and other partial pressure problems

Ten Examples

This law was discovered by John Dalton in 1801.

For any pure gas (let's use helium), PV = nRT holds true. Therefore, P is directly proportional to n if V and T remain constant. As n goes up, so would P. Or the reverse.

Suppose you were to double the moles of helium gas present. What would happen?

Answer: the gas pressure doubles.

However, suppose the new quantity of gas added was a DIFFERENT gas. Suppose that, instead of helium, you added neon.

What would happen to the pressure?

Answer: the pressure doubles, same as before.

Dalton's Law immediately follows from this example since each gas is causing 50% of the pressure. Summing their two pressures gives the total pressure.

Written as an equation, it looks like this:

P_{He}+ P_{Ne}= P_{total}

Dalton's Law of Partial Pressures: each gas in a mixture creates pressure as if the other gases were not present. The total pressure is the sum of the pressures created by the gases in the mixture.

P_{total}= P_{1}+ P_{2}+ P_{3}+ .... + P_{n}

Where n is the total number of gases in the mixture.

The only necessity is that the two gases do not interact in some chemical fashion, such as reacting with each other.

The pressure each gas exerts in mixture is called its partial pressure.

The most common use of Dalton's Law seen in high school is with water vapor.

A common method of collecting gas during an experiment is by trapping it "over water." An inverted bottle filled with water sits in a water bath. A tube from the reaction vessel conducts the gas into the bottle where it bubbles to the top and displaces water, which runs out the mouth of the bottle into the water bath.

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.

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. Usually the textbook will have an abbreviated table with more complete tables in reference manuals like "The Handbook of Chemistry and Physics."

Here is an example: 0.750 L of a gas is collected over water at 23.0°C with a total pressure of 99.75 kPa. What is the pressure of the dry gas? Look up the vapor pressure data here.

For the solution, we look up the vapor pressure and find it to be 2.8104.

We then perform a subtraction:

99.75 kPa − 2.8104 kPa = 96.9396 kPa

We then round off to four significant figures for the final answer:

96.94 kPa

Another common concept that crops up in a Dalton's Law context is mole fraction.

Suppose you had equal moles of two different gases in a mixture. Then the mole fraction for each would be 0.50.

The mole fraction for each gas is simply the moles of that gas divided by the total moles in the mixture.

Seems simple enough. How does it relate to Dalton's Law?

Answer: the mole fraction also gives the fraction of the total pressue each gas contributes. So if the mole fraction for a gas was 0.50, then it would contribute 50% of the total pressure. If the mole fraction of a gas was 0.15, then its partial pressure would be 0.15 times the total pressure.

The reverse is also true. If you divided the partial pressure of a gas by the total pressure, you would get the mole fraction for that gas. (I hope you know enough by now that the two pressures would have to be in the same units!)

By the way, mole fractions are unitless numbers. The mole (or pressure) units cancel out.

**Example #1:** 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?

**Solution:**

1) Using symbols, I write Dalton's Law of Partial Pressures:

P_{tot}= P_{O2}+ P_{CO2}+ P_{He}

2) The sum of all the partial pressures in the container gives us the total pressure:

P_{tot}= 2.00 atm + 3.00 atm + 4.00 atm = 9.00 atm

**Example #2:** The partial pressure of F_{2} in a mixture of gases where the total pressure is 1.00 atm is 300. torr. What is the mole fraction of F_{2}?

**Solution:**

Let us use 760. torr for 1.00 atm300. torr / 760. torr = 0.395

**Example #3:** 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?

**Solution:**

P_{tot}= P_{O2}+ P_{H2O}96.00 kPa = P

_{O2}+ 12.344 kPaP

_{O2}= 83.66 kPa

**Example #4:** 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?

**Solution:**

1) First, a statement of Dalton's Law:

P_{tot}= P_{N2}+ P_{H2O}

2) Next is a little bit of work on P_{H2O}:

We have to look up the vapor pressure of water at 40.0 °C. We do so and find it to be 7.3814 kPa. For the purposes of this problem, we need the value in mmHg:(7.3814 kPa) (760.0 mmHg / 101.325 kPa) = 55.4 mmHg

By the way, I consulted a list that gave the values in kPa. There are many vapor pressure of water data sets available on the Internet and many of them include mmHg values. Here is an example of one.

3) Determine the partial pressure of the nitrogen:

760.0 atm = P_{N2}+ 55.4 mmHgP

_{N2}= 704.6 mmHg

4) Note that the problem says 'atmospheric pressure.' However, is the total pressure inside the container truly equal to the pressure outside the container? We must assume yes, because assuming no means the problem cannot be solved. And the intent of the writer is to give a problem that can be solved, not one that cannot be solved.

**Example #5:** A mixture of 2 moles of H_{2}, 3 moles of NH_{3}, 4 moles of CO_{2}, and 5 moles of N_{2} exert a total pressure of 800. torr. What is the partial pressure of each gas?

**Solution:**

1) Determine the total number of moles of gas:

2 + 3 + 4 + 5 = 14 mol

2) Determine the partial pressure of each gas:

H_{2}---> (800. torr) (2/14) = 114 torr

NH_{3}---> (800. torr) (3/14) = 171 torr

CO_{2}---> (800. torr) (4/14) = 228 torr

N_{2}---> (800. torr) (5/14) = 286 torr

3) Comment:

Note that 114 + 171 + 228 + 286 equals 799, so we have a bit of error due to rounding. The usual solution is to determine the last value by subtraction rather than by direct calculation.

**Example #6:** If you place 3.00 mol of N_{2} and 4.00 mol of O_{2} in a 35.0 L container at 25.0 °C, what will the pressure of the resulting mixture of gases be?

**Solution:**

1) The total pressure in the container is the sum of the partial pressures of the gases (Dalton's Law):

3.00 mol + 4.00 mol = 7.00 mol

2) The Ideal Gas Law is used to calculate the total pressure:

PV = nRT(P) (35.0 L) = (7.00 mol) (0.08206 L atm / mol K) (273 K)

P = 4.48 atm

**Example #7:** A gas mixture contains hydrogn, helium, neon and argon. The total pressure of the mixture is 93.6 kPa. The partial pressures of helium, neon and argon are 15.4 kPa, 25.7 kPa, and 35.6 kPa, respectively. What is the pressure extended by the hydrogen?

**Solution:**

Use Dalton's Law of Partial Pressures:93.6 − (15.4 + 25.7 + 35.6) = 16.9 kPa

**Example #8:** A mixture of 14.0 grams of hydrogen, 84.0 grams of nitrogen, and 2.00 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?

**Solution:**

1) Determine the moles of each gas:

nitrogen ---> 84.0 g / 28.014 g/mol = 2.9985 mol

hydrogen ---> 14.0 g / 1.008 g/mol = 13.8889 mol

oxygen ---> 2.00 mol

2) Add up the moles for the total number of moles:

2.9985 mol + 13.8889 mol + 2.00 mol = 18.8874 mol

3) Divide each mole amount by the total moles to obtain the mole fraction for each gas:

nitrogen ---> 2.9985 mol / 18.8874 mol = 0.15876

hydrogen ---> 13.8889 mol / 18.8874 mol = 0.735353

oxygen ---> 2.00 mol / 18.8874 mol = 0.10589Notice this:

0.15876 + 0.735353 + 0.10589 = 1.000003

It should, of course, equal 1. The fact that it is not is due, of course, to rounding. In many case, the value is slightly below 1, say something like 0.999997.

The manner for dealing with this is to use subtraction to obtain the final value, as opposed to doing the division. I'll use subtraction in the next example.

Note: I could have also just rounded off the hydrogen value to 0.73535.

4) We will use the oxygen information in two ratio and proportions to determine the partial pressures of nitrogen and hydrogen.

Nitrogen:

0.10589 0.15876 ––––––––––– = ––––––– 78.00 mmHg x x = 116.94 mmHg

Hydrogen:

0.10589 0.735353 ––––––––––– = –––––––– 78.00 mmHg x x = 541.67 mmHg

5) The total pressure is arrived at by adding the partial pressures:

541.67 + 116.94 + 78.00 = 736.61 mmHg

**Example #9:** A mixture of gases contains 2.14 g of N_{2}, 5.85 g of H_{2}, and 4.18 g of NH_{3}. If the total pressure of the mixture is 4.58 atm, what is the partial pressure of each component?

**Solution:**

1) Determine the moles of each gas:

nitrogen ---> 2.14 g / 28.014 g/mol = 0.07639 mol

hydrogen ---> 5.85 g / 1.008 g/mol = 5.80357 mol

ammonia ---> 4.18 g / 17.031 g/mol = 0.245435 mol

2) Add up the moles for the total number of moles:

0.07639 mol + 5.80357 mol + 0.245435 mol = 6.125435 mol

3) Divide each mole amount by the total moles:

nitrogen ---> 0.07639 mol / 6.125435 mol = 0.01248

hydrogen ---> 5.80357 mol / 6.125435 mol = 0.94745

ammonia ---> 0.245435 mol / 6.125435 mol = 0.04007This determines the mole fraction of each gas. Also, I actually obtained the nitrogen mole fraction by subtraction. Like this:

1 - [0.94745 + 0.04007] = 0.01248

4) Multiply the mole fraction of each gas times the total pressure. Each answer is the partial pressure of the gas.

nitrogen ---> 4.58 atm - [4.34 + 0.184] = 0.056 atm <--- note 'by subtraction' technique

hydrogen ---> (4.58 atm) (0.94745) = 4.34 atm

ammonia ---> (4.58 atm) (0.04007) = 0.184 atm

**Example #10:** 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?

**Solution:**

Consider only the helium. The argon needs to be double the moles of He in order to have double the mole fraction. This is because the mole fraction of each gas is proportional to the partial pressure of the gas.Therefore, 4.00 moles of argon are required. 4.00 mol times 40.0 g/mol yields 160. g for the answer.

Notice the mole fractions after the addition of 4.00 mol:

N

_{2}---> 2/8 = 0.25

He ---> 2/8 = 0.25

Ar ---> 4/8 = 0.50

**Bonus Example:** 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.

**Solution:**

The temperature and pressure remain constant, therefore Avogadro's Hypothesis applies:

two samples of gas of equal volume, at the same temperature and pressure, contain the same number of molecules.

What this means is that helium, with 30.0% of the volume, also has 30% of the molecules. This means that helium's mole fraction is 0.300. Consequently:

(4.00 atm) (0.300) = 1.20 atmis the partial pressure of helium.

The partial pressure of argon is obtained by subtraction:

4.00 atm − 1.20 atm = 2.80 atm