Dalton's Law of Partial Pressure and Mole Fractions

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Dalton’s Law states that "The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone."

P_total = P₁ + P₂ + . . . P_n

P_t is the total pressure of a sample which contains a mixture of gases
P₁, P₂, P₃, etc. are the partial pressures (in the same units) of the gases in the mixture

However, there is an unavoidable problem. The gas saturates with water vapor and now the total pressure inside the bottle is the sum of two pressures - the gas itself and the added water vapor.

WE DO NOT WANT THE WATER VAPOR PRESSURE.

So we get rid of it by subtraction.

P_{dry gas} = P_total - P_{water vapor}

This means we must get the water vapor pressure from somewhere.

We get it from a table because the water vapor pressure depends only on the temperature, NOT how big the container is or the pressure of the other gas.

Temp (^oC)	Vapor Pressure (mmHg)	Temp (^oC)	Vapor Pressure (mmHg)
-10	2.15	40	55.3
0	4.58	60	149.4
5	6.54	80	355.1
10	9.21	95	634
11	9.84	96	658
12	10.52	97	682
13	11.23	98	707
14	11.99	99	733
15	12.79	100	760
20	17.54	101	788
25	23.76	110	1074.6
30	31.8	120	1489
37	47.07	200	11659

Regents Questions

Example 1 A sample of hydrogen gas is collected over water at 14.0 ^oC. The pressure of the resultant mixture is 113.0 kPa. What is the pressure that is exerted by the dry hydrogen alone?

P_{dry
gas}=113.0 kPa(760mmHg/101.3 kPa)=847.8mmHg

P_{water
vapor}= From Table=11.99mmHg

P_total=?

P_{dry gas} = P_total - P_{water
vapor}

P_{dry gas} = 847.8mmHg - 11.99mmHg

P_{dry gas} =835.8mmHg

Example 2 A mixture of oxygen, hydrogen and nitrogen gases exerts a total pressure of 278 kPa. If the partial pressures of the oxygen and the hydrogen are 112 kPa and 101 kPa respectively, what would be the partial pressure exerted by the nitrogen.

P_O2=112 kPa

P_N2= ?

P_H2=101 kPa

P_total=278 kPa

P_total = P_O2 + P_N2 + P_H2

278 kPa = 112 kPa + 101 kPa + P_nitrogen

P_nitrogen = 278 kPa - (112 kPa + 101 kPa)

P_nitrogen = 65 kPa

Questions	Highlight to Reveal Answers
3. A mixture of neon and argon gases exerts a total pressure of 2.39 atm. The partial pressure of the neon alone is 1.84 atm, what is the partial pressure of the argon?	P_total = P_Ne + P_Ar 2.39atm=1.4 atm +P_Ar P_Ar =0.55atm
4. A 450 cm³ sample of hydrogen is collect over water at 12^oC. The pressure of the hydrogen and water vapor mixture is 78.5 kPa. What is the partial pressure of the dry hydrogen gas?	P_H2=78.5kPa(760 mmHg/101.3kPa)=589mmHg P_total = P_H2 + P_H2O P_total = 589mmHg + 10.52mmHg = 600.mmHg
5. 888 cm³ of oxygen are collected over water with a temperature of 25 ^oC. The total pressure of the gases is 55.8 kPa. What is the partial pressure of the dry gas?	P_total =55.8kPa(760 mmHg/101.3kPa)=419mmHg P_total = P_H2 + P_H2O 419mmHg= P_O2 + 23.76mmHg P_O2 = 395mmHg

AP 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:

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

where:
$x i$	= mole fraction of any individual gas component in a gas mixture
$P i$	= partial pressure of any individual gas component in a gas mixture
$n i$	= moles of any individual gas component in a gas mixture
$n$	= total moles of the gas mixture
$P$	= 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. P_He = 0.300 x 4.00 atm = 1.20 atm. P_Ar = 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

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