas so the change in entropy of the universe is zero, correct?
$endgroup$
– khaled014z
Nov 24 '18 at 16:59
$begingroup$
@khaled014z Yes, as long as it is reversible. Keep in mind that all real processes are irreversible and that the reversible process is an idealization. For the total entropy change to actually be zero, the temperature between the system and surroundings would have to be the same. But if there is no difference in temperature, you can not have heat transfer.
$endgroup$
– Bob D
Nov 24 '18 at 17:07
$begingroup$
@khaled014z You can prove it to yourself by letting the temperature of the surroundings be greater than the system. Then the temperature in the denominator for the surroundings will be greater than the temperature in the denominator of the system. When you add the two entropies (system + surroundings for the expansion) you will get a number greater that zero.
$endgroup$
– Bob D
Nov 24 '18 at 17:12
1
$begingroup$
@khaled014z The increase in entropy of the surroundings will be greater than the decrease in entropy of the coffee for a total entropy change greater than zero. The process is irreversible. Let Q be the heat transfer from the coffee to the surroundings and let the transfer be isothermal (no change in temp of the coffee and the surroundings). Let the temperature of the coffee be $T_H$ and the temperature of the surroundings be $T_L$. Then the total entropy change will be $frac{-Q}{T_H} + frac{+Q}{T_L}$. You can see that for all $T_{H}$ > $T_L$ the total entropy change will be >0.
$endgroup$
– Bob D
Nov 24 '18 at 18:05
1
$begingroup$
@khaled014z Another way to think about irreversibility and entropy increase. The transfer of heat from the coffee to the surroundings occurs spontaneously. Can you envision the transfer of the same heat from the surroundings to the coffee happening spontaneously?
$endgroup$
– Bob D
Nov 24 '18 at 18:40
|
show 2 more comments
$begingroup$
It helped indeed, just one thing I need to be confirmed, if you have a reversible process then when you expand the gas then compress it back to the original state, the change of entropy of the system is zero and the change of entropy of the surroundings is also zero, but, if I don't reverse the path, say I only expand the gas so the change in entropy of the universe is zero, correct?
$endgroup$
– khaled014z
Nov 24 '18 at 16:59
$begingroup$
@khaled014z Yes, as long as it is reversible. Keep in mind that all real processes are irreversible and that the reversible process is an idealization. For the total entropy change to actually be zero, the temperature between the system and surroundings would have to be the same. But if there is no difference in temperature, you can not have heat transfer.
$endgroup$
– Bob D
Nov 24 '18 at 17:07
$begingroup$
@khaled014z You can prove it to yourself by letting the temperature of the surroundings be greater than the system. Then the temperature in the denominator for the surroundings will be greater than the temperature in the denominator of the system. When you add the two entropies (system + surroundings for the expansion) you will get a number greater that zero.
$endgroup$
– Bob D
Nov 24 '18 at 17:12
1
$begingroup$
@khaled014z The increase in entropy of the surroundings will be greater than the decrease in entropy of the coffee for a total entropy change greater than zero. The process is irreversible. Let Q be the heat transfer from the coffee to the surroundings and let the transfer be isothermal (no change in temp of the coffee and the surroundings). Let the temperature of the coffee be $T_H$ and the temperature of the surroundings be $T_L$. Then the total entropy change will be $frac{-Q}{T_H} + frac{+Q}{T_L}$. You can see that for all $T_{H}$ > $T_L$ the total entropy change will be >0.
$endgroup$
– Bob D
Nov 24 '18 at 18:05
1
$begingroup$
@khaled014z Another way to think about irreversibility and entropy increase. The transfer of heat from the coffee to the surroundings occurs spontaneously. Can you envision the transfer of the same heat from the surroundings to the coffee happening spontaneously?
$endgroup$
– Bob D
Nov 24 '18 at 18:40
$begingroup$
It helped indeed, just one thing I need to be confirmed, if you have a reversible process then when you expand the gas then compress it back to the original state, the change of entropy of the system is zero and the change of entropy of the surroundings is also zero, but, if I don't reverse the path, say I only expand the gas so the change in entropy of the universe is zero, correct?
$endgroup$
– khaled014z
Nov 24 '18 at 16:59
$begingroup$
It helped indeed, just one thing I need to be confirmed, if you have a reversible process then when you expand the gas then compress it back to the original state, the change of entropy of the system is zero and the change of entropy of the surroundings is also zero, but, if I don't reverse the path, say I only expand the gas so the change in entropy of the universe is zero, correct?
$endgroup$
– khaled014z
Nov 24 '18 at 16:59
$begingroup$
@khaled014z Yes, as long as it is reversible. Keep in mind that all real processes are irreversible and that the reversible process is an idealization. For the total entropy change to actually be zero, the temperature between the system and surroundings would have to be the same. But if there is no difference in temperature, you can not have heat transfer.
$endgroup$
– Bob D
Nov 24 '18 at 17:07
$begingroup$
@khaled014z Yes, as long as it is reversible. Keep in mind that all real processes are irreversible and that the reversible process is an idealization. For the total entropy change to actually be zero, the temperature between the system and surroundings would have to be the same. But if there is no difference in temperature, you can not have heat transfer.
$endgroup$
– Bob D
Nov 24 '18 at 17:07
$begingroup$
@khaled014z You can prove it to yourself by letting the temperature of the surroundings be greater than the system. Then the temperature in the denominator for the surroundings will be greater than the temperature in the denominator of the system. When you add the two entropies (system + surroundings for the expansion) you will get a number greater that zero.
$endgroup$
– Bob D
Nov 24 '18 at 17:12
$begingroup$
@khaled014z You can prove it to yourself by letting the temperature of the surroundings be greater than the system. Then the temperature in the denominator for the surroundings will be greater than the temperature in the denominator of the system. When you add the two entropies (system + surroundings for the expansion) you will get a number greater that zero.
$endgroup$
– Bob D
Nov 24 '18 at 17:12
1
1
$begingroup$
@khaled014z The increase in entropy of the surroundings will be greater than the decrease in entropy of the coffee for a total entropy change greater than zero. The process is irreversible. Let Q be the heat transfer from the coffee to the surroundings and let the transfer be isothermal (no change in temp of the coffee and the surroundings). Let the temperature of the coffee be $T_H$ and the temperature of the surroundings be $T_L$. Then the total entropy change will be $frac{-Q}{T_H} + frac{+Q}{T_L}$. You can see that for all $T_{H}$ > $T_L$ the total entropy change will be >0.
$endgroup$
– Bob D
Nov 24 '18 at 18:05
$begingroup$
@khaled014z The increase in entropy of the surroundings will be greater than the decrease in entropy of the coffee for a total entropy change greater than zero. The process is irreversible. Let Q be the heat transfer from the coffee to the surroundings and let the transfer be isothermal (no change in temp of the coffee and the surroundings). Let the temperature of the coffee be $T_H$ and the temperature of the surroundings be $T_L$. Then the total entropy change will be $frac{-Q}{T_H} + frac{+Q}{T_L}$. You can see that for all $T_{H}$ > $T_L$ the total entropy change will be >0.
$endgroup$
– Bob D
Nov 24 '18 at 18:05
1
1
$begingroup$
@khaled014z Another way to think about irreversibility and entropy increase. The transfer of heat from the coffee to the surroundings occurs spontaneously. Can you envision the transfer of the same heat from the surroundings to the coffee happening spontaneously?
$endgroup$
– Bob D
Nov 24 '18 at 18:40
$begingroup$
@khaled014z Another way to think about irreversibility and entropy increase. The transfer of heat from the coffee to the surroundings occurs spontaneously. Can you envision the transfer of the same heat from the surroundings to the coffee happening spontaneously?
$endgroup$
– Bob D
Nov 24 '18 at 18:40
|
show 2 more comments
$begingroup$
Reversible just like you said doesn't mean that there is no exchange, it just mean that if you go back there would be no change. In the formula you are using everything depends on your final state. If the final state is the same as the initial state (f = i) then you would find that $Delta S = 0$. But if you do not go back then it is expected that the entropy has changed.
If you separate the path in two part the first part will have $delta Q > 0$ for instance and when you go back you will have $delta Q < 0$ so the integral will be null.
answered Nov 24 '18 at 15:33
Q.ReindeersonQ.Reindeerson
3396
$endgroup$
$begingroup$
So far so good, what if I now have an irreversible path from i to f, now I will not go back to i, how would I know that this process is irreversible without going back? Why is change in entropy restricted only on which it has to be calculated along a reversible path? Am I making sense?
$endgroup$
– khaled014z
Nov 24 '18 at 15:54
$begingroup$
Reversibility is just a property. You define that a path is reversible because $textbf{if}$ you chose to go back then there would not have any change. It is just the (kind of) the definition. It is an abstract situation that you don't encounter in everyday life. I'm not sure that i'm getting the "Why is change in entropy restricted only on which it has to be calculated along a reversible path?"
$endgroup$
– Q.Reindeerson
Nov 24 '18 at 16:07
$begingroup$
I mean that, entropy is a state function, which depends only on initial and final points, then why is entropy defined as $Delta S = int_i^f{frac{dQ}{T}}$ if the process is reversible, why can't we calculate the change in entropy on an irreversible path too?
$endgroup$
– khaled014z
Nov 24 '18 at 16:09
2
$begingroup$
@khaled014z you can calculate the change in entropy for an irreversible process, you just need to add the entropy generation term to your equation.
$endgroup$
– Drew
Nov 24 '18 at 17:26
add a comment |
$begingroup$
Reversible just like you said doesn't mean that there is no exchange, it just mean that if you go back there would be no change. In the formula you are using everything depends on your final state. If the final state is the same as the initial state (f = i) then you would find that $Delta S = 0$. But if you do not go back then it is expected that the entropy has changed.
If you separate the path in two part the first part will have $delta Q > 0$ for instance and when you go back you will have $delta Q < 0$ so the integral will be null.
answered Nov 24 '18 at 15:33
Q.ReindeersonQ.Reindeerson
3396
$endgroup$
$begingroup$
So far so good, what if I now have an irreversible path from i to f, now I will not go back to i, how would I know that this process is irreversible without going back? Why is change in entropy restricted only on which it has to be calculated along a reversible path? Am I making sense?
$endgroup$
– khaled014z
Nov 24 '18 at 15:54
$begingroup$
Reversibility is just a property. You define that a path is reversible because $textbf{if}$ you chose to go back then there would not have any change. It is just the (kind of) the definition. It is an abstract situation that you don't encounter in everyday life. I'm not sure that i'm getting the "Why is change in entropy restricted only on which it has to be calculated along a reversible path?"
$endgroup$
– Q.Reindeerson
Nov 24 '18 at 16:07
$begingroup$
I mean that, entropy is a state function, which depends only on initial and final points, then why is entropy defined as $Delta S = int_i^f{frac{dQ}{T}}$ if the process is reversible, why can't we calculate the change in entropy on an irreversible path too?
$endgroup$
– khaled014z
Nov 24 '18 at 16:09
2
$begingroup$
@khaled014z you can calculate the change in entropy for an irreversible process, you just need to add the entropy generation term to your equation.
$endgroup$
– Drew
Nov 24 '18 at 17:26
add a comment |
$begingroup$
Reversible just like you said doesn't mean that there is no exchange, it just mean that if you go back there would be no change. In the formula you are using everything depends on your final state. If the final state is the same as the initial state (f = i) then you would find that $Delta S = 0$. But if you do not go back then it is expected that the entropy has changed.
If you separate the path in two part the first part will have $delta Q > 0$ for instance and when you go back you will have $delta Q < 0$ so the integral will be null.
answered Nov 24 '18 at 15:33
Q.ReindeersonQ.Reindeerson
3396
$endgroup$
Reversible just like you said doesn't mean that there is no exchange, it just mean that if you go back there would be no change. In the formula you are using everything depends on your final state. If the final state is the same as the initial state (f = i) then you would find that $Delta S = 0$. But if you do not go back then it is expected that the entropy has changed.
If you separate the path in two part the first part will have $delta Q > 0$ for instance and when you go back you will have $delta Q < 0$ so the integral will be null.
answered Nov 24 '18 at 15:33
Q.ReindeersonQ.Reindeerson
3396
edited Nov 24 '18 at 15:42
answered Nov 24 '18 at 15:33
Q.ReindeersonQ.Reindeerson
3396
answered Nov 24 '18 at 15:33
Q.ReindeersonQ.Reindeerson
3396
answered Nov 24 '18 at 15:33
Q.ReindeersonQ.Reindeerson
3396
3396
$begingroup$
So far so good, what if I now have an irreversible path from i to f, now I will not go back to i, how would I know that this process is irreversible without going back? Why is change in entropy restricted only on which it has to be calculated along a reversible path? Am I making sense?
$endgroup$
– khaled014z
Nov 24 '18 at 15:54
$begingroup$
Reversibility is just a property. You define that a path is reversible because $textbf{if}$ you chose to go back then there would not have any change. It is just the (kind of) the definition. It is an abstract situation that you don't encounter in everyday life. I'm not sure that i'm getting the "Why is change in entropy restricted only on which it has to be calculated along a reversible path?"
$endgroup$
– Q.Reindeerson
Nov 24 '18 at 16:07
$begingroup$
I mean that, entropy is a state function, which depends only on initial and final points, then why is entropy defined as $Delta S = int_i^f{frac{dQ}{T}}$ if the process is reversible, why can't we calculate the change in entropy on an irreversible path too?
$endgroup$
– khaled014z
Nov 24 '18 at 16:09
2
$begingroup$
@khaled014z you can calculate the change in entropy for an irreversible process, you just need to add the entropy generation term to your equation.
$endgroup$
– Drew
Nov 24 '18 at 17:26
add a comment |
$begingroup$
So far so good, what if I now have an irreversible path from i to f, now I will not go back to i, how would I know that this process is irreversible without going back? Why is change in entropy restricted only on which it has to be calculated along a reversible path? Am I making sense?
$endgroup$
– khaled014z
Nov 24 '18 at 15:54
$begingroup$
Reversibility is just a property. You define that a path is reversible because $textbf{if}$ you chose to go back then there would not have any change. It is just the (kind of) the definition. It is an abstract situation that you don't encounter in everyday life. I'm not sure that i'm getting the "Why is change in entropy restricted only on which it has to be calculated along a reversible path?"
$endgroup$
– Q.Reindeerson
Nov 24 '18 at 16:07
$begingroup$
I mean that, entropy is a state function, which depends only on initial and final points, then why is entropy defined as $Delta S = int_i^f{frac{dQ}{T}}$ if the process is reversible, why can't we calculate the change in entropy on an irreversible path too?
$endgroup$
– khaled014z
Nov 24 '18 at 16:09
2
$begingroup$
@khaled014z you can calculate the change in entropy for an irreversible process, you just need to add the entropy generation term to your equation.
$endgroup$
– Drew
Nov 24 '18 at 17:26
$begingroup$
So far so good, what if I now have an irreversible path from i to f, now I will not go back to i, how would I know that this process is irreversible without going back? Why is change in entropy restricted only on which it has to be calculated along a reversible path? Am I making sense?
$endgroup$
– khaled014z
Nov 24 '18 at 15:54
$begingroup$
So far so good, what if I now have an irreversible path from i to f, now I will not go back to i, how would I know that this process is irreversible without going back? Why is change in entropy restricted only on which it has to be calculated along a reversible path? Am I making sense?
$endgroup$
– khaled014z
Nov 24 '18 at 15:54
$begingroup$
Reversibility is just a property. You define that a path is reversible because $textbf{if}$ you chose to go back then there would not have any change. It is just the (kind of) the definition. It is an abstract situation that you don't encounter in everyday life. I'm not sure that i'm getting the "Why is change in entropy restricted only on which it has to be calculated along a reversible path?"
$endgroup$
– Q.Reindeerson
Nov 24 '18 at 16:07
$begingroup$
Reversibility is just a property. You define that a path is reversible because $textbf{if}$ you chose to go back then there would not have any change. It is just the (kind of) the definition. It is an abstract situation that you don't encounter in everyday life. I'm not sure that i'm getting the "Why is change in entropy restricted only on which it has to be calculated along a reversible path?"
$endgroup$
– Q.Reindeerson
Nov 24 '18 at 16:07
$begingroup$
I mean that, entropy is a state function, which depends only on initial and final points, then why is entropy defined as $Delta S = int_i^f{frac{dQ}{T}}$ if the process is reversible, why can't we calculate the change in entropy on an irreversible path too?
$endgroup$
– khaled014z
Nov 24 '18 at 16:09
$begingroup$
I mean that, entropy is a state function, which depends only on initial and final points, then why is entropy defined as $Delta S = int_i^f{frac{dQ}{T}}$ if the process is reversible, why can't we calculate the change in entropy on an irreversible path too?
$endgroup$
– khaled014z
Nov 24 '18 at 16:09
2
2
$begingroup$
@khaled014z you can calculate the change in entropy for an irreversible process, you just need to add the entropy generation term to your equation.
$endgroup$
– Drew
Nov 24 '18 at 17:26
$begingroup$
@khaled014z you can calculate the change in entropy for an irreversible process, you just need to add the entropy generation term to your equation.
$endgroup$
– Drew
Nov 24 '18 at 17:26
add a comment |
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$begingroup$
How would you add heat? From some body at a higher temperature? Then how would you get it back to that object?
$endgroup$
– Pieter
Nov 24 '18 at 15:42
$begingroup$
I want to add heat from a higher temperature object then lose the heat back to a lower temperature object, I want to focus on only the state of the system
$endgroup$
– khaled014z
Nov 24 '18 at 15:56
$begingroup$
Then you changed the entropy of the surroundings.
$endgroup$
– Pieter
Nov 24 '18 at 16:12
$begingroup$
But reversibility implies no change in entropy, so the only way in which the entropy of the surroundings are changed is that the process itself I mentioned is not reversible, correct?
$endgroup$
– khaled014z
Nov 24 '18 at 16:14
1
$begingroup$
@khaled014z, reversibility does not imply no change in entropy, it implies no entropy generation. You can have a reversible entropy change, and it's given by $Delta S = int_i^f({frac{dQ}{T}})_{rev}$. Note the $rev$ subscript, as this is integrated along a reversible path. Entropy is a state function, so this same $Delta S$ will also apply for an irreversible interaction between 2 states. For an irreversible interaction, you have $Delta S = int_i^f{frac{dQ}{T}} + S_{gen}$, where we no longer integrate the entropy change term $int_i^f{frac{dQ}{T}}$ along a reversible path.
$endgroup$
– Drew
Nov 24 '18 at 18:52