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Is the equation dG = Vdp − SdT valid only for a reversible process? Can it be applied for an irreversible one too?
Difference between reversible and irreversible thermodynamic processWhat is the cause for thermodynamic reversible and irreversible process?Temperature dependence of reaction enthalpyCalculate Work Done for Reversible and Irreversible Adiabatic processhow to mathematically predict reaction resultsDetermination of the integrated rate equation for a reactionDenominator in rate law?Calculating entropy change: reversible vs irreversible process
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
In the derivation, we write:
$$
beginalign
G &= H - TS &quad&to&quad mathrm dG &= mathrm dH - T,mathrm dS - S,mathrm dT \
H &= U + pV &quad&to&quad mathrm dH &= mathrm dU + p,mathrm dV + V,mathrm dp \
U &= Q + W &quad&to&quad mathrm dU &= mathrm dQ - p,mathrm dV \
& & & & &= T,mathrm dS - p,mathrm dV
endalign
$$
So solving these gives
$$mathrm dG = V,mathrm dp - S,mathrm dT$$
But what we actually write is
$$mathrm dH = mathrm dU + p_mathrmint,mathrm dV + V,mathrm dp$$
And in
$$mathrm dU = mathrm dQ - p,mathrm dV = T,mathrm dS - p_mathrmext,mathrm dV$$
so on solving, it should be
$$mathrm dH = T,mathrm dS - (p_mathrmext - p_mathrmint)mathrm dV + V,mathrm dp$$
And only in reversible process, should it be valid to write
$$p_mathrmext - p_mathrmint = 0$$
But my teacher argued that because this is a state function, it should be valid for all types of processes. I'm a bit confused about how his claim is valid. Can someone please explain what he means or even if he is correct or not?
physical-chemistry thermodynamics
edited 8 hours ago
andselisk♦
21.9k8 gold badges77 silver badges147 bronze badges
asked 19 hours ago
DivMitDivMit
5441 silver badge11 bronze badges
$endgroup$
add a comment |
$begingroup$
In the derivation, we write:
$$
beginalign
G &= H - TS &quad&to&quad mathrm dG &= mathrm dH - T,mathrm dS - S,mathrm dT \
H &= U + pV &quad&to&quad mathrm dH &= mathrm dU + p,mathrm dV + V,mathrm dp \
U &= Q + W &quad&to&quad mathrm dU &= mathrm dQ - p,mathrm dV \
& & & & &= T,mathrm dS - p,mathrm dV
endalign
$$
So solving these gives
$$mathrm dG = V,mathrm dp - S,mathrm dT$$
But what we actually write is
$$mathrm dH = mathrm dU + p_mathrmint,mathrm dV + V,mathrm dp$$
And in
$$mathrm dU = mathrm dQ - p,mathrm dV = T,mathrm dS - p_mathrmext,mathrm dV$$
so on solving, it should be
$$mathrm dH = T,mathrm dS - (p_mathrmext - p_mathrmint)mathrm dV + V,mathrm dp$$
And only in reversible process, should it be valid to write
$$p_mathrmext - p_mathrmint = 0$$
But my teacher argued that because this is a state function, it should be valid for all types of processes. I'm a bit confused about how his claim is valid. Can someone please explain what he means or even if he is correct or not?
physical-chemistry thermodynamics
edited 8 hours ago
andselisk♦
21.9k8 gold badges77 silver badges147 bronze badges
asked 19 hours ago
DivMitDivMit
5441 silver badge11 bronze badges
$endgroup$
$begingroup$
Since you've already posted several (good) questions, I allow myself to add a couple of notes on formatting: please do not add hyphens after colons and stick with uniform notations, e.g. don't mix "$p$" and "$P$" for pressure (I personally prefer lowercase $p$, while $P$ is ambiguous and is also used for power).
$endgroup$
– andselisk♦
18 hours ago
add a comment |
$begingroup$
In the derivation, we write:
$$
beginalign
G &= H - TS &quad&to&quad mathrm dG &= mathrm dH - T,mathrm dS - S,mathrm dT \
H &= U + pV &quad&to&quad mathrm dH &= mathrm dU + p,mathrm dV + V,mathrm dp \
U &= Q + W &quad&to&quad mathrm dU &= mathrm dQ - p,mathrm dV \
& & & & &= T,mathrm dS - p,mathrm dV
endalign
$$
So solving these gives
$$mathrm dG = V,mathrm dp - S,mathrm dT$$
But what we actually write is
$$mathrm dH = mathrm dU + p_mathrmint,mathrm dV + V,mathrm dp$$
And in
$$mathrm dU = mathrm dQ - p,mathrm dV = T,mathrm dS - p_mathrmext,mathrm dV$$
so on solving, it should be
$$mathrm dH = T,mathrm dS - (p_mathrmext - p_mathrmint)mathrm dV + V,mathrm dp$$
And only in reversible process, should it be valid to write
$$p_mathrmext - p_mathrmint = 0$$
But my teacher argued that because this is a state function, it should be valid for all types of processes. I'm a bit confused about how his claim is valid. Can someone please explain what he means or even if he is correct or not?
physical-chemistry thermodynamics
edited 8 hours ago
andselisk♦
21.9k8 gold badges77 silver badges147 bronze badges
asked 19 hours ago
DivMitDivMit
5441 silver badge11 bronze badges
$endgroup$
In the derivation, we write:
$$
beginalign
G &= H - TS &quad&to&quad mathrm dG &= mathrm dH - T,mathrm dS - S,mathrm dT \
H &= U + pV &quad&to&quad mathrm dH &= mathrm dU + p,mathrm dV + V,mathrm dp \
U &= Q + W &quad&to&quad mathrm dU &= mathrm dQ - p,mathrm dV \
& & & & &= T,mathrm dS - p,mathrm dV
endalign
$$
So solving these gives
$$mathrm dG = V,mathrm dp - S,mathrm dT$$
But what we actually write is
$$mathrm dH = mathrm dU + p_mathrmint,mathrm dV + V,mathrm dp$$
And in
$$mathrm dU = mathrm dQ - p,mathrm dV = T,mathrm dS - p_mathrmext,mathrm dV$$
so on solving, it should be
$$mathrm dH = T,mathrm dS - (p_mathrmext - p_mathrmint)mathrm dV + V,mathrm dp$$
And only in reversible process, should it be valid to write
$$p_mathrmext - p_mathrmint = 0$$
But my teacher argued that because this is a state function, it should be valid for all types of processes. I'm a bit confused about how his claim is valid. Can someone please explain what he means or even if he is correct or not?
physical-chemistry thermodynamics
physical-chemistry thermodynamics
edited 8 hours ago
andselisk♦
21.9k8 gold badges77 silver badges147 bronze badges
asked 19 hours ago
DivMitDivMit
5441 silver badge11 bronze badges
edited 8 hours ago
andselisk♦
21.9k8 gold badges77 silver badges147 bronze badges
asked 19 hours ago
DivMitDivMit
5441 silver badge11 bronze badges
edited 8 hours ago
andselisk♦
21.9k8 gold badges77 silver badges147 bronze badges
edited 8 hours ago
andselisk♦
21.9k8 gold badges77 silver badges147 bronze badges
edited 8 hours ago
andselisk♦
21.9k8 gold badges77 silver badges147 bronze badges
21.9k8 gold badges77 silver badges147 bronze badges
asked 19 hours ago
DivMitDivMit
5441 silver badge11 bronze badges
asked 19 hours ago
DivMitDivMit
5441 silver badge11 bronze badges
asked 19 hours ago
DivMitDivMit
5441 silver badge11 bronze badges
5441 silver badge11 bronze badges
$begingroup$
Since you've already posted several (good) questions, I allow myself to add a couple of notes on formatting: please do not add hyphens after colons and stick with uniform notations, e.g. don't mix "$p$" and "$P$" for pressure (I personally prefer lowercase $p$, while $P$ is ambiguous and is also used for power).
$endgroup$
– andselisk♦
18 hours ago
add a comment |
$begingroup$
Since you've already posted several (good) questions, I allow myself to add a couple of notes on formatting: please do not add hyphens after colons and stick with uniform notations, e.g. don't mix "$p$" and "$P$" for pressure (I personally prefer lowercase $p$, while $P$ is ambiguous and is also used for power).
$endgroup$
– andselisk♦
18 hours ago
$begingroup$
Since you've already posted several (good) questions, I allow myself to add a couple of notes on formatting: please do not add hyphens after colons and stick with uniform notations, e.g. don't mix "$p$" and "$P$" for pressure (I personally prefer lowercase $p$, while $P$ is ambiguous and is also used for power).
$endgroup$
– andselisk♦
18 hours ago
$begingroup$
Since you've already posted several (good) questions, I allow myself to add a couple of notes on formatting: please do not add hyphens after colons and stick with uniform notations, e.g. don't mix "$p$" and "$P$" for pressure (I personally prefer lowercase $p$, while $P$ is ambiguous and is also used for power).
$endgroup$
– andselisk♦
18 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The easiest way to avoid confusion is to start from the expression for the $pV$ work associated with a reversible process:
$$mathrm dw_mathrmrev = -p_mathrmext,mathrm dV = -p,mathrm dV$$
where $p$ is the internal pressure of the system, and here the system is subjected to $pV$ work only. Yes, it is absolutely true that for reversible $pV$ work the condition $p_mathrmext = p$ holds. While you came to that conclusion, you did so in a flawed way because your expressions for $mathrm dH$ are not correct.
The above definition of reversible $pV$ work leads to the expression for $U$ in terms of $pV$ work and entropy $(mathrm dU = -p,mathrm dV + T,mathrm dS),$ and ultimately, through the definition $H = U + pV,$ to the expression $mathrm dH = V,mathrm dp + T,mathrm dS.$
But my teacher argued that because this is a state function, it should be valid for all types of processes.
The laws of thermodynamics hold irrespective of path. Provided the end points are the same, two different paths will result in the same changes in state variables.
In practice, when attempting to determine the change in a state variable by evaluating a path integral, you want to choose a convenient path. This is usually a reversible one because then the work can be related to equation of state variables, and the entropy change can be related to the heat via the second law.
edited 8 hours ago
andselisk♦
21.9k8 gold badges77 silver badges147 bronze badges
answered 14 hours ago
Buck ThornBuck Thorn
5,3832 gold badges6 silver badges33 bronze badges
$endgroup$
$begingroup$
As far as I can understand is , you solve for reversible. How is my expression for dH wrong ?? . Because for work we are talking about external pressure. But for U, H its internal pressure . How can you just cancel that out if process is irreversible??
$endgroup$
– DivMit
6 hours ago
$begingroup$
The expressions following where you write "but what we actually write..." are "incorrect" ($dH=dU+p_intdV +Vdp$ and the expression for dU). Well, depending on what you assume. As soon as you insert $dQ = TdS$ you implicitly assume reversibility, so $p_int=p_ext$. As for the expression for H, you are right, it involves the internal pressure, but again, you fall back to this implicit assumption $p_int=p_ext$ as soon as you insert an equilibrium expression implying $dQ = TdS$. Try using the first law in its "raw" form instead and see that it also works.
$endgroup$
– Buck Thorn
5 hours ago
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
The easiest way to avoid confusion is to start from the expression for the $pV$ work associated with a reversible process:
$$mathrm dw_mathrmrev = -p_mathrmext,mathrm dV = -p,mathrm dV$$
where $p$ is the internal pressure of the system, and here the system is subjected to $pV$ work only. Yes, it is absolutely true that for reversible $pV$ work the condition $p_mathrmext = p$ holds. While you came to that conclusion, you did so in a flawed way because your expressions for $mathrm dH$ are not correct.
The above definition of reversible $pV$ work leads to the expression for $U$ in terms of $pV$ work and entropy $(mathrm dU = -p,mathrm dV + T,mathrm dS),$ and ultimately, through the definition $H = U + pV,$ to the expression $mathrm dH = V,mathrm dp + T,mathrm dS.$
But my teacher argued that because this is a state function, it should be valid for all types of processes.
The laws of thermodynamics hold irrespective of path. Provided the end points are the same, two different paths will result in the same changes in state variables.
In practice, when attempting to determine the change in a state variable by evaluating a path integral, you want to choose a convenient path. This is usually a reversible one because then the work can be related to equation of state variables, and the entropy change can be related to the heat via the second law.
edited 8 hours ago
andselisk♦
21.9k8 gold badges77 silver badges147 bronze badges
answered 14 hours ago
Buck ThornBuck Thorn
5,3832 gold badges6 silver badges33 bronze badges
$endgroup$
$begingroup$
As far as I can understand is , you solve for reversible. How is my expression for dH wrong ?? . Because for work we are talking about external pressure. But for U, H its internal pressure . How can you just cancel that out if process is irreversible??
$endgroup$
– DivMit
6 hours ago
$begingroup$
The expressions following where you write "but what we actually write..." are "incorrect" ($dH=dU+p_intdV +Vdp$ and the expression for dU). Well, depending on what you assume. As soon as you insert $dQ = TdS$ you implicitly assume reversibility, so $p_int=p_ext$. As for the expression for H, you are right, it involves the internal pressure, but again, you fall back to this implicit assumption $p_int=p_ext$ as soon as you insert an equilibrium expression implying $dQ = TdS$. Try using the first law in its "raw" form instead and see that it also works.
$endgroup$
– Buck Thorn
5 hours ago
add a comment |
$begingroup$
The easiest way to avoid confusion is to start from the expression for the $pV$ work associated with a reversible process:
$$mathrm dw_mathrmrev = -p_mathrmext,mathrm dV = -p,mathrm dV$$
where $p$ is the internal pressure of the system, and here the system is subjected to $pV$ work only. Yes, it is absolutely true that for reversible $pV$ work the condition $p_mathrmext = p$ holds. While you came to that conclusion, you did so in a flawed way because your expressions for $mathrm dH$ are not correct.
The above definition of reversible $pV$ work leads to the expression for $U$ in terms of $pV$ work and entropy $(mathrm dU = -p,mathrm dV + T,mathrm dS),$ and ultimately, through the definition $H = U + pV,$ to the expression $mathrm dH = V,mathrm dp + T,mathrm dS.$
But my teacher argued that because this is a state function, it should be valid for all types of processes.
The laws of thermodynamics hold irrespective of path. Provided the end points are the same, two different paths will result in the same changes in state variables.
In practice, when attempting to determine the change in a state variable by evaluating a path integral, you want to choose a convenient path. This is usually a reversible one because then the work can be related to equation of state variables, and the entropy change can be related to the heat via the second law.
edited 8 hours ago
andselisk♦
21.9k8 gold badges77 silver badges147 bronze badges
answered 14 hours ago
Buck ThornBuck Thorn
5,3832 gold badges6 silver badges33 bronze badges
$endgroup$
$begingroup$
As far as I can understand is , you solve for reversible. How is my expression for dH wrong ?? . Because for work we are talking about external pressure. But for U, H its internal pressure . How can you just cancel that out if process is irreversible??
$endgroup$
– DivMit
6 hours ago
$begingroup$
The expressions following where you write "but what we actually write..." are "incorrect" ($dH=dU+p_intdV +Vdp$ and the expression for dU). Well, depending on what you assume. As soon as you insert $dQ = TdS$ you implicitly assume reversibility, so $p_int=p_ext$. As for the expression for H, you are right, it involves the internal pressure, but again, you fall back to this implicit assumption $p_int=p_ext$ as soon as you insert an equilibrium expression implying $dQ = TdS$. Try using the first law in its "raw" form instead and see that it also works.
$endgroup$
– Buck Thorn
5 hours ago
add a comment |
$begingroup$
The easiest way to avoid confusion is to start from the expression for the $pV$ work associated with a reversible process:
$$mathrm dw_mathrmrev = -p_mathrmext,mathrm dV = -p,mathrm dV$$
where $p$ is the internal pressure of the system, and here the system is subjected to $pV$ work only. Yes, it is absolutely true that for reversible $pV$ work the condition $p_mathrmext = p$ holds. While you came to that conclusion, you did so in a flawed way because your expressions for $mathrm dH$ are not correct.
The above definition of reversible $pV$ work leads to the expression for $U$ in terms of $pV$ work and entropy $(mathrm dU = -p,mathrm dV + T,mathrm dS),$ and ultimately, through the definition $H = U + pV,$ to the expression $mathrm dH = V,mathrm dp + T,mathrm dS.$
But my teacher argued that because this is a state function, it should be valid for all types of processes.
The laws of thermodynamics hold irrespective of path. Provided the end points are the same, two different paths will result in the same changes in state variables.
In practice, when attempting to determine the change in a state variable by evaluating a path integral, you want to choose a convenient path. This is usually a reversible one because then the work can be related to equation of state variables, and the entropy change can be related to the heat via the second law.
edited 8 hours ago
andselisk♦
21.9k8 gold badges77 silver badges147 bronze badges
answered 14 hours ago
Buck ThornBuck Thorn
5,3832 gold badges6 silver badges33 bronze badges
$endgroup$
The easiest way to avoid confusion is to start from the expression for the $pV$ work associated with a reversible process:
$$mathrm dw_mathrmrev = -p_mathrmext,mathrm dV = -p,mathrm dV$$
where $p$ is the internal pressure of the system, and here the system is subjected to $pV$ work only. Yes, it is absolutely true that for reversible $pV$ work the condition $p_mathrmext = p$ holds. While you came to that conclusion, you did so in a flawed way because your expressions for $mathrm dH$ are not correct.
The above definition of reversible $pV$ work leads to the expression for $U$ in terms of $pV$ work and entropy $(mathrm dU = -p,mathrm dV + T,mathrm dS),$ and ultimately, through the definition $H = U + pV,$ to the expression $mathrm dH = V,mathrm dp + T,mathrm dS.$
But my teacher argued that because this is a state function, it should be valid for all types of processes.
The laws of thermodynamics hold irrespective of path. Provided the end points are the same, two different paths will result in the same changes in state variables.
In practice, when attempting to determine the change in a state variable by evaluating a path integral, you want to choose a convenient path. This is usually a reversible one because then the work can be related to equation of state variables, and the entropy change can be related to the heat via the second law.
edited 8 hours ago
andselisk♦
21.9k8 gold badges77 silver badges147 bronze badges
answered 14 hours ago
Buck ThornBuck Thorn
5,3832 gold badges6 silver badges33 bronze badges
edited 8 hours ago
andselisk♦
21.9k8 gold badges77 silver badges147 bronze badges
edited 8 hours ago
andselisk♦
21.9k8 gold badges77 silver badges147 bronze badges
edited 8 hours ago
andselisk♦
21.9k8 gold badges77 silver badges147 bronze badges
21.9k8 gold badges77 silver badges147 bronze badges
answered 14 hours ago
Buck ThornBuck Thorn
5,3832 gold badges6 silver badges33 bronze badges
answered 14 hours ago
Buck ThornBuck Thorn
5,3832 gold badges6 silver badges33 bronze badges
answered 14 hours ago
Buck ThornBuck Thorn
5,3832 gold badges6 silver badges33 bronze badges
5,3832 gold badges6 silver badges33 bronze badges
$begingroup$
As far as I can understand is , you solve for reversible. How is my expression for dH wrong ?? . Because for work we are talking about external pressure. But for U, H its internal pressure . How can you just cancel that out if process is irreversible??
$endgroup$
– DivMit
6 hours ago
$begingroup$
The expressions following where you write "but what we actually write..." are "incorrect" ($dH=dU+p_intdV +Vdp$ and the expression for dU). Well, depending on what you assume. As soon as you insert $dQ = TdS$ you implicitly assume reversibility, so $p_int=p_ext$. As for the expression for H, you are right, it involves the internal pressure, but again, you fall back to this implicit assumption $p_int=p_ext$ as soon as you insert an equilibrium expression implying $dQ = TdS$. Try using the first law in its "raw" form instead and see that it also works.
$endgroup$
– Buck Thorn
5 hours ago
add a comment |
$begingroup$
As far as I can understand is , you solve for reversible. How is my expression for dH wrong ?? . Because for work we are talking about external pressure. But for U, H its internal pressure . How can you just cancel that out if process is irreversible??
$endgroup$
– DivMit
6 hours ago
$begingroup$
The expressions following where you write "but what we actually write..." are "incorrect" ($dH=dU+p_intdV +Vdp$ and the expression for dU). Well, depending on what you assume. As soon as you insert $dQ = TdS$ you implicitly assume reversibility, so $p_int=p_ext$. As for the expression for H, you are right, it involves the internal pressure, but again, you fall back to this implicit assumption $p_int=p_ext$ as soon as you insert an equilibrium expression implying $dQ = TdS$. Try using the first law in its "raw" form instead and see that it also works.
$endgroup$
– Buck Thorn
5 hours ago
$begingroup$
As far as I can understand is , you solve for reversible. How is my expression for dH wrong ?? . Because for work we are talking about external pressure. But for U, H its internal pressure . How can you just cancel that out if process is irreversible??
$endgroup$
– DivMit
6 hours ago
$begingroup$
As far as I can understand is , you solve for reversible. How is my expression for dH wrong ?? . Because for work we are talking about external pressure. But for U, H its internal pressure . How can you just cancel that out if process is irreversible??
$endgroup$
– DivMit
6 hours ago
$begingroup$
The expressions following where you write "but what we actually write..." are "incorrect" ($dH=dU+p_intdV +Vdp$ and the expression for dU). Well, depending on what you assume. As soon as you insert $dQ = TdS$ you implicitly assume reversibility, so $p_int=p_ext$. As for the expression for H, you are right, it involves the internal pressure, but again, you fall back to this implicit assumption $p_int=p_ext$ as soon as you insert an equilibrium expression implying $dQ = TdS$. Try using the first law in its "raw" form instead and see that it also works.
$endgroup$
– Buck Thorn
5 hours ago
$begingroup$
The expressions following where you write "but what we actually write..." are "incorrect" ($dH=dU+p_intdV +Vdp$ and the expression for dU). Well, depending on what you assume. As soon as you insert $dQ = TdS$ you implicitly assume reversibility, so $p_int=p_ext$. As for the expression for H, you are right, it involves the internal pressure, but again, you fall back to this implicit assumption $p_int=p_ext$ as soon as you insert an equilibrium expression implying $dQ = TdS$. Try using the first law in its "raw" form instead and see that it also works.
$endgroup$
– Buck Thorn
5 hours ago
add a comment |
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$begingroup$
Since you've already posted several (good) questions, I allow myself to add a couple of notes on formatting: please do not add hyphens after colons and stick with uniform notations, e.g. don't mix "$p$" and "$P$" for pressure (I personally prefer lowercase $p$, while $P$ is ambiguous and is also used for power).
$endgroup$
– andselisk♦
18 hours ago