I don't know, what's given. So I can't help to find an answer to your question. One way to derive the caloric equation of state, i.e., the internal energy [itex]U[/itex] is to use the canonical partition sum
[tex]Z=\frac{z^N}{N!}[/tex]
with the single-particle partition function
[tex]z=\frac{V^*}{\Lambda^3} \exp \left (-\frac{\phi}{2 T} \right),[/tex]
where [itex]V^*=V-2 \pi N d^3/3[/itex] is the available volume (geometrical volume minus excluded volume due to the hard-sphere model for the molecules), and [itex]\phi[/itex] is the interaction energy of the particles
[tex]\phi=\frac{N}{V} \int_d^\infty \mathrm{d} r \; U_{\text{rel}}(r) 4 \pi r^2,[/tex]
with the effective two-body potential
[tex]U_{\text{rel}}(r)=\begin{cases}<br />
\infty & \text{for} \quad r \leq d ,\\<br />
-\alpha (d/r)^6 & \text{for} \quad r>d.<br />
\end{cases}[/tex]
The Helmholtz free energy is given by
[tex]A(T,V,N)=-T \ln Z,[/tex]
and from this you can derive all thermodynamical quantities from the usual thermodynamic relations like
[tex]p=-\left (\frac{\partial A}{\partial V} \right)_{T,N}[/tex]
etc. The internal energy is given by the usual Legendre transformation
[tex]U=A+T S.[/tex]