Consider the adjacent experimental set-up shown, in which a gas is collected in
ID: 995225 • Letter: C
Question
Consider the adjacent experimental set-up shown, in which a gas is collected in a tube, and the tube is inverted into a vessel that contains a nonvolatile liquid. The nonvolatile liquid in this experiment has a density of 1.33 g/mL. If the atmospheric pressure is 763 torr, and if the difference in the levels of the liquid surfaces is 172 mm, what is the pressure of the gas in atm? Hint: in the figure, Pa is the pressure of the atmosphere and Pg is the pressure of the gas. Also, the density of mercury is 13.6 g/mL.
Explanation / Answer
Outside the tube, the total pressure is the atmospheric pressure, which is 763 torr.
Pressure outside (Po):
Po = Pa = 763 torr
Convert to Pascal:
Po = 763 torr . (133.322 Pa/1 torr)
Po = 101724.7 Pa
Inside the tube, the total pressure is the one from the gas collected and the column of nonvolatile liquid contained in it.
Pi = Pg + pgh
Pi, pressure inside the tube
Pg, gas pressure
p, liquid density (1.33 g/ml = 1330 kg/m3)
g, gravity constant (9.806 m/s2)
h, difference in the levels of the liquid (172 mm = 0.172 m)
The total pressure inside the tube should be equal to the total pressure outside the tube.
Pi = Po
Po = Pg + pgh
Clear Pg:
Pg = Po - pgh
Pg = 101724.7 Pa - (1330 kg/m3).(9.806 m/s2).(0.172 m)
Pg = 101724.7 Pa - 2243.22 Pa
Pg = 99481.5 Pa
Convert to atmosphere:
Pg = 99481.5 Pa . (1 atm/101325 Pa)
Pg = 0.9818 atm