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 Partial Pressure and Mole Fractions

The mole fraction of an individual gas component in an ideal gas mixture can be expressed in terms of the component's partial pressure or the moles of the component:

$x_{\mathrm{i}} = \frac{P_{\mathrm{i}}}{P} = \frac{n_{\mathrm{i}}}{n}$

and the partial pressure of an individual gas component in an ideal gas can be obtained using this expression:

$P_{\mathrm{i}} = x_{\mathrm{i}} \cdot P$

xi where: = mole fraction of any individual gas component in a gas mixture = partial pressure of any individual gas component in a gas mixture = moles of any individual gas component in a gas mixture = total moles of the gas mixture = pressure of the gas mixture

The mole fraction of a gas component in a gas mixture is equal to the volumetric fraction of that component in a gas mixture.

AP Dalton's Law Problems

65. A container holds three gases: oxygen, carbon dioxide, and helium. The partial pressures of the three gases are 2.00 atm, 3.00 atm, and 4.00 atm, respectively. What is the total pressure inside the container? 65. 9.00 atm.
66. A container with two gases, helium and argon, is 30.0% by volume helium. Calculate the partial pressure of helium and argon if the total pressure inside the container is 4.00 atm. 66. PHe = 0.300 x 4.00 atm = 1.20 atm. PAr = 4.00 - 1.20
67. If 60.0 L of nitrogen is collected over water at 40.0 °C when the atmospheric pressure is 760.0 mm Hg, what is the partial pressure of the nitrogen? 67. 760.0 mmHg minus 55.3 mmHg
68. 80.0 liters of oxygen is collected over water at 50.0 °C. The atmospheric pressure in the room is 96.00 kPa. What is the partial pressure of the oxygen? 68. 96.00 kPa minus 12.33 kPa
69. A tank contains 480.0 grams of oxygen and 80.00 grams of helium at a total pressure of 7.00 atmospheres. Calculate the following.
a) How many moles of O2 are in the tank?
b) How many moles of He are in the tank?
c) Total moles of gas in tank.
d) Mole fraction of O2.
e) Mole fraction of He.
f) Partial pressure of O2.
g) Partial pressure of He.

69.
a) 480.0 g O2 / 32.0 g/mol
b) 80.00 g He / 4.00 g/mol
c) 35.0 moles
d) 15.0 mol O2 / 35.0 mol
e) 20.0 mol He / 35.0 mol
f) 7.00 atm x 0.4286
g) 7.00 atm x 0.5714

Keep in mind that once one partial pressure is calculated, the other can be arrived at by subtraction, if so desired.

70. A tank contains 5.00 moles of O2, 3.00 moles of neon, 6.00 moles of H2S, and 4.00 moles of argon at a total pressure of 1620.0 mm Hg. Complete the following table

O2	Ne	H2S	Ar	Total
Moles							18.00

Mole fraction						1

Pressure fraction					1

Partial Pressure	                                1620.0

70. Complete the following table
 O2 Ne H2S Ar Total Moles 5.00 3.00 6.00 9.00 18.00 Mole fraction 5/18 = 0.278 3/18 = 0.167 6/18 = 0.333 4/18 = 0.222 1 Pressure fraction 0.278 0.167 0.333 0.222 1 Partial Pressure 1620 x 0.278 = 450.36 1620 x 0.167 = 270.54 1620 x 0.0.333 = 539.46 1620 x 0.222 = 359.64 1620.0

71. A mixture of 14.0 grams of hydrogen, 84.0 grams of nitrogen, and 2.0 moles of oxygen are placed in a flask. When the partial pressure of the oxygen is 78.00 mm of mercury, what is the total pressure in the flask? 71. (14.0 g / 2.00 g/mol) + (84.0 g /28.0 g/mol) + (2.0 moles) = 12.0 moles total

2.0/ 12.0 = 0.167 of the total pressure. 78.00 is to 0.167 as the total pressure is to one, so 468 mmHg is the answer.

72. A flask contains 2.00 moles of nitrogen and 2.00 moles of helium. How many grams of argon must be pumped into the flask in order to make the partial pressure of argon twice that of helium? 72. 4.00 moles
 Consider the flask apparatus in the following diagram, which contains 2.00 L of H2 at a pressure of 409 torr and 1.00 L of N2 at an unknown pressure. If the total pressure in the flasks is 340. torr after the stopcock is opened, determine the initial pressure of N2 in the 1.00 L flask. Highlight to reveal Answer---->            since mole fraction can also be a volume fraction                                                   X1=1L/3L =0.33 X2 2L/3L=0.66 409torr Ptot =XP1 +XP2 340torr=0.33P1 + 0.66 (409 torr) 70torr= 0.33P1 212 torr=P1