Calculating entropy change: reversible vs irreversible process
$begingroup$
Since the change in internal energy and enthalpy, which are equal to the heats for a constant-volume and constant-pressure process, respectively, are state functions, the heats for a reversible v.s. irreversible process should be equal. Thus, the entropy change of the surroundings which equals the quotient of the heat and temperature are equal for reversible v.s. irreversible processes.
As Engel's Physical Chemistry states it:
Because H and U are state functions, the amount of heat entering the surroundings is independent of the path; q is the same whether the transfer occurs reversibly or irreversibly.
However, when the book proceeds to evaluating the change in entropy for the surroundings for a reversible vs irreersible process, the q used for evaluating the change in entropy of surroundings are not equal for the reversible v.s. irreversible process, which contradicts what the textbook previously stated.
The question:
One mole of an ideal gas at 300. K is reversibly and isothermally compressed from a volume of 25.0 L to a volume of 10.0 L. Because the water bath thermal reservoir in the surroundings is very large, T remains essentially constant at 300. K during the process.
Calculation of change in entropy for reversible path:
Calculation for irreversible path:
It seems that the book used the actual heat for the irreversible process. However, I don't understand why the entropy's from the two calculations are not equal. Also, I fail to understand why the change in entropy for the surroundings are not equal for the two process for which the final and initial states are equal --> change in entropy is a state function and should be the same regardless of path.
References
Engel, Thomas, and Philip Reid. Physical Chemistry. San Francisco: Pearson Benjamin Cummings, 2006.
physical-chemistry thermodynamics
asked Apr 3 at 16:12
EthiopiusEthiopius
4061114
$endgroup$
add a comment |
$begingroup$
Since the change in internal energy and enthalpy, which are equal to the heats for a constant-volume and constant-pressure process, respectively, are state functions, the heats for a reversible v.s. irreversible process should be equal. Thus, the entropy change of the surroundings which equals the quotient of the heat and temperature are equal for reversible v.s. irreversible processes.
As Engel's Physical Chemistry states it:
Because H and U are state functions, the amount of heat entering the surroundings is independent of the path; q is the same whether the transfer occurs reversibly or irreversibly.
However, when the book proceeds to evaluating the change in entropy for the surroundings for a reversible vs irreersible process, the q used for evaluating the change in entropy of surroundings are not equal for the reversible v.s. irreversible process, which contradicts what the textbook previously stated.
The question:
One mole of an ideal gas at 300. K is reversibly and isothermally compressed from a volume of 25.0 L to a volume of 10.0 L. Because the water bath thermal reservoir in the surroundings is very large, T remains essentially constant at 300. K during the process.
Calculation of change in entropy for reversible path:
Calculation for irreversible path:
It seems that the book used the actual heat for the irreversible process. However, I don't understand why the entropy's from the two calculations are not equal. Also, I fail to understand why the change in entropy for the surroundings are not equal for the two process for which the final and initial states are equal --> change in entropy is a state function and should be the same regardless of path.
References
Engel, Thomas, and Philip Reid. Physical Chemistry. San Francisco: Pearson Benjamin Cummings, 2006.
physical-chemistry thermodynamics
asked Apr 3 at 16:12
EthiopiusEthiopius
4061114
$endgroup$
$begingroup$
Update: I am starting to get the feeling that (somehow) the final state of the surroundings after an irreversible process is not the same as the final state of the surroundings after a reversible process. Something (for example, friction between the piston and walls of container) make the final states of surroundings different for irreversible vs reversible --> change in S is not the same. However, since the initial and final T and V are the same for the system in reversible and irreversible, the change in entropy is the same for reversible vs irreversible for system. Is this a valid argument?
$endgroup$
– Ethiopius
Apr 3 at 17:09
1
$begingroup$
Entropy change of your system will be the same for both the reversible and irreversible path. However, the entropy of the surroundings will not be the same , as you have seen. The system goes from the same state A to the same state B for both the reversible and irreversible paths, the surroundings are not in the same state after an irreversible process as they would be after a reversible one. @Ethiopius
$endgroup$
– Tyberius
Apr 3 at 17:31
1
$begingroup$
@Tyberius What you mean is that the surroundings are not in the same state after an irreversible process imposed on the system as they would be after a reversible process imposed on the system between its same two end states.
$endgroup$
– Chet Miller
Apr 3 at 23:01
1
$begingroup$
@ChetMiller That is a clearer description. The way I had phrased I was open to misinterpretation.
$endgroup$
– Tyberius
Apr 3 at 23:03
add a comment |
$begingroup$
Since the change in internal energy and enthalpy, which are equal to the heats for a constant-volume and constant-pressure process, respectively, are state functions, the heats for a reversible v.s. irreversible process should be equal. Thus, the entropy change of the surroundings which equals the quotient of the heat and temperature are equal for reversible v.s. irreversible processes.
As Engel's Physical Chemistry states it:
Because H and U are state functions, the amount of heat entering the surroundings is independent of the path; q is the same whether the transfer occurs reversibly or irreversibly.
However, when the book proceeds to evaluating the change in entropy for the surroundings for a reversible vs irreersible process, the q used for evaluating the change in entropy of surroundings are not equal for the reversible v.s. irreversible process, which contradicts what the textbook previously stated.
The question:
One mole of an ideal gas at 300. K is reversibly and isothermally compressed from a volume of 25.0 L to a volume of 10.0 L. Because the water bath thermal reservoir in the surroundings is very large, T remains essentially constant at 300. K during the process.
Calculation of change in entropy for reversible path:
Calculation for irreversible path:
It seems that the book used the actual heat for the irreversible process. However, I don't understand why the entropy's from the two calculations are not equal. Also, I fail to understand why the change in entropy for the surroundings are not equal for the two process for which the final and initial states are equal --> change in entropy is a state function and should be the same regardless of path.
References
Engel, Thomas, and Philip Reid. Physical Chemistry. San Francisco: Pearson Benjamin Cummings, 2006.
physical-chemistry thermodynamics
asked Apr 3 at 16:12
EthiopiusEthiopius
4061114
$endgroup$
Since the change in internal energy and enthalpy, which are equal to the heats for a constant-volume and constant-pressure process, respectively, are state functions, the heats for a reversible v.s. irreversible process should be equal. Thus, the entropy change of the surroundings which equals the quotient of the heat and temperature are equal for reversible v.s. irreversible processes.
As Engel's Physical Chemistry states it:
Because H and U are state functions, the amount of heat entering the surroundings is independent of the path; q is the same whether the transfer occurs reversibly or irreversibly.
However, when the book proceeds to evaluating the change in entropy for the surroundings for a reversible vs irreersible process, the q used for evaluating the change in entropy of surroundings are not equal for the reversible v.s. irreversible process, which contradicts what the textbook previously stated.
The question:
One mole of an ideal gas at 300. K is reversibly and isothermally compressed from a volume of 25.0 L to a volume of 10.0 L. Because the water bath thermal reservoir in the surroundings is very large, T remains essentially constant at 300. K during the process.
Calculation of change in entropy for reversible path:
Calculation for irreversible path:
It seems that the book used the actual heat for the irreversible process. However, I don't understand why the entropy's from the two calculations are not equal. Also, I fail to understand why the change in entropy for the surroundings are not equal for the two process for which the final and initial states are equal --> change in entropy is a state function and should be the same regardless of path.
References
Engel, Thomas, and Philip Reid. Physical Chemistry. San Francisco: Pearson Benjamin Cummings, 2006.
physical-chemistry thermodynamics
physical-chemistry thermodynamics
asked Apr 3 at 16:12
EthiopiusEthiopius
4061114
asked Apr 3 at 16:12
EthiopiusEthiopius
4061114
edited Apr 3 at 17:10
Ethiopius
asked Apr 3 at 16:12
EthiopiusEthiopius
4061114
asked Apr 3 at 16:12
EthiopiusEthiopius
4061114
asked Apr 3 at 16:12
EthiopiusEthiopius
4061114
4061114
$begingroup$
Update: I am starting to get the feeling that (somehow) the final state of the surroundings after an irreversible process is not the same as the final state of the surroundings after a reversible process. Something (for example, friction between the piston and walls of container) make the final states of surroundings different for irreversible vs reversible --> change in S is not the same. However, since the initial and final T and V are the same for the system in reversible and irreversible, the change in entropy is the same for reversible vs irreversible for system. Is this a valid argument?
$endgroup$
– Ethiopius
Apr 3 at 17:09
1
$begingroup$
Entropy change of your system will be the same for both the reversible and irreversible path. However, the entropy of the surroundings will not be the same , as you have seen. The system goes from the same state A to the same state B for both the reversible and irreversible paths, the surroundings are not in the same state after an irreversible process as they would be after a reversible one. @Ethiopius
$endgroup$
– Tyberius
Apr 3 at 17:31
1
$begingroup$
@Tyberius What you mean is that the surroundings are not in the same state after an irreversible process imposed on the system as they would be after a reversible process imposed on the system between its same two end states.
$endgroup$
– Chet Miller
Apr 3 at 23:01
1
$begingroup$
@ChetMiller That is a clearer description. The way I had phrased I was open to misinterpretation.
$endgroup$
– Tyberius
Apr 3 at 23:03
add a comment |
$begingroup$
Update: I am starting to get the feeling that (somehow) the final state of the surroundings after an irreversible process is not the same as the final state of the surroundings after a reversible process. Something (for example, friction between the piston and walls of container) make the final states of surroundings different for irreversible vs reversible --> change in S is not the same. However, since the initial and final T and V are the same for the system in reversible and irreversible, the change in entropy is the same for reversible vs irreversible for system. Is this a valid argument?
$endgroup$
– Ethiopius
Apr 3 at 17:09
1
$begingroup$
Entropy change of your system will be the same for both the reversible and irreversible path. However, the entropy of the surroundings will not be the same , as you have seen. The system goes from the same state A to the same state B for both the reversible and irreversible paths, the surroundings are not in the same state after an irreversible process as they would be after a reversible one. @Ethiopius
$endgroup$
– Tyberius
Apr 3 at 17:31
1
$begingroup$
@Tyberius What you mean is that the surroundings are not in the same state after an irreversible process imposed on the system as they would be after a reversible process imposed on the system between its same two end states.
$endgroup$
– Chet Miller
Apr 3 at 23:01
1
$begingroup$
@ChetMiller That is a clearer description. The way I had phrased I was open to misinterpretation.
$endgroup$
– Tyberius
Apr 3 at 23:03
$begingroup$
Update: I am starting to get the feeling that (somehow) the final state of the surroundings after an irreversible process is not the same as the final state of the surroundings after a reversible process. Something (for example, friction between the piston and walls of container) make the final states of surroundings different for irreversible vs reversible --> change in S is not the same. However, since the initial and final T and V are the same for the system in reversible and irreversible, the change in entropy is the same for reversible vs irreversible for system. Is this a valid argument?
$endgroup$
– Ethiopius
Apr 3 at 17:09
$begingroup$
Update: I am starting to get the feeling that (somehow) the final state of the surroundings after an irreversible process is not the same as the final state of the surroundings after a reversible process. Something (for example, friction between the piston and walls of container) make the final states of surroundings different for irreversible vs reversible --> change in S is not the same. However, since the initial and final T and V are the same for the system in reversible and irreversible, the change in entropy is the same for reversible vs irreversible for system. Is this a valid argument?
$endgroup$
– Ethiopius
Apr 3 at 17:09
1
1
$begingroup$
Entropy change of your system will be the same for both the reversible and irreversible path. However, the entropy of the surroundings will not be the same , as you have seen. The system goes from the same state A to the same state B for both the reversible and irreversible paths, the surroundings are not in the same state after an irreversible process as they would be after a reversible one. @Ethiopius
$endgroup$
– Tyberius
Apr 3 at 17:31
$begingroup$
Entropy change of your system will be the same for both the reversible and irreversible path. However, the entropy of the surroundings will not be the same , as you have seen. The system goes from the same state A to the same state B for both the reversible and irreversible paths, the surroundings are not in the same state after an irreversible process as they would be after a reversible one. @Ethiopius
$endgroup$
– Tyberius
Apr 3 at 17:31
1
1
$begingroup$
@Tyberius What you mean is that the surroundings are not in the same state after an irreversible process imposed on the system as they would be after a reversible process imposed on the system between its same two end states.
$endgroup$
– Chet Miller
Apr 3 at 23:01
$begingroup$
@Tyberius What you mean is that the surroundings are not in the same state after an irreversible process imposed on the system as they would be after a reversible process imposed on the system between its same two end states.
$endgroup$
– Chet Miller
Apr 3 at 23:01
1
1
$begingroup$
@ChetMiller That is a clearer description. The way I had phrased I was open to misinterpretation.
$endgroup$
– Tyberius
Apr 3 at 23:03
$begingroup$
@ChetMiller That is a clearer description. The way I had phrased I was open to misinterpretation.
$endgroup$
– Tyberius
Apr 3 at 23:03
add a comment |
1 Answer
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$begingroup$
You are confused about how to calculate the entropy change for a system and surroundings that have experienced an irreversible process. For the irreversible path, you correctly calculated the final state using the first law of thermodynamics. From this point on, you must totally forget about the irreversible path and, instead, focus exclusively on the initial and final end states. You must separate the system from the surroundings, and then devise an alternative reversible path for each of them separately that takes each from its initial state to its final state. You did that correctly for the system, and determined its correct entropy change (-7.62 J/K). This is the entropy change for the system both for the reversible path you devised as well as for the actual irreversible path.
In the case of the surroundings, during the irreversible process, its internal energy increased by 3.741 kJ at the constant temperature of 300 K. So, for its alternative reversible path, to achieve this same internal energy change at 300 K, the amount of reversible heat it would have to absorb would be the same 3.741 kJ. So, the change in entropy of the surroundings in the irreversible process was 12.47 J/K.
I know that this might all sound a little confusing. So here is a more detailed explanation of the fundamentals as well as a cookbook (foolproof) recipe for determining the change in entropy of the system and surroundings for any arbitrary irreversible process on a closed system (including worked examples of the methodology): https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
answered Apr 3 at 17:13
Chet MillerChet Miller
6,5261613
$endgroup$
add a comment |
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You are confused about how to calculate the entropy change for a system and surroundings that have experienced an irreversible process. For the irreversible path, you correctly calculated the final state using the first law of thermodynamics. From this point on, you must totally forget about the irreversible path and, instead, focus exclusively on the initial and final end states. You must separate the system from the surroundings, and then devise an alternative reversible path for each of them separately that takes each from its initial state to its final state. You did that correctly for the system, and determined its correct entropy change (-7.62 J/K). This is the entropy change for the system both for the reversible path you devised as well as for the actual irreversible path.
In the case of the surroundings, during the irreversible process, its internal energy increased by 3.741 kJ at the constant temperature of 300 K. So, for its alternative reversible path, to achieve this same internal energy change at 300 K, the amount of reversible heat it would have to absorb would be the same 3.741 kJ. So, the change in entropy of the surroundings in the irreversible process was 12.47 J/K.
I know that this might all sound a little confusing. So here is a more detailed explanation of the fundamentals as well as a cookbook (foolproof) recipe for determining the change in entropy of the system and surroundings for any arbitrary irreversible process on a closed system (including worked examples of the methodology): https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
answered Apr 3 at 17:13
Chet MillerChet Miller
6,5261613
$endgroup$
add a comment |
$begingroup$
You are confused about how to calculate the entropy change for a system and surroundings that have experienced an irreversible process. For the irreversible path, you correctly calculated the final state using the first law of thermodynamics. From this point on, you must totally forget about the irreversible path and, instead, focus exclusively on the initial and final end states. You must separate the system from the surroundings, and then devise an alternative reversible path for each of them separately that takes each from its initial state to its final state. You did that correctly for the system, and determined its correct entropy change (-7.62 J/K). This is the entropy change for the system both for the reversible path you devised as well as for the actual irreversible path.
In the case of the surroundings, during the irreversible process, its internal energy increased by 3.741 kJ at the constant temperature of 300 K. So, for its alternative reversible path, to achieve this same internal energy change at 300 K, the amount of reversible heat it would have to absorb would be the same 3.741 kJ. So, the change in entropy of the surroundings in the irreversible process was 12.47 J/K.
I know that this might all sound a little confusing. So here is a more detailed explanation of the fundamentals as well as a cookbook (foolproof) recipe for determining the change in entropy of the system and surroundings for any arbitrary irreversible process on a closed system (including worked examples of the methodology): https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
answered Apr 3 at 17:13
Chet MillerChet Miller
6,5261613
$endgroup$
add a comment |
$begingroup$
You are confused about how to calculate the entropy change for a system and surroundings that have experienced an irreversible process. For the irreversible path, you correctly calculated the final state using the first law of thermodynamics. From this point on, you must totally forget about the irreversible path and, instead, focus exclusively on the initial and final end states. You must separate the system from the surroundings, and then devise an alternative reversible path for each of them separately that takes each from its initial state to its final state. You did that correctly for the system, and determined its correct entropy change (-7.62 J/K). This is the entropy change for the system both for the reversible path you devised as well as for the actual irreversible path.
In the case of the surroundings, during the irreversible process, its internal energy increased by 3.741 kJ at the constant temperature of 300 K. So, for its alternative reversible path, to achieve this same internal energy change at 300 K, the amount of reversible heat it would have to absorb would be the same 3.741 kJ. So, the change in entropy of the surroundings in the irreversible process was 12.47 J/K.
I know that this might all sound a little confusing. So here is a more detailed explanation of the fundamentals as well as a cookbook (foolproof) recipe for determining the change in entropy of the system and surroundings for any arbitrary irreversible process on a closed system (including worked examples of the methodology): https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
answered Apr 3 at 17:13
Chet MillerChet Miller
6,5261613
$endgroup$
You are confused about how to calculate the entropy change for a system and surroundings that have experienced an irreversible process. For the irreversible path, you correctly calculated the final state using the first law of thermodynamics. From this point on, you must totally forget about the irreversible path and, instead, focus exclusively on the initial and final end states. You must separate the system from the surroundings, and then devise an alternative reversible path for each of them separately that takes each from its initial state to its final state. You did that correctly for the system, and determined its correct entropy change (-7.62 J/K). This is the entropy change for the system both for the reversible path you devised as well as for the actual irreversible path.
In the case of the surroundings, during the irreversible process, its internal energy increased by 3.741 kJ at the constant temperature of 300 K. So, for its alternative reversible path, to achieve this same internal energy change at 300 K, the amount of reversible heat it would have to absorb would be the same 3.741 kJ. So, the change in entropy of the surroundings in the irreversible process was 12.47 J/K.
I know that this might all sound a little confusing. So here is a more detailed explanation of the fundamentals as well as a cookbook (foolproof) recipe for determining the change in entropy of the system and surroundings for any arbitrary irreversible process on a closed system (including worked examples of the methodology): https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
answered Apr 3 at 17:13
Chet MillerChet Miller
6,5261613
answered Apr 3 at 17:13
Chet MillerChet Miller
6,5261613
answered Apr 3 at 17:13
Chet MillerChet Miller
6,5261613
answered Apr 3 at 17:13
Chet MillerChet Miller
6,5261613
6,5261613
add a comment |
add a comment |
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$begingroup$
Update: I am starting to get the feeling that (somehow) the final state of the surroundings after an irreversible process is not the same as the final state of the surroundings after a reversible process. Something (for example, friction between the piston and walls of container) make the final states of surroundings different for irreversible vs reversible --> change in S is not the same. However, since the initial and final T and V are the same for the system in reversible and irreversible, the change in entropy is the same for reversible vs irreversible for system. Is this a valid argument?
$endgroup$
– Ethiopius
Apr 3 at 17:09
1
$begingroup$
Entropy change of your system will be the same for both the reversible and irreversible path. However, the entropy of the surroundings will not be the same , as you have seen. The system goes from the same state A to the same state B for both the reversible and irreversible paths, the surroundings are not in the same state after an irreversible process as they would be after a reversible one. @Ethiopius
$endgroup$
– Tyberius
Apr 3 at 17:31
1
$begingroup$
@Tyberius What you mean is that the surroundings are not in the same state after an irreversible process imposed on the system as they would be after a reversible process imposed on the system between its same two end states.
$endgroup$
– Chet Miller
Apr 3 at 23:01
1
$begingroup$
@ChetMiller That is a clearer description. The way I had phrased I was open to misinterpretation.
$endgroup$
– Tyberius
Apr 3 at 23:03