Question

If 60 L of nitrogen is collected over water at 40°c when the atmospheric perssure is 760mm Hg, what is the partial pressure of the nitrogen?

Answer 1

You need to quote `"the saturated vapour pressure"`...at the given temperature.

Explanation:

We know the total pressure. We do not know the saturated vapour pressure of WATER at the given temperature. These data should have been supplied with the question. We can find the pressure exerted by the gas when we subtract the saturated vapour pressure. The difference will be small but it will still be non-zero.

This site gives `P_"SVP"=55.4*"mm Hg"` at `40` `""^@C`...of course, `P_"SVP"=760*"mm Hg"` at `100` `""^@C`...why so? `P_"SVP"` is often quoted in terms of `"mm Hg"` by reason of convenience given that a mercury column is STILL the best way to measure pressure in a laboratory, despite the best efforts of the health and safety Nazis.

And so ……… we got …….

`P_"measured"=P_"dinitrogen"+P_"SVP"=P_"dinitrogen"+55.4*mm*Hg`...

`P_"dinitrogen"=(760-55.4)*mm*Hg=704.6*mm*Hg`...

And thus, given a volume, and a pressure, and a temperature....we can address the molar quantity and mass of the collected gas...given `PV=nRT`...

`n=((704.6*mm*Hg)/(760.0*mm*Hg*atm^-1)xx60*L)/(0.0821*(L*atm)/(K*mol)xx283.15*K)=2.39*mol`.

Here all units cancel out EXCEPT for `mol`, as is appropriate...

And so we can out the mass of the collected dinitrogen....