proportionality between Gibbs free energy and number of particles











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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:



enter image description here



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)?










share|cite|improve this question
























  • How can your substance have free energy if there are no molecules present? (N=0)
    – Chester Miller
    5 hours ago










  • @ChesterMiller That's also a fair point.
    – Sha Vuklia
    5 hours ago















up vote
5
down vote

favorite
1












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:



enter image description here



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)?










share|cite|improve this question
























  • How can your substance have free energy if there are no molecules present? (N=0)
    – Chester Miller
    5 hours ago










  • @ChesterMiller That's also a fair point.
    – Sha Vuklia
    5 hours ago













up vote
5
down vote

favorite
1









up vote
5
down vote

favorite
1






1





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:



enter image description here



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)?










share|cite|improve this question















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:



enter image description here



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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 6 hours ago

























asked 6 hours ago









Sha Vuklia

453215




453215












  • How can your substance have free energy if there are no molecules present? (N=0)
    – Chester Miller
    5 hours ago










  • @ChesterMiller That's also a fair point.
    – Sha Vuklia
    5 hours ago


















  • How can your substance have free energy if there are no molecules present? (N=0)
    – Chester Miller
    5 hours ago










  • @ChesterMiller That's also a fair point.
    – Sha Vuklia
    5 hours ago
















How can your substance have free energy if there are no molecules present? (N=0)
– Chester Miller
5 hours ago




How can your substance have free energy if there are no molecules present? (N=0)
– Chester Miller
5 hours ago












@ChesterMiller That's also a fair point.
– Sha Vuklia
5 hours ago




@ChesterMiller That's also a fair point.
– Sha Vuklia
5 hours ago










2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










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.






share|cite|improve this answer





















  • Ah, that's genious. Thanks:)
    – Sha Vuklia
    5 hours ago










  • Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
    – Sha Vuklia
    5 hours ago




















up vote
5
down vote













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.






share|cite|improve this answer





















  • 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!
    – Sha Vuklia
    49 mins ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










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.






share|cite|improve this answer





















  • Ah, that's genious. Thanks:)
    – Sha Vuklia
    5 hours ago










  • Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
    – Sha Vuklia
    5 hours ago

















up vote
1
down vote



accepted










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.






share|cite|improve this answer





















  • Ah, that's genious. Thanks:)
    – Sha Vuklia
    5 hours ago










  • Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
    – Sha Vuklia
    5 hours ago















up vote
1
down vote



accepted







up vote
1
down vote



accepted






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 6 hours ago









probably_someone

15.8k12554




15.8k12554












  • Ah, that's genious. Thanks:)
    – Sha Vuklia
    5 hours ago










  • Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
    – Sha Vuklia
    5 hours ago




















  • Ah, that's genious. Thanks:)
    – Sha Vuklia
    5 hours ago










  • Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
    – Sha Vuklia
    5 hours ago


















Ah, that's genious. Thanks:)
– Sha Vuklia
5 hours ago




Ah, that's genious. Thanks:)
– Sha Vuklia
5 hours ago












Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
– Sha Vuklia
5 hours ago






Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
– Sha Vuklia
5 hours ago












up vote
5
down vote













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.






share|cite|improve this answer





















  • 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!
    – Sha Vuklia
    49 mins ago















up vote
5
down vote













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.






share|cite|improve this answer





















  • 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!
    – Sha Vuklia
    49 mins ago













up vote
5
down vote










up vote
5
down vote









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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









GiorgioP

1,059212




1,059212












  • 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!
    – Sha Vuklia
    49 mins ago


















  • 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!
    – Sha Vuklia
    49 mins ago
















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!
– Sha Vuklia
49 mins ago




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!
– Sha Vuklia
49 mins ago


















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