Honors Chemistry - Ideal Gas Law WS 3 - 4

Name: ______Period: ______

PV = nRT 1atm = 760 mmHg = 760 torr = 14.7 psi = 101.3 kPa STP = 1atm, 0ºC

1.)How many moles of gas are contained in 890.0 mL at 21.0 °C and 750.0 mm Hg pressure? (0.037)

2.)How many moles of gas would be present in a gas trapped within a 100.0 mL vessel at 25.0 °C at a pressure of 2.50 atmospheres? (0.010)

3.)How many moles of a gas would be present in a gas trapped within a 37.0 liter vessel at 80.00 °C at a pressure of 2.50 atm? (3.19)

4.)If the number of moles of a gas are doubled at the same temperature and pressure, will the volume increase or decrease?

5.)What volume will 1.27 moles of helium gas occupy at STP? (28.4)

6.)At what pressure would 0.150 mole of nitrogen gas at 23.0 °C occupy 8.90 L? (0.41)

7.)What volume would 0.70 moles of NO2 gas occupy at 3.12 atm and 18.0 °C? (5.3)

8.)Find the volume of 2.40 mol of gas whose temperature is 50.0 °C and whose pressure is 2.00 atm. (31.8)

9.)Determine the number of moles of Krypton contained in a 3.25 liter gas tank at 5.80 atm and 25.5 °C. If the gas is Oxygen instead of Krypton, will the answer be the same? Why or why not? (0.77)

10.)At what Celsius temperature will 0.654 moles of neon gas occupy 12.30 liters at 1.95 atmospheres? (174)

11.)Determine the volume in liters occupied by 2.34 grams of carbon dioxide gas at STP. (1.19)

12.)Determine the number of grams of carbon dioxide in a 450.6 mL tank at 1.80 atm and -50.5 °C. Determine the number of grams of oxygen that the same container will contain under the same temperature and pressure. (1.94, 1.41)

13.)20.83 g. of a gas occupies 4.167 L at 79.97 kPa at 30.0 °C. What is its molar mass? (157.8)

14.)At STP 150.0 mL of an unknown gas has a mass of 0.250 gram. Calculate its molar mass. (37.3)

15.)0.226 g of a gas occupies 250.0 mL at STP. What is its molar mass? What gas is it? Hint - calculate molar mass of the gas (20.18, Ne)

Gas Law Problems- Ideal Gas Law Answers

  1. n = PV / RT

n = [ (750.0 mmHg / 760.0 mmHg atm¯1) (0.890 L) ] / (0.08206 L atm mol¯1 K¯1) (294.0 K)

Please note the division of 750 by 760. That is done in order to convert the pressure from mmHg to atm., because the value for R contains atm. as the pressure unit. If we used mmHg, the pressure units would not cancel and we need to have them cancel because we require mol. only to be in the answer.

  1. P = nRT / V

P = [ (1.09 g / 2.02 g mol¯1) (0.08206 L atm mol¯1 K¯1) (293.0 K) ] / 2.00 L

Multiply the answer (which is in atm) by 760.0 mmHg atm¯1 to get mmHg

  1. V = nRT / P

V = [ (3.00 mol) (0.08206 L atm mol¯1 K¯1) (297.0 K) ] / (762.4 mmHg / 760.0 mmHg atm¯1

  1. V = nRT / P

V = [ (20.0 g / 40.0 g mol¯1) (0.08206 L atm mol¯1 K¯1) (273.0 K) ] / (1.00 atm)

  1. n = PV / RT

n = [ (2.50 atm) (0.1000 L) ] / [ (0.08206 L atm mol¯1 K¯1) (298.0 K) ]

  1. n = PV / RT

n = [ (2.50 atm) (37.0 L) ] / [ (0.08206 L atm mol¯1 K¯1) (353.0 K) ]

  1. The volume would increase. In fact, it would double.
  2. V = nRT / P

V = [ (1.27 mol) (0.08206 L atm mol¯1 K¯1) (273.0 K) ] / 1.00 atm or (22.4 L / 1.00 mol) = (x / 1.27 mol)

  1. P = nRT / V

P = [ (0.150 mol) (0.08206 L atm mol¯1 K¯1) (296.0 K) ] / 8.90 L

  1. V = nRT / P

V = [ (32.0 g / 46.0 g mol¯1) (0.08206 L atm mol¯1 K¯1) (291.0 K) ] / 3.12 atm

  1. V = nRT / P

V = [ 2.40 mol) (0.08206 L atm mol¯1 K¯1) (323.0 K) ] / 2.00 atm

  1. n = PV / RT

n = [ (60.0 atm) (7.50 L) ] / [ (0.08206 L atm mol¯1 K¯1) (308.5 K) ]

Divide 35.44 g by the number of moles calculated to get the molecular weight.

  1. n = PV / RT

n = [ (100.0 atm) (5.00 L) ] / [ (0.08206 L atm mol¯1 K¯1) (308.0 K) ]

  1. n = PV / RT

n = [ (5.80 atm) (3.25 L) ] / [ (0.08206 L atm mol¯1 K¯1) (298.5 K) ]

The moles of gas would be the same if the gas was switched to oxygen. Since the temperature and pressure would be the same, the same volume willcontain the same number of molecules of gas, i.e. moles of gas. This is Avogadro's Hypothesis.

  1. n = PV / RT

n = [ (0.4506 atm) (1.80 L) ] / [ (0.08206 L atm mol¯1 K¯1) (222.5 K) ]

This calculates the number of moles of CO2. Multiply the moles by the molecular weight of CO2 to get the grams. Under the same set of conditions, the moles of oxygen would be the same, so multiply the calculated moles by the molecular weight of O2 to get the grams.

  1. V = nRT / P

V = [ (2.34 g / 44.0 g mol¯1) (0.08206 L atm mol¯1 K¯1) (273.0 K) ] / 1.00 atm

  1. n = PV / RT

n = [ (1.00 atm) (56.2 L) ] / [ (0.08206 L atm mol¯1 K¯1) (273.0 K) ] Multiply the moles by the atomic weight of Ar to get the grams.

  1. T = PV / nR T = [ (1.95 atm) (12.30 L) ] / [ (0.654 mol) (0.08206 L atm mol¯1 K¯1) ]

Since one mole of gas occupies 22.4 L at STP, the molecular weight of the gas is 30.6 g mol¯1

  1. 11.2 L at STP is one-half molar volume, so there is 0.5 mol of gas present. Therefore, the molecular weight is 80.0 g mol¯1
  2. This problem, as well as the two just above can be solved with PV = nRT. You would solve for n, the number of moles. Then you would divide the grams given by the mole calculated.

Since it is at STP, we can also use a ratio method (see prob. 111)

(19.2 L / 12.0 g) = (22.4 L / x )

  1. n = PV / RT

n = [ (700.0 mmHg / 760.0 mmHg atm¯1) (48.0 L) ] / [ (0.08206 L atm mol¯1 K¯1) (293.0 K) ]

Then, divide the grams given (96.0) by the moles just calculated above. This will be the molecular weight.

  1. n = PV / RT

n = [ (79.97 kPa / 101.325 kPa atm¯1) (4.167 L) ] / [ (0.08206 L atm mol¯1 K¯1) (303.0 K) ]

Then, divide the grams given (20.83) by the moles just calculated above. This will be the molecular weight.

  1. (3.00 L / 9.50 g) = (22.4 L / x )
  2. (0.250 L / 1.00 g) = (22.4 L / x )
  3. (0.150 L / 0.250 g) = (22.4 L / x )
  4. n = PV / RT

n = [ (0.890 atm) (4.50 L) ] / [ (0.08206 L atm mol¯1 K¯1) (293.5 K) ]

Then, divide the grams given (1.089) by the moles just calculated above. This will be the molecular weight.

  1. n = PV / RT

n = [ (1.00 atm) (0.2500 L) ] / [ (0.08206 L atm mol¯1 K¯1) (273.0 K) ]

Then, divide the grams given (0.190) by the moles just calculated above. This will be the molecular weight.

  1. If 9.006 grams of a gas are enclosed in a 50.00 liter vessel at 273.15 K and 2.000 atmospheres of pressure, what is the molar mass of the gas? What gas is this?

n = PV / RT

n = [ (2.000 atm) (50.00 L) ] / [ (0.08206 L atm mol¯1 K¯1) (273.15 K) ]

Then, divide the grams given (9.006) by the moles just calculated above. This will be the molecular weight.

The answer (2.019 g mol¯1) is approximately that of hydrogen gas, H2

  1. 0.08206 L atm mol¯1 K¯1

The gas constant.

a. 14.0 g / 4.00 g mol¯1 = 3.50 mol

b. 49.0 g / 28.0 g mol¯1 = 1.75 mol

c. 3.50 mol / 5.25 mol

d. 1.75 mol / 5.25 mol

e. Use PV = nRT to determine the total pressure in the container.

P = nRT / V

P = [ (5.25 mol) (0.08206 L atm mol¯1 K¯1) (258.0 K) ] / 50.00 L

The total pressure times the mole fraction of He will give helium's partial pressure.

f. The total pressure times the mole fraction of N2 will give nitrogen's partial pressure. Since it is the only other value, you could subtract helium's answer from the total.

g. Already done to answer part e.

h. V = nRT / P

V = [ (5.25 mol) (0.08206 L atm mol¯1 K¯1) (273.0 K) ] / 1.00 atm

|| Copyright © 1998 by John L. Park