Why must the Gibbs Free Energy be proportional to N?

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SUMMARY

The Gibbs Free Energy (G) must be proportional to the number of particles (N) because it is an extensive property. In the discussion, a counterexample using G(T, P, N) = (TPN)^{1/3} illustrates that while G can be extensive, it does not satisfy the proportionality to N under certain transformations. The distinction between extensive and intensive variables is crucial; extensive variables scale with the system size, while intensive variables like temperature (T) and pressure (P) do not. Thus, for G to be extensive, it must depend on extensive variables only.

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aliens123
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In the textbook Thermal Physics by Daniel Schroeder he says the following:
5.35.PNG


However, I don't follow this argument. Let's say that G was the following:
$$G(T, P, N) = (TPN)^{1/3}$$
Then
$$G(\lambda T, \lambda P, \lambda N) = \lambda G$$
So $$G$$ is extensive, but $$G \not \propto N.$$
 
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aliens123 said:
Let's say that G was the following:
$$G(T, P, N) = (TPN)^{1/3}$$
Then
$$G(\lambda T, \lambda P, \lambda N) = \lambda G$$
So $$G$$ is extensive, but $$G \not \propto N.$$
In this example ##G## is not extensive since ##G(T,P,\lambda N)\ne \lambda G(T,P,N)##
 
Dale said:
In this example ##G## is not extensive since ##G(T,P,\lambda N)\ne \lambda G(T,P,N)##
Isn't the definition of extensive
$$Q(\lambda q_1, \lambda q_2, ..., \lambda q_n ) = \lambda Q(q_1, q_2, ..., q_n)?$$
 
aliens123 said:
Isn't the definition of extensive
$$Q(\lambda q_1, \lambda q_2, ..., \lambda q_n ) = \lambda Q(q_1, q_2, ..., q_n)?$$
The ##q_i## need to be extensive here. You are looking at ##P## and ##T## which are intensive. If you double the amount of stuff, temperature and pressure don't double. Have a look at the rabbits on the page before the excerpt you used in the OP.
 
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No, you have to multiply only the extensive variables with ##\lambda##. The intensive variables don't change when scaling the "system size". That's the definition of extensive vs. intensive variables.
 

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