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 $$\Phi = -p’(T,V, \mu)V?$$


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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.