proportionality between Gibbs free energy and number of particles
$begingroup$
We have the thermodynamic identity for the Gibbs free energy (for a pure system):
$$
dG=-SdT+VdP+mu dN.
$$
Now if we keep $T$ and $P$ fixed, we get
$$
mu=left(frac{partial G}{partial N}right)_{T,P}.
$$
Using the argument that $T,P$ are intensive quantities (which we keep fixed), while $G$ is extensive, if follows that $Delta G=muDelta N$. However, my book actually claims that
$$
G=mu N.
$$
Now, to me this seems odd, because $G$ still depends on $S,T,V,P$. I know that we're assuming that $T$ and $P$ are fixed, but I would think that we have
$$
G=mu N+x,
$$
where $x$ is some function of $S,T,V,P$. Why isn't this the case? I would think that we can't just set this $x$ to zero for every system, because that way we wouldn't be able to compare two systems where this $x$ would technically be different? Or can it be set to zero, and why?
Here is the relevant text from my book:
EDIT
Hm, so I just looked at this question:
Prove that $G=mu N$ and independance of $mu$ on $N$
It seems then that it's $mu$ which depends on $T,P$, and we're using the fact that $G$ is an extensive quantity. I guess that makes sense. What I do wonder about it how we know that the density stays fixed? Because that's what we need too. So I'm basically confused about the relation
$$
G(T,P,alpha N)=alpha G(T,P,N),
$$
which should be the mathematical phrasing that $G$ is extensive. However, I don't understand why we can "forget" about the volume? We'd still need that
$$
rho=N/V=text{constant}. (1)
$$
Where can we find the guarantee that this is the case, and why isn't it expressed in (1)?
thermodynamics
asked Dec 11 '18 at 11:49
Sha VukliaSha Vuklia
474316
$endgroup$
add a comment |
$begingroup$
We have the thermodynamic identity for the Gibbs free energy (for a pure system):
$$
dG=-SdT+VdP+mu dN.
$$
Now if we keep $T$ and $P$ fixed, we get
$$
mu=left(frac{partial G}{partial N}right)_{T,P}.
$$
Using the argument that $T,P$ are intensive quantities (which we keep fixed), while $G$ is extensive, if follows that $Delta G=muDelta N$. However, my book actually claims that
$$
G=mu N.
$$
Now, to me this seems odd, because $G$ still depends on $S,T,V,P$. I know that we're assuming that $T$ and $P$ are fixed, but I would think that we have
$$
G=mu N+x,
$$
where $x$ is some function of $S,T,V,P$. Why isn't this the case? I would think that we can't just set this $x$ to zero for every system, because that way we wouldn't be able to compare two systems where this $x$ would technically be different? Or can it be set to zero, and why?
Here is the relevant text from my book:
EDIT
Hm, so I just looked at this question:
Prove that $G=mu N$ and independance of $mu$ on $N$
It seems then that it's $mu$ which depends on $T,P$, and we're using the fact that $G$ is an extensive quantity. I guess that makes sense. What I do wonder about it how we know that the density stays fixed? Because that's what we need too. So I'm basically confused about the relation
$$
G(T,P,alpha N)=alpha G(T,P,N),
$$
which should be the mathematical phrasing that $G$ is extensive. However, I don't understand why we can "forget" about the volume? We'd still need that
$$
rho=N/V=text{constant}. (1)
$$
Where can we find the guarantee that this is the case, and why isn't it expressed in (1)?
thermodynamics
asked Dec 11 '18 at 11:49
Sha VukliaSha Vuklia
474316
$endgroup$
$begingroup$
How can your substance have free energy if there are no molecules present? (N=0)
$endgroup$
– Chester Miller
Dec 11 '18 at 12:58
$begingroup$
@ChesterMiller That's also a fair point.
$endgroup$
– Sha Vuklia
Dec 11 '18 at 12:58
add a comment |
$begingroup$
We have the thermodynamic identity for the Gibbs free energy (for a pure system):
$$
dG=-SdT+VdP+mu dN.
$$
Now if we keep $T$ and $P$ fixed, we get
$$
mu=left(frac{partial G}{partial N}right)_{T,P}.
$$
Using the argument that $T,P$ are intensive quantities (which we keep fixed), while $G$ is extensive, if follows that $Delta G=muDelta N$. However, my book actually claims that
$$
G=mu N.
$$
Now, to me this seems odd, because $G$ still depends on $S,T,V,P$. I know that we're assuming that $T$ and $P$ are fixed, but I would think that we have
$$
G=mu N+x,
$$
where $x$ is some function of $S,T,V,P$. Why isn't this the case? I would think that we can't just set this $x$ to zero for every system, because that way we wouldn't be able to compare two systems where this $x$ would technically be different? Or can it be set to zero, and why?
Here is the relevant text from my book:
EDIT
Hm, so I just looked at this question:
Prove that $G=mu N$ and independance of $mu$ on $N$
It seems then that it's $mu$ which depends on $T,P$, and we're using the fact that $G$ is an extensive quantity. I guess that makes sense. What I do wonder about it how we know that the density stays fixed? Because that's what we need too. So I'm basically confused about the relation
$$
G(T,P,alpha N)=alpha G(T,P,N),
$$
which should be the mathematical phrasing that $G$ is extensive. However, I don't understand why we can "forget" about the volume? We'd still need that
$$
rho=N/V=text{constant}. (1)
$$
Where can we find the guarantee that this is the case, and why isn't it expressed in (1)?
thermodynamics
asked Dec 11 '18 at 11:49
Sha VukliaSha Vuklia
474316
$endgroup$
We have the thermodynamic identity for the Gibbs free energy (for a pure system):
$$
dG=-SdT+VdP+mu dN.
$$
Now if we keep $T$ and $P$ fixed, we get
$$
mu=left(frac{partial G}{partial N}right)_{T,P}.
$$
Using the argument that $T,P$ are intensive quantities (which we keep fixed), while $G$ is extensive, if follows that $Delta G=muDelta N$. However, my book actually claims that
$$
G=mu N.
$$
Now, to me this seems odd, because $G$ still depends on $S,T,V,P$. I know that we're assuming that $T$ and $P$ are fixed, but I would think that we have
$$
G=mu N+x,
$$
where $x$ is some function of $S,T,V,P$. Why isn't this the case? I would think that we can't just set this $x$ to zero for every system, because that way we wouldn't be able to compare two systems where this $x$ would technically be different? Or can it be set to zero, and why?
Here is the relevant text from my book:
EDIT
Hm, so I just looked at this question:
Prove that $G=mu N$ and independance of $mu$ on $N$
It seems then that it's $mu$ which depends on $T,P$, and we're using the fact that $G$ is an extensive quantity. I guess that makes sense. What I do wonder about it how we know that the density stays fixed? Because that's what we need too. So I'm basically confused about the relation
$$
G(T,P,alpha N)=alpha G(T,P,N),
$$
which should be the mathematical phrasing that $G$ is extensive. However, I don't understand why we can "forget" about the volume? We'd still need that
$$
rho=N/V=text{constant}. (1)
$$
Where can we find the guarantee that this is the case, and why isn't it expressed in (1)?
thermodynamics
thermodynamics
asked Dec 11 '18 at 11:49
Sha VukliaSha Vuklia
474316
asked Dec 11 '18 at 11:49
Sha VukliaSha Vuklia
474316
edited Dec 11 '18 at 12:35
Sha Vuklia
asked Dec 11 '18 at 11:49
Sha VukliaSha Vuklia
474316
asked Dec 11 '18 at 11:49
Sha VukliaSha Vuklia
474316
asked Dec 11 '18 at 11:49
Sha VukliaSha Vuklia
474316
474316
$begingroup$
How can your substance have free energy if there are no molecules present? (N=0)
$endgroup$
– Chester Miller
Dec 11 '18 at 12:58
$begingroup$
@ChesterMiller That's also a fair point.
$endgroup$
– Sha Vuklia
Dec 11 '18 at 12:58
add a comment |
$begingroup$
How can your substance have free energy if there are no molecules present? (N=0)
$endgroup$
– Chester Miller
Dec 11 '18 at 12:58
$begingroup$
@ChesterMiller That's also a fair point.
$endgroup$
– Sha Vuklia
Dec 11 '18 at 12:58
$begingroup$
How can your substance have free energy if there are no molecules present? (N=0)
$endgroup$
– Chester Miller
Dec 11 '18 at 12:58
$begingroup$
How can your substance have free energy if there are no molecules present? (N=0)
$endgroup$
– Chester Miller
Dec 11 '18 at 12:58
$begingroup$
@ChesterMiller That's also a fair point.
$endgroup$
– Sha Vuklia
Dec 11 '18 at 12:58
$begingroup$
@ChesterMiller That's also a fair point.
$endgroup$
– Sha Vuklia
Dec 11 '18 at 12:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Any thermodynamic system has an equation of state, which in this case is of the form $f(P,V,T,N)=0$. Fixing $T$ and $P$ means that $V$ is completely determined by $N$.
The only way this reasoning would fail is if there was an equation of state that didn't involve the volume at all. I have never seen a physical system that had such an equation of state, and I strongly suspect such a system would be unphysical.
answered Dec 11 '18 at 12:45
probably_someoneprobably_someone
17.4k12757
$endgroup$
$begingroup$
Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
$endgroup$
– Sha Vuklia
Dec 11 '18 at 12:53
add a comment |
$begingroup$
The question contains a very common confusion about which independent variables a thermodynamic potential depends on.
$G$ is not a function of $S,T,V,P$. Actually the natural independent variables $G$ depends on can be obtained just looking at the differential: they are $T,P,N$. Each coefficient of the differential form $dG$ should be intended as a function of $T,P,N$, for a fluid one-component system.
Therefore, one would expect to have $mu=mu(P,T,N)$. However, for a normal thermodynamic system, thermodynamic potentials are expected to be extensive, i.e., in this context, $G$ is expected to be a homogeneous function of degree one of its extensive argument $N$. Formally,
$$
G(T,P,alpha N) = alpha G(T,P,N)
$$
should hold for all positive values of $alpha$.
Thus, its is enough to take $alpha=1/N$ (allowed since $N>0$) to get
$$
G(T,P,1) = frac{G(T,P,N)}{N}
$$
i.e. $ G(T,P,N) = N G(T,P,1) $, where $ G(T,P,1) $ has no dependence on N.
On the other hand, $mu = left.frac{partial{G}}{partial{N}}right|_{T,P}$, and we arrive to the conclusion, since: $mu=G(T,P,1)$ is clearly independent on $N$.
This derivation makes clear that a key ingredient to get the result is the extensiveness of the Gibbs free energy, which is granted for large (macroscopic) thermodynamic systems, but could fail for finite systems made by a small number of particles.
answered Dec 11 '18 at 14:37
GiorgioPGiorgioP
2,843321
$endgroup$
$begingroup$
Oh, that's so exactly what I needed. I'll keep the other answer as the "accepted one" (as it answered my edited question) - but many, many thanks for the insights you provided, and +1 of course!
$endgroup$
– Sha Vuklia
Dec 11 '18 at 17:56
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Any thermodynamic system has an equation of state, which in this case is of the form $f(P,V,T,N)=0$. Fixing $T$ and $P$ means that $V$ is completely determined by $N$.
The only way this reasoning would fail is if there was an equation of state that didn't involve the volume at all. I have never seen a physical system that had such an equation of state, and I strongly suspect such a system would be unphysical.
answered Dec 11 '18 at 12:45
probably_someoneprobably_someone
17.4k12757
$endgroup$
$begingroup$
Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
$endgroup$
– Sha Vuklia
Dec 11 '18 at 12:53
add a comment |
$begingroup$
Any thermodynamic system has an equation of state, which in this case is of the form $f(P,V,T,N)=0$. Fixing $T$ and $P$ means that $V$ is completely determined by $N$.
The only way this reasoning would fail is if there was an equation of state that didn't involve the volume at all. I have never seen a physical system that had such an equation of state, and I strongly suspect such a system would be unphysical.
answered Dec 11 '18 at 12:45
probably_someoneprobably_someone
17.4k12757
$endgroup$
$begingroup$
Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
$endgroup$
– Sha Vuklia
Dec 11 '18 at 12:53
add a comment |
$begingroup$
Any thermodynamic system has an equation of state, which in this case is of the form $f(P,V,T,N)=0$. Fixing $T$ and $P$ means that $V$ is completely determined by $N$.
The only way this reasoning would fail is if there was an equation of state that didn't involve the volume at all. I have never seen a physical system that had such an equation of state, and I strongly suspect such a system would be unphysical.
answered Dec 11 '18 at 12:45
probably_someoneprobably_someone
17.4k12757
$endgroup$
Any thermodynamic system has an equation of state, which in this case is of the form $f(P,V,T,N)=0$. Fixing $T$ and $P$ means that $V$ is completely determined by $N$.
The only way this reasoning would fail is if there was an equation of state that didn't involve the volume at all. I have never seen a physical system that had such an equation of state, and I strongly suspect such a system would be unphysical.
answered Dec 11 '18 at 12:45
probably_someoneprobably_someone
17.4k12757
answered Dec 11 '18 at 12:45
probably_someoneprobably_someone
17.4k12757
answered Dec 11 '18 at 12:45
probably_someoneprobably_someone
17.4k12757
answered Dec 11 '18 at 12:45
probably_someoneprobably_someone
17.4k12757
17.4k12757
$begingroup$
Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
$endgroup$
– Sha Vuklia
Dec 11 '18 at 12:53
add a comment |
$begingroup$
Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
$endgroup$
– Sha Vuklia
Dec 11 '18 at 12:53
$begingroup$
Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
$endgroup$
– Sha Vuklia
Dec 11 '18 at 12:53
$begingroup$
Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
$endgroup$
– Sha Vuklia
Dec 11 '18 at 12:53
add a comment |
$begingroup$
The question contains a very common confusion about which independent variables a thermodynamic potential depends on.
$G$ is not a function of $S,T,V,P$. Actually the natural independent variables $G$ depends on can be obtained just looking at the differential: they are $T,P,N$. Each coefficient of the differential form $dG$ should be intended as a function of $T,P,N$, for a fluid one-component system.
Therefore, one would expect to have $mu=mu(P,T,N)$. However, for a normal thermodynamic system, thermodynamic potentials are expected to be extensive, i.e., in this context, $G$ is expected to be a homogeneous function of degree one of its extensive argument $N$. Formally,
$$
G(T,P,alpha N) = alpha G(T,P,N)
$$
should hold for all positive values of $alpha$.
Thus, its is enough to take $alpha=1/N$ (allowed since $N>0$) to get
$$
G(T,P,1) = frac{G(T,P,N)}{N}
$$
i.e. $ G(T,P,N) = N G(T,P,1) $, where $ G(T,P,1) $ has no dependence on N.
On the other hand, $mu = left.frac{partial{G}}{partial{N}}right|_{T,P}$, and we arrive to the conclusion, since: $mu=G(T,P,1)$ is clearly independent on $N$.
This derivation makes clear that a key ingredient to get the result is the extensiveness of the Gibbs free energy, which is granted for large (macroscopic) thermodynamic systems, but could fail for finite systems made by a small number of particles.
answered Dec 11 '18 at 14:37
GiorgioPGiorgioP
2,843321
$endgroup$
$begingroup$
Oh, that's so exactly what I needed. I'll keep the other answer as the "accepted one" (as it answered my edited question) - but many, many thanks for the insights you provided, and +1 of course!
$endgroup$
– Sha Vuklia
Dec 11 '18 at 17:56
add a comment |
$begingroup$
The question contains a very common confusion about which independent variables a thermodynamic potential depends on.
$G$ is not a function of $S,T,V,P$. Actually the natural independent variables $G$ depends on can be obtained just looking at the differential: they are $T,P,N$. Each coefficient of the differential form $dG$ should be intended as a function of $T,P,N$, for a fluid one-component system.
Therefore, one would expect to have $mu=mu(P,T,N)$. However, for a normal thermodynamic system, thermodynamic potentials are expected to be extensive, i.e., in this context, $G$ is expected to be a homogeneous function of degree one of its extensive argument $N$. Formally,
$$
G(T,P,alpha N) = alpha G(T,P,N)
$$
should hold for all positive values of $alpha$.
Thus, its is enough to take $alpha=1/N$ (allowed since $N>0$) to get
$$
G(T,P,1) = frac{G(T,P,N)}{N}
$$
i.e. $ G(T,P,N) = N G(T,P,1) $, where $ G(T,P,1) $ has no dependence on N.
On the other hand, $mu = left.frac{partial{G}}{partial{N}}right|_{T,P}$, and we arrive to the conclusion, since: $mu=G(T,P,1)$ is clearly independent on $N$.
This derivation makes clear that a key ingredient to get the result is the extensiveness of the Gibbs free energy, which is granted for large (macroscopic) thermodynamic systems, but could fail for finite systems made by a small number of particles.
answered Dec 11 '18 at 14:37
GiorgioPGiorgioP
2,843321
$endgroup$
$begingroup$
Oh, that's so exactly what I needed. I'll keep the other answer as the "accepted one" (as it answered my edited question) - but many, many thanks for the insights you provided, and +1 of course!
$endgroup$
– Sha Vuklia
Dec 11 '18 at 17:56
add a comment |
$begingroup$
The question contains a very common confusion about which independent variables a thermodynamic potential depends on.
$G$ is not a function of $S,T,V,P$. Actually the natural independent variables $G$ depends on can be obtained just looking at the differential: they are $T,P,N$. Each coefficient of the differential form $dG$ should be intended as a function of $T,P,N$, for a fluid one-component system.
Therefore, one would expect to have $mu=mu(P,T,N)$. However, for a normal thermodynamic system, thermodynamic potentials are expected to be extensive, i.e., in this context, $G$ is expected to be a homogeneous function of degree one of its extensive argument $N$. Formally,
$$
G(T,P,alpha N) = alpha G(T,P,N)
$$
should hold for all positive values of $alpha$.
Thus, its is enough to take $alpha=1/N$ (allowed since $N>0$) to get
$$
G(T,P,1) = frac{G(T,P,N)}{N}
$$
i.e. $ G(T,P,N) = N G(T,P,1) $, where $ G(T,P,1) $ has no dependence on N.
On the other hand, $mu = left.frac{partial{G}}{partial{N}}right|_{T,P}$, and we arrive to the conclusion, since: $mu=G(T,P,1)$ is clearly independent on $N$.
This derivation makes clear that a key ingredient to get the result is the extensiveness of the Gibbs free energy, which is granted for large (macroscopic) thermodynamic systems, but could fail for finite systems made by a small number of particles.
answered Dec 11 '18 at 14:37
GiorgioPGiorgioP
2,843321
$endgroup$
The question contains a very common confusion about which independent variables a thermodynamic potential depends on.
$G$ is not a function of $S,T,V,P$. Actually the natural independent variables $G$ depends on can be obtained just looking at the differential: they are $T,P,N$. Each coefficient of the differential form $dG$ should be intended as a function of $T,P,N$, for a fluid one-component system.
Therefore, one would expect to have $mu=mu(P,T,N)$. However, for a normal thermodynamic system, thermodynamic potentials are expected to be extensive, i.e., in this context, $G$ is expected to be a homogeneous function of degree one of its extensive argument $N$. Formally,
$$
G(T,P,alpha N) = alpha G(T,P,N)
$$
should hold for all positive values of $alpha$.
Thus, its is enough to take $alpha=1/N$ (allowed since $N>0$) to get
$$
G(T,P,1) = frac{G(T,P,N)}{N}
$$
i.e. $ G(T,P,N) = N G(T,P,1) $, where $ G(T,P,1) $ has no dependence on N.
On the other hand, $mu = left.frac{partial{G}}{partial{N}}right|_{T,P}$, and we arrive to the conclusion, since: $mu=G(T,P,1)$ is clearly independent on $N$.
This derivation makes clear that a key ingredient to get the result is the extensiveness of the Gibbs free energy, which is granted for large (macroscopic) thermodynamic systems, but could fail for finite systems made by a small number of particles.
answered Dec 11 '18 at 14:37
GiorgioPGiorgioP
2,843321
answered Dec 11 '18 at 14:37
GiorgioPGiorgioP
2,843321
answered Dec 11 '18 at 14:37
GiorgioPGiorgioP
2,843321
answered Dec 11 '18 at 14:37
GiorgioPGiorgioP
2,843321
2,843321
$begingroup$
Oh, that's so exactly what I needed. I'll keep the other answer as the "accepted one" (as it answered my edited question) - but many, many thanks for the insights you provided, and +1 of course!
$endgroup$
– Sha Vuklia
Dec 11 '18 at 17:56
add a comment |
$begingroup$
Oh, that's so exactly what I needed. I'll keep the other answer as the "accepted one" (as it answered my edited question) - but many, many thanks for the insights you provided, and +1 of course!
$endgroup$
– Sha Vuklia
Dec 11 '18 at 17:56
$begingroup$
Oh, that's so exactly what I needed. I'll keep the other answer as the "accepted one" (as it answered my edited question) - but many, many thanks for the insights you provided, and +1 of course!
$endgroup$
– Sha Vuklia
Dec 11 '18 at 17:56
$begingroup$
Oh, that's so exactly what I needed. I'll keep the other answer as the "accepted one" (as it answered my edited question) - but many, many thanks for the insights you provided, and +1 of course!
$endgroup$
– Sha Vuklia
Dec 11 '18 at 17:56
add a comment |
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$begingroup$
How can your substance have free energy if there are no molecules present? (N=0)
$endgroup$
– Chester Miller
Dec 11 '18 at 12:58
$begingroup$
@ChesterMiller That's also a fair point.
$endgroup$
– Sha Vuklia
Dec 11 '18 at 12:58