**Problem #1:** A mixture of 40.0 g of oxygen and 40.0 g of helium has a total pressure of 0.900 atm. What is the partial pressure of each gas?

**Solution:**

1) Calculate moles of each gas:

He: 40.0 g / 4.0026 g/mol = 9.9935 mol

O_{2}: 40.0 g / 31.9988 g/mol = 1.25005 mol

2) Calculate mole fraction:

He: 9.9935 mol / 11.24355 mol = 0.88882

3) Calculate partial pressure:

He: 0.900 atm times 0.88882 = 0.79938 atm = 0.800 atm (to three sig figs)

O_{2}: 0.900 minus 0.800 = 0.100 atm

**Problem #2:** If a gas is collected over water, what corrections need to be made when calculating the volume of the dry gas at STP?

**Solution:**

The pressure of the water vapor needs to be removed, by subtraction. This changes the gas pressure down a bit, but leaves the volume and temperature unaffected.The gas with the water vapor is often called "wet," whereas the gas after the water vapor pressure has been removed is called the "dry" gas.

**Problem #3:** Nitrogen is collected over water at 40.0 °C. What is the partial pressure of nitrogen if the total pressure is 99.42 kPa?

**Solution:**

1) The vapor pressure of water at 40.0 °C is looked up and found to be 7.38 kPa.

2) Dalton's Law of Partial Pressure is used:

P_{tot}= P_{N2}+ P_{H2O}99.42 kPa = P

_{N2}+ 7.38 kPaP

_{N2}= 92.04 kPa

**Problem #4:** Three gases (8.00 g of methane, CH_{4}, 18.0 g of ethane, C_{2}H_{6}, and an unknown amount of propane, C_{3}H_{8}) were added to the same 10.0 L container. At 23.0 °C, the total pressure in the container was measured to be 4.43 atm. Calculate the partial pressure of each gas in the container.

**Solution:**

1) Calculate the moles of methane and ethane that are present:

methane ---> 8.00 g / 16.0 g/mol = 0.500 mol

ethane ---> 18.0 g / 30.0 g/mol = 0.600 mol

2) Using PV = nRT, determine the total moles present:

(4.43 atm) (10.0 L) = (n) (0.08206 L atm / mol K) (300. K)n = 1.80 mol

3) Determine the partial pressure of methane and ethane:

methane ---> (4.43 atm) (0.500 mol / 1.80 mol) = 1.23 atm

ethane ---> (4.43 atm) (0.600 mol / 1.80 mol) = 1.48 atm

4) Determine the partial pressure of propane:

4.43 − (1.23 + 1.48) = 1.72 atm

5) Note that there is no need to determine the moles of propane (since we can determine its partial pressure by subtraction), although we could if desired:

1.80 − (0.500 + 0.600) = 0.700 moland thence, to a partial pressure calculation:

(4.43 atm) (0.700 mol / 1.80 mol) = 1.72 atm

**Problem #5:** A gaseous mixture of O_{2} and N_{2} contains 32.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture, if the total pressure is 785.0 mmHg?

**Solution:**

1) Assume 100.0 g of the mixture is present. Using 32.8% means these amounts are present in the mixture:

32.8 g of nitrogen

67.2 g of oxygen

2) Convert the two masses to moles:

nitrogen ---> 32.8 g / 28.014 g/mol = 1.17 mol

oxygen ---> 67.2 g / 16.00 g/mol = 4.20

3) Get the total mount of moles:

1.17 + 4.20 = 5.37 mol

4) Determine the partial pressure of oxygen in the mixture:

(785.0 mmHg) (4.20 mol / 5.37 mol) = 614.0 mmHg

**Problem #6:** A 1.50 L bulb containing He at 155 torr is connected by a valve to a 2.00 L bulb containing CH_{4} at 245 torr. The valve between the two bulbs is opened and the two gases mix.

(a) What is the partial pressure (torr) of He and CH_{4}?

(b) What is the mole fraction of He?

**Solution #1:**

1) Partial pressure of He

P_{1}V_{1}= P_{2}V_{2}(155 torr) (1.50 L) = (x) (3.50 L)

x = 66.4 torr

2) Partial pressure of CH_{4}:

(245 torr) (2.00 L) = (y) (3.50 L)y = 140. torr

3) Total pressure:

66.4 torr + 140. torr = 206.4 torr

4) Mole fraction of He:

66.4/206.4 = 0.322

**Solution #2:**

1) What would the methane pressure be in 1.5 liters?

P_{1}V_{1}= P_{2}V_{2}(245 torr) (2.00 L) = (x) (1.50 L)

x = 326.7 torr

2) Total pressure in 1.50 L:

155 torr + 326.7 torr = 481.7 torr

3) Mole fraction of helium:

155 torr / 481.7 torr = 0.322 (to three sig figs)

4) Allow the 1.50 L to expand to 3.50 L

P_{1}V_{1}= P_{2}V_{2}(481.7 torr) (1.50 L) = (y) (3.50 L)

y = 206.4 torr

5) Determine partial pressures:

He ---> (206.4 torr) (0.322) = 66.5 torr

Ne ---> 206.4 − 66.5 = 139.9 torr

**Problem #7:** A mixture of He and O_{2} gases used by deep-sea divers. If the pressure of the gas a diver inhales is 8.0 atm, what percent of the mixture should be O_{2}, if the partial pressure of O_{2} is to be the same as what the diver would ordinarily breathe at sea level?

**Solution:**

O_{2}is 21% of the atmosphere, so its partial pressure is 0.21 atmThe He/O

_{2}mix is at 8 atm, so:(0.21 atm / 8 atm) times 100 = 2.625%

If our He/O

_{2}mix contains 2.625% O_{2}, the diver will be breathing O_{2}at 0.21 atm in the total 8 atm being breathed.

**Problem #8:** A diver breathes a helium-oxygen mixture with an oxygen mole fraction of 0.050. What must the total pressure be for the partial pressure of oxygen to be 0.21 atm?

**Solution:**

Since the mole fraction of the O_{2}is 0.050, it contributes 0.050 (or, if you will, 5.0%) of the total pressure.Thus total pressure is 0.21/0.050 = 4.2 atm

**Problem #9:** A sample of 1.43 g of helium and an unweighed quantity of O_{2} are mixed in a flask at room temperature. The partial pressure of helium in the flask is 42.5 torr, and the partial pressure of oxygen is 158 torr. What mass of O_{2} is in the sample?

**Solution:**

1) Mole fraction of oxygen:

158 / 200.5 = 0.788

2) Mole fraction of He:

1.000 minus 0.788 = 0.212

3) Moles of helium present:

1.43 g / 4.0026 g/mol = 0.35727 mol (I will keep a few guard digits)0.35727 mol is to 0.212 as x is 0.788

x = 1.32796 mol of O

_{2}1.32796 mol times 31.9988 g/mol = 42.5 g (to three sig figs)

**Problem #10:** The mass percent of a three component gas sample is 22.70% O_{2}, 21.00% C_{2}H_{2}F_{4} and 56.30% C_{6}H_{6}. Calculate the partial pressure (atm) of C_{2}H_{2}F_{4} if the total pressure of the sample is 1444 torr.

**Solution:**

1) Asume 100 g of the sample is present. This allows the percentages to be converted to mass. Calculate moles of each substance:

22.70 g / 16.00 g/mol = 1.41875 mol

21.00 g / 102.03 g/mol = 0.205822 mol

56.30 g / 78.1134 g/mol = 0.720747 mol

2) Determine mole fraction of C_{2}H_{2}F_{4}:

0.205822 mol / (1.41875 mol + 0.205822 mol + 0.720747 mol) = 0.0877586

3) Determine partial pressure of C_{2}H_{2}F_{4}:

1444 torr times 0.0877586 = 126.7234 torr

4) Convert to atm:

126.7234 torr / 760.0 torr/atm = 0.1667 atm (to four sig figs)

**Problem #11:** A student has stored 100.0 mL of neon gas over water on a day when the temperature is 27.0 °C. The barometer in the room reads 99.10 kPa. What is the pressure of the neon gas?

**Solution:**

1) The key phrase is "over water." This means that the 100.0 mL of neon gas also contains water vapor. That means this problem uses Dalton's Law:

P_{tot}= P_{Ne}+ P_{H2O}Our unknown will be P

_{Ne}.

2) P_{tot} = 99.10 kPa

The pressure of the gas collected over the water is the same as the pressure in the room.

3) P_{H2O}
= 3.57 kPa

The values for the vapor pressure of water at various temperatures have been tabulated. I used this table of the values.

4) We are now ready to solve:

P_{tot}= P_{Ne}+ P_{H2O}99.10 kPa = P

_{Ne}+ 3.57 kPaP

_{Ne}= 95.53 kPa

**Bonus Problem:** A closed vessel contains 80.0 g of O_{2}, 14.0 g of N_{2}, and 44.0 g of CO_{2}.The total pressure in the vessel is 200. kPa. What is the partial pressure (in kPa) exerted by the oxygen in the mixture?

**Solution:**

A comment: the question asks for the partial pressure of the oxygen, which should be taken to mean O_{2}. The oxygen present in the CO_{2} is NOT part of the solution to the problem.

You might think this a weird thing to comment on, but I was asked about it in class a number of years ago.

1) Determine moles of each gas:

oxygen ---> 80.0 g / 32.0 g/mol = 2.50 mol

nitrogen ---> 14.0 g / 28.0 g/mol = 0.500 mol

carbon dioxide ---> 44.0 g / 44.0 g/mol = 1.00 moltotal moles of gas ---> 4.00 mol

2) Determine the mole fraction of the oxygen:

2.50 mol / 4.00 mol = 0.625

3) Determine the partial pressure of the oxygen gas:

(200. kPa) (0.625) = 125 kPa