Question

# A 0.75 L bottle is cleaned, dried, and closed in a room where the air is...

A 0.75 L bottle is cleaned, dried, and closed in a room where the air is 20°C and 47% relative humidity (that is, the water vapor in the air is 0.47 of the equilibrium vapor pressure at 20°C). The bottle is brought outside and stored at 0.0°C. What mass of water condenses inside the bottle?

Here we can use ideal gas equation PV = nRT to calculate mass at 20 deg

From table we know, PH2O = 17.5 torr @ 20 deg C

R = gas constant = 62.3636 L torr / (deg K * mol)

n = PV/RT = (17.5 torr)(0.75 L)/(62.3636 L torr/[deg K*mol])(273 + 20 K) = 7.1829-4 mol

This would be for 100% relative humidity, but we have 47% RH, so it is 7.1829-4 mol * 00.47 = 3.3759-4 mol

So the vapor pressure falls to 4.6 torr at 0 deg. What is the mol of water that can be held in gas phase at 100% humidity?

n = PV/RT = (4.6 torr)(0.75 L)/(62.3636 L torr/[deg K*mol])(273 + 0 K) = 2.02640-4 mol

This means that (3.3759-4 - 2.02640-4 mol) = 0.00013 mol

Since there are 18.02 g H2O

So, 18.02 x 0.00013 mol H2O, then 0.0023 g = 2.3 mg of liquid should form.

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