Question

The temperature during freeze-drying of a tissue sample cannot exceed -10 °C. The vapor pressure of ice at 273.16 K is 624 Pa.
What is the maximum pressure at which freeze-drying can be carried out? (in Pa)

Answer 1

We have to make use of the Clausius Clapeyron equation here, but in this case since we are dealing with freeze drying, so we have to deal with sublimation.

During freeze drying, we first make the water present in the sample to get converted to solid state (ice) by decreasing the temperature at a constant pressure. In this case when the temparature reaches -10⁰C, we then start deceasing the pressure, which causes the ice to sublime and get converted to vapors.

So,

dHsub = dHf + dHvap = 6.01+40.656 = 46.666 kJ/mole

Now, using the Clausius Clapeyron equation:

ln(P1/P2) = (dHsub/R)*(1/T2 - 1/T1)

The data given to us is:

P1 = 624 Pa, T1 = 273.16 K

T2 = 273.16-10 = 263.16 K, R = 8.314 J/(mol.K)

dHsub = 46670 J/mol

Put the values:

ln(624/P2) = (46666/8.314)*(1/263.16 - 1/273.16)

Solve to get:

P2 = 285.81 Pa