Answer:
[tex]P_{N2}[/tex] = 704.7 mm Hg
Explanation:
We are given that an experiment collected 60.0L of nitrogen gas over water at 40.0 degrees Celsius, and the atmospheric pressure was 760.0 mm Hg. We want to find the partial pressure of the nitrogen gas.
Recall that when we collect a gas over water, the pressure that is recorded is actually a sum of the partial pressure of water (dependent on the temperature) and the partial pressure of the gas collected.
In this case, this means that our total pressure (760.0 mm Hg) is really just a sum of [tex]P_{H2O}[/tex] and [tex]P_{N2}[/tex].
Feel free to use a table or graph that contains the partial pressure of water at various temperatures in order to determine what the partial pressure of water is at 40.0 degrees C. In this case, it will be 55.32 mm Hg.
We now know that:
[tex]P_T = P_{H2O} + P_{N2}[/tex]
Where [tex]P_T[/tex] is 760.0 mm Hg and [tex]P_{H2O[/tex] is 55.32 mm Hg.
So:
760.0 mm Hg = 55.32 mm Hg + [tex]P_{N2}[/tex]
Subtract 55.32 mm Hg from both sides.
704.7 mm Hg = [tex]P_{N2}[/tex]
