= Solution
Use the normal boiling point as the reference, $T_B=373.15\,\mathrm K$ and $P_B=101325\,\mathrm{Pa}$. At $T=313.15\,\mathrm K$,
$$
P_{\rm sat}\simeq101325\exp\left[\frac{41000}{8.314}\left(\frac1{373.15}-\frac1{313.15}\right)\right]=\boxed{8.05\times10^3\,\mathrm{Pa}}.
$$
This is approximately 0.0795 atmosphere. It uses the supplied constant <enthalpy of vaporization> approximation across a 60 K interval; it should not be presented as an exact tabulated <vapor pressure>. Absolute <temperature>, rather than Celsius values, belongs in the exponential.
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