proportionality between Gibbs free energy and number of particles

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6














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
$$
mu=left(fracpartial Gpartial Nright)_T,P.
$$

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
$$
rho=N/V=textconstant. (1)
$$

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
    Dec 11 at 12:58










  • @ChesterMiller That's also a fair point.
    – Sha Vuklia
    Dec 11 at 12:58















6














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
$$
mu=left(fracpartial Gpartial Nright)_T,P.
$$

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
$$
rho=N/V=textconstant. (1)
$$

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
    Dec 11 at 12:58










  • @ChesterMiller That's also a fair point.
    – Sha Vuklia
    Dec 11 at 12:58













6












6








6


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
$$
mu=left(fracpartial Gpartial Nright)_T,P.
$$

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
$$
rho=N/V=textconstant. (1)
$$

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
$$
mu=left(fracpartial Gpartial Nright)_T,P.
$$

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
$$
rho=N/V=textconstant. (1)
$$

Where can we find the guarantee that this is the case, and why isn't it expressed in (1)?







thermodynamics






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edited Dec 11 at 12:35

























asked Dec 11 at 11:49









Sha Vuklia

477216




477216











  • How can your substance have free energy if there are no molecules present? (N=0)
    – Chester Miller
    Dec 11 at 12:58










  • @ChesterMiller That's also a fair point.
    – Sha Vuklia
    Dec 11 at 12:58
















  • How can your substance have free energy if there are no molecules present? (N=0)
    – Chester Miller
    Dec 11 at 12:58










  • @ChesterMiller That's also a fair point.
    – Sha Vuklia
    Dec 11 at 12:58















How can your substance have free energy if there are no molecules present? (N=0)
– Chester Miller
Dec 11 at 12:58




How can your substance have free energy if there are no molecules present? (N=0)
– Chester Miller
Dec 11 at 12:58












@ChesterMiller That's also a fair point.
– Sha Vuklia
Dec 11 at 12:58




@ChesterMiller That's also a fair point.
– Sha Vuklia
Dec 11 at 12:58










2 Answers
2






active

oldest

votes


















2














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
    Dec 11 at 12:51










  • Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
    – Sha Vuklia
    Dec 11 at 12:53



















5














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
$$
G(T,P,1) = fracG(T,P,N)N
$$

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, $mu = left.fracpartialGpartialNright|_T,P$, 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
    Dec 11 at 17:56










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














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
    Dec 11 at 12:51










  • Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
    – Sha Vuklia
    Dec 11 at 12:53
















2














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
    Dec 11 at 12:51










  • Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
    – Sha Vuklia
    Dec 11 at 12:53














2












2








2






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 Dec 11 at 12:45









probably_someone

16.3k12655




16.3k12655











  • Ah, that's genious. Thanks:)
    – Sha Vuklia
    Dec 11 at 12:51










  • Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
    – Sha Vuklia
    Dec 11 at 12:53

















  • Ah, that's genious. Thanks:)
    – Sha Vuklia
    Dec 11 at 12:51










  • Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
    – Sha Vuklia
    Dec 11 at 12:53
















Ah, that's genious. Thanks:)
– Sha Vuklia
Dec 11 at 12:51




Ah, that's genious. Thanks:)
– Sha Vuklia
Dec 11 at 12:51












Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
– Sha Vuklia
Dec 11 at 12:53





Oh, and $V$ is usually (if not always) proportional to $N$, so $rho$ will always be fixed.
– Sha Vuklia
Dec 11 at 12:53












5














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
$$
G(T,P,1) = fracG(T,P,N)N
$$

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, $mu = left.fracpartialGpartialNright|_T,P$, 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
    Dec 11 at 17:56















5














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
$$
G(T,P,1) = fracG(T,P,N)N
$$

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, $mu = left.fracpartialGpartialNright|_T,P$, 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
    Dec 11 at 17:56













5












5








5






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
$$
G(T,P,1) = fracG(T,P,N)N
$$

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, $mu = left.fracpartialGpartialNright|_T,P$, 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
$$
G(T,P,1) = fracG(T,P,N)N
$$

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, $mu = left.fracpartialGpartialNright|_T,P$, 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 Dec 11 at 14:37









GiorgioP

1,369212




1,369212











  • 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
    Dec 11 at 17:56
















  • 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
    Dec 11 at 17:56















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
Dec 11 at 17:56




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
Dec 11 at 17:56

















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