# Why one-particle irreducible functional is closely related to pressure (electroweak phase transition)?

Consider the following system: the SM lagrangian somewhat below the EW transition, where we keep only bilinear terms, only the heaviest fermion -- $$t$$-quark, and plus the potential terms for a VEV $$\phi$$: $$\mathcal{L} = \mathcal{L}_{\text{kinetic}}-V(\phi) - \frac{(m_{h}^{\phi})^{2}}{2}h^{2} + (m_{W}^{\phi})^{2}|W|^{2}+\frac{(m_{Z}^{\phi})^{2}}{2}Z^{2} - m_{t}^{\phi}\bar{t}t,$$ where $$m_{X}^{\phi} \sim \phi$$ is mass.

Having this Lagrangian, by integrating out fields $$h,W,Z,t$$ one may compute the thermal one-loop one-particle irreducible effective action $$\Gamma[\phi] = \int d^{4}x V_{\text{eff}}(\phi)$$. After tediuos dealing with thermal QFT, one arrives just to the conclusion that $$\tag 1 V_{\text{eff}}(\phi) = V(\phi) - \sum_{i}P_{i},$$ where $$P_{i}$$ denotes pressure of $$i$$-th species (free Fermi/Bose particles). My question is the following: is it possible to predict the result $$(1)$$ without calculating $$\Gamma$$ explicitly?

My attempt.

I expect that this follows from thermodynamic relations. E.g., $$P =T\frac{\partial\log(Z)}{\partial V},$$ where $$Z$$ is the partition function $$Z = \text{Tr}[e^{-H - \mu N}]$$ The only question is how to relate $$Z$$ to $$\Gamma$$. Let us introduce the generating functional $$Z$$: $$\mathcal{Z}[j] = \text{Tr}[T_{c} e^{-H - \mu N}e^{i\int d^{4}x\phi j}]$$ We have $$\mathcal{Z}[j] = \exp[-W[j]], \quad W[j] = \Gamma[\phi] - \int d^{4}x \frac{\delta W}{\delta j}j$$ It seems to be $$Z = \left( \mathcal{Z}[j]\right)_{j = 0} = e^{-\Gamma[\phi]}$$ Is it correct?