The pressure is given by

\[
\frac{P}{n k T}=1-\frac{n}{2 d k T} \int r g(r) \frac{d u}{d r} d \boldsymbol{r}
\]

where \(g(r)=y(r) e^{-\beta u(r)}\). This gives

\[
\frac{P}{n k T}=1-\frac{n}{2 d k T} \int r y(r) \frac{d u}{d r} e^{-\beta u(r)} d \boldsymbol{r}=1+\frac{n}{2 d} \int r y(r) \frac{d}{d r}\left(e^{-\beta u(r)}ight) d \boldsymbol{r},
\]

For the case of hard spheres,

\[
\frac{d}{d r}\left(e^{-\beta u(r)}ight)=\delta(r-D)
\]

so

\[
\frac{P}{n k T}=1+\frac{n D^{d}}{2 d} \Omega_{d} y(D)
\]

where \(\Omega_{d}\) is the area of the \(d\)-dimensional unit sphere. For hard spheres \(y(D)=g\left(D^{+}ight)\). In three dimensions \(\eta=\pi n D^{3} / 6\) and \(\Omega_{3}=4 \pi\), so \(P /(n k T)=1+4 \eta g\left(D^{+}ight)\).