Saturation Vapour Pressure over Ice (Murphy and Koop)

This is the most water vapour that air can hold in balance with a flat surface of ice, written as a pressure: $p_{ice}$, in pascals, at a temperature $T$ in kelvins. Past it, frost or ice cloud forms. Below freezing it is lower than the pressure over liquid water at the same temperature, so air can be saturated for ice while it is still dry for liquid water. That is why relative humidity with respect to ice is a quantity of its own: the vapour pressure divided by this number. A contrail persists, or a cirrus cloud grows, where it passes 100 percent.

The equation is a fit. Murphy and Koop (2005) fitted it to the best laboratory data and to the Clapeyron equation, and it is their equation 7: $p_{ice} = \exp\left(9.550426 - \frac{5723.265}{T} + 3.53068\ln T - 0.00728332\,T\right)$. It holds from 110 kelvins up to the melting point. The exponential means the answer changes by many orders of magnitude over that range, and the paper's Table C1, values recommended for checking computer codes, shows how many: 6.106e-6 Pa at 150 K, 0.0053975 Pa at 180 K, 0.70202 Pa at 210 K, 27.272 Pa at 240 K, 611.154 Pa at 273.15 K and 611.657 Pa at the triple point, 273.16 K.

Try 150 K: the equation gives 6.106e-6 Pa, the first line of the table. At 240 K, which is -33.15 degrees Celsius, it gives 27.272 Pa. The page also solves the equation for temperature, and 27.272 Pa takes you back to 240 K. That inverse is the idea of the frost point: the temperature at which air holding a given vapour pressure is saturated over ice, the counterpart of the dew point over liquid water. The triple point, 273.16 K, is 0.01 degrees Celsius.

The equation is the standard one for ice only. For supercooled liquid water, and for liquid water above freezing, use the page for the liquid, which is the other half of every humidity calculation near freezing. Enter the temperature in kelvins; the answer is in pascals, and the page shows it in whichever pressure unit you pick.