WikiMechanics

Let us assess the difference between the enthalpy and the internal energy in P. Recall that enthalpy is defined as

$\begin{align} H \equiv \sum_{\zeta =1}^{10} \Delta n^{\zeta} U^{\zeta} \end{align}$

where $\zeta$ is the quark index, and $\Delta n^{\zeta}$ notes the net number of $\zeta$-type quarks in P

$\Delta n^{\zeta} \equiv n^{\bar{\zeta}} - n^{\zeta}$

And, as discussed earlier, the internal energy is conserved so that

$\begin{align} U = \sum_{\zeta =1}^{10} N^{\zeta} U^{\zeta} \end{align}$

where $N^{\zeta}$ notes the total number of $\zeta$-type quarks in P

$N^{\zeta} \equiv n^{\bar{\zeta}} + n^{\zeta}$

Then

$\begin{align} H-U &= \left[ \sum_{\zeta=1}^{10} \left( n^{\bar{\zeta}} - n^{\zeta} \right) U^{\sf{\zeta}} \right] - \left[ \sum_{\zeta=1}^{10} \left( n^{\bar{\zeta}} + n^{\zeta} \right) U^{\sf{\zeta}} \right] \\ &= -2 \sum_{\zeta=1}^{10} n^{\zeta} U^{\sf{\zeta}} \end{align}$

Let the energy of a typical ordinary-quark be defined as

$\begin{align} \tilde{ U } _{\! \sf{o}} \equiv \sum_{\zeta=1}^{16} \frac{ n^{\zeta} }{ N_{\sf{o}}} U^{\sf{\zeta}} \end{align}$

where $N_{ \sf{o}}$ notes the total number of all types of ordinary-quarks in P

$\begin{align} N_{ \sf{o}} \equiv \sum_{\zeta =1}^{16} n^{\zeta} \end{align}$

Then

$H-U = -2 \, N_{\sf{o}} \tilde{ U } _{\! \sf{o}}$

or in terms of the pressure and volume