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Let $\Phi$ be the grand potential. The typical argument goes that $$\Phi(T, \lambda V, \mu) = \lambda\Phi(T, V, \mu) \implies \Phi = -p(T, \mu)V$$ why do we rule out the possibility that $$\Phi = -p’(T,V, \mu)V?$$

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Let's suppose that $$\Phi = -p(T, \mu)V + f(T, \mu)$$

When one uses $\Phi(T, \lambda V, \mu) = \lambda\Phi(T, V, \mu)$ with $\lambda = 0$, one reaches $$\Phi(T,0,\mu) = 0$$

So $$0 = \Phi(T,0,\mu) = -p(T,\mu)\times 0 + f(T,\mu) = f(T, \mu)$$

So $$f(T,\mu) = 0$$

In the end, $$ \Phi = -p(T, \mu)V $$

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