 Solution

The Clapeyron equation for molar latent heat $L$ is $dp/dT=L/[T(V_{\mathrm{vap}}-V_{\mathrm{liq}})]$. Neglect the liquid molar volume, treat the vapor as an ideal gas, and approximate $L$ as constant over the interval. ThenUsing one atmosphere as $101325\mathrm{Pa}$ givesThis is the effective molar enthalpy of vaporization inferred from the Integrated Clausius-Clapeyron relation. If $L$ varies with temperature, the inferred value is weighted by $1/T^2$, not an unweighted arithmetic mean. Real helium is not exactly an ideal gas over the whole interval, and its liquid changes superfluid state within it, so this is an estimate under the stated approximations rather than a precise temperature-independent material constant.