I'd be tempted to use the vapor pressure as a function of temperature, which is very well calibrated, then use the Sakur-Tetrode formula for the entropy, from which you can get the chemical potential.

I guess, the chemical potential of liquid He-4 at zero temperature is about -7.17 K. At least 7.17 K is just the energy you need to add to the system to evaporate one atom to vacuum (evaporation heat). If the chemical potential of the liquid were positive, it would be unstable with respect to evaporation because the chemical potential of a gas (at least of an ideal gas) tends to 0 at T=0 from below (the formula is \mu = kT\ln(n\lambda^2), where \lambda = \sqrt{2\pi\hbar^2/mkT} is the thermal deBroglie wavelength). Above zero, there appear phonons in liquid helium, which add to its chemical potential. The next are rotons, etc. For details, I recommend the article by Vitaly Syvokon "Influence of the superfluid transition on the adsorption of thin helium films" Low Temp. Phys. 32, 48 (2006); http://dx.doi.org/10.1063/1.2161928 and references therein. There you can find even the plot of the chemical potential (see the attached file).

Dr Lobb's suggestion is certainly useful if we are dealing with Helium gas as an ideal gas.

However, my question is more involved since liquid helium-4 is neither an ideal gas nor an ordinary liquid. It is a quantum fluid of strongly interacting bosons. Somehow I missed to mention this aspect in my question.

My question arises out of a conclusion of convensional theories of liquid heleium-4 that different number of He atoms even at T=0 occupy single particle states of different momenta (including p=0). The distribution of bosons over different states is controlled by

$n(E) = 1/[\exp{(E-\mu)/kT}  - 1 ]$

representing the probability for a particle to occupy a state of energy $E$ with $\mu$ being the chemical potential. In what follows from its standard definition, $\mu$ at T=0 can be total energy per He-4 atom which is estimated to be around -7 K and we note that a -ve value of $\mu$ increases $n(E)$ in comparision to that found for $\mu = 0$ (non-interacting boson case); this does not agree with conventional theories of liquid helium-4 which unanimously conclude that $n(E=0)$ decreases because the particles in the liquid basically interact repulsively. This suggests that $\mu$ has positive value and it could be represented by zero point energy estimated to be around 13 K. However, when we adopt this value for $\mu$, $n(E)$ assumes -ve value for all $E > \mu$ indicating that no He atom can occupy a state of energy $E >13$ K which again contradicts the particle distribution concluded by these theories because condensation will not be in any state of energy < 13 K. So what is the right $\mu$ about -7 K or 0 K or 13 K to estimate the correct nature of $n(E)$. In summary, with the discusssion posted by Dr Safonov, things are clear. Thanks Drs Lobb and Safonov.