Gases

Gas Laws Boyle’s Law states that the pressure of a fixed mass of gas at a constant temperature is inversely proportional to its volume: P ∝ 1/V P = pressure; V = volume So, PV = constant

Charles’ Law states that the volume of a fixed mass of gas at constant pressure is directly proportional to its temperature (in Kelvin): V ∝ T V = volume; T = temperature (in K) So, V/T = constant (K = oC + 273)

Gay-Lussac’s Law states that the pressure of a fixed mass of gas at constant volume is proportional to the temperature. P ∝ T So, P = contsant T

Gases which obey these laws are called ideal gases Gases which obey these laws are called ideal gases. By combining them, the equation of state for an ideal gas is obtained: P1V1 = P2V2 T1 T2 A gas has a volume V1 at a temperature T1 and pressure P1. If the conditions are changed to a pressure P2 and a temperature T2, the new volume can be calculated from the equation.

We also get PV = constant T for a fixed mass of gas Avagadro’s Law states that equal volumes of gases, measured at the same temperature and pressure, contain equal numbers of molecules. So, the constant is the same for one mole of all gases If we assume 1 mole of gas, the constant is the (ideal or universal) gas constant, R. R = 8.31 J K-1 mol-1 And PV = RT

For n moles of gas, the equation becomes: PV = nRT the ideal gas equation P in Pa (N m-2) V in m3 T in K R in J K-1 mol-1 Room temperature is taken to be 20oC i.e. __________K Pressure is taken to be 1 atm i.e. 101 325 Pa (often rounded to 100 000 Pa)

The ideal gas equation and relative molecular mass. If we know the mass of a gas, as well at its temperature, pressure and volume, we can calculate its Mr.

1. What volume would 2.00 moles of hydrogen occupy at room temperature and pressure? P = V = n = R = T = PV = nRT ∴

1. What volume would 2.00 moles of hydrogen occupy at room temperature and pressure? P = 100 000 Pa V = ? n = 2.00 R = 8.31 J K-1 mol-1 T = (20 + 273) = 293 K PV = nRT ∴

1. What volume would 2.00 moles of hydrogen occupy at room temperature and pressure? P = 100 000 Pa V = ? n = 2.00 R = 8.31 J K-1 mol-1 T = (20 + 273) = 293 K PV = nRT ∴ V = nRT = P

1. What volume would 2.00 moles of hydrogen occupy at room temperature and pressure? P = 100 000 Pa V = ? n = 2.00 R = 8.31 J K-1 mol-1 T = (20 + 273) = 293 K PV = nRT ∴ V = nRT = 2.00 x 8.31 x 293 P 100 000

1. What volume would 2.00 moles of hydrogen occupy at room temperature and pressure? P = 100 000 Pa V = ? n = 2.00 R = 8.31 J K-1 mol-1 T = (20 + 273) = 293 K PV = nRT ∴ V = nRT = 2.00 x 8.31 x 293 P 100 000 V = 0.0487 m3 = 48.7 dm3

2. If 5.00 moles of helium is contained at room temperature in a cylinder with a volume of 1.5 m3, what would be the pressure? P = V = n = R = T = PV = nRT ∴

2. If 5.00 moles of helium is contained at room temperature in a cylinder with a volume of 1.5 m3, what would be the pressure? P = ? V = 1.5 m3 n = 5.00 R = 8.31 J K-1 mol-1 T = (20 + 273) = 293 K PV = nRT ∴

2. If 5.00 moles of helium is contained at room temperature in a cylinder with a volume of 1.5 m3, what would be the pressure? P = ? V = 1.5 m3 n = 5.00 R = 8.31 J K-1 mol-1 T = (20 + 273) = 293 K PV = nRT ∴ P = nRT = V

2. If 5.00 moles of helium is contained at room temperature in a cylinder with a volume of 1.5 m3, what would be the pressure? P = ? V = 1.5 m3 n = 5.00 R = 8.31 J K-1 mol-1 T = (20 + 273) = 293 K PV = nRT ∴ P = nRT = 5.00 x 8.31 x 293 V 1.5

2. If 5.00 moles of helium is contained at room temperature in a cylinder with a volume of 1.5 m3, what would be the pressure? P = ? V = 1.5 m3 n = 5.00 R = 8.31 J K-1 mol-1 T = (20 + 273) = 293 K PV = nRT ∴ P = nRT = 5.00 x 8.31 x 293 V 1.5 P = 8120 Pa

3. At what temperature would 0 3. At what temperature would 0.50 moles of fluorine be if it were contained in a vessel of volume 0.01 m3 at standard pressure? P = V = n = R = T = PV = nRT ∴

3. At what temperature would 0 3. At what temperature would 0.50 moles of fluorine be if it were contained in a vessel of volume 0.01 m3 at standard pressure? P = 100 000 Pa V = 0.01 m3 n = 0.50 R = 8.31 J K-1 mol-1 T = ? PV = nRT ∴

3. At what temperature would 0 3. At what temperature would 0.50 moles of fluorine be if it were contained in a vessel of volume 0.01 m3 at standard pressure? P = 100 000 Pa V = 0.01 m3 n = 0.50 R = 8.31 J K-1 mol-1 T = ? PV = nRT ∴ T = PV = nR

3. At what temperature would 0 3. At what temperature would 0.50 moles of fluorine be if it were contained in a vessel of volume 0.01 m3 at standard pressure? P = 100 000 Pa V = 0.01 m3 n = 0.50 R = 8.31 J K-1 mol-1 T = ? PV = nRT ∴ T = PV = 100 000 x 0.01 nR 0.5 x 8.31

3. At what temperature would 0 3. At what temperature would 0.50 moles of fluorine be if it were contained in a vessel of volume 0.01 m3 at standard pressure? P = 100 000 Pa V = 0.01 m3 n = 0.50 R = 8.31 J K-1 mol-1 T = ? PV = nRT ∴ T = PV = 100 000 x 0.01 nR 0.5 x 8.31 T = 241 K

4. At room temperature and pressure: a) How many moles of gas are there in 20 m3 of air? P = V = n = R = T = PV = nRT ∴

4. At room temperature and pressure: a) How many moles of gas are there in 20 m3 of air? P = 100 000 Pa V = 20 m3 n = ? R = 8.31 J K-1 mol-1 T = (20 + 273) = 293 K PV = nRT ∴

4. At room temperature and pressure: a) How many moles of gas are there in 20 m3 of air? P = 100 000 Pa V = 20 m3 n = ? R = 8.31 J K-1 mol-1 T = (20 + 273) = 293 K PV = nRT ∴ n = PV = RT

4. At room temperature and pressure: a) How many moles of gas are there in 20 m3 of air? P = 100 000 Pa V = 20 m3 n = ? R = 8.31 J K-1 mol-1 T = (20 + 273) = 293 K PV = nRT ∴ n = PV = 100 000 x 20 RT 8.31 x 293

4. At room temperature and pressure: a) How many moles of gas are there in 20 m3 of air? P = 100 000 Pa V = 20 m3 n = ? R = 8.31 J K-1 mol-1 T = (20 + 273) = 293 K PV = nRT ∴ n = PV = 100 000 x 20 RT 8.31 x 293 n = 821 moles

4. At room temperature and pressure: a) How many moles of gas are there in 20 m3 of air? P = 100 000 Pa V = 20 m3 n = ? R = 8.31 J K-1 mol-1 T = (20 + 273) = 293 K PV = nRT ∴ n = PV = 100 000 x 20 RT 8.31 x 293 n = 821 moles

b) How many moles of nitrogen would there be in 100 m3 of air? P = V = n = R = T = PV = nRT ∴

b) How many moles of nitrogen would there be in 100 m3 of air? P = 100 000 Pa V = 100 m3 n = ? R = 8.31 J K-1 mol-1 T = (20 + 273) = 293 K PV = nRT ∴

b) How many moles of nitrogen would there be in 100 m3 of air? P = 100 000 Pa V = 100 m3 n = ? R = 8.31 J K-1 mol-1 T = (20 + 273) = 293 K PV = nRT ∴ n = PV = RT

b) How many moles of nitrogen would there be in 100 m3 of air? P = 100 000 Pa V = 100 m3 n = ? R = 8.31 J K-1 mol-1 T = (20 + 273) = 293 K PV = nRT ∴ n = PV = 100 000 x 100 RT 8.31 x 293 n = 4107 moles of ‘air’

b) How many moles of nitrogen would there be in 100 m3 of air? P = 100 000 Pa V = 100 m3 n = ? R = 8.31 J K-1 mol-1 T = (20 + 273) = 293 K PV = nRT ∴ n = PV = 100 000 x 100 RT 8.31 x 293 n = 4107 moles of ‘air’ So, moles of N2 = 4107 x 79 / 100 = 3245

5. Calculate the relative molecular mass of a gas if 50 5. Calculate the relative molecular mass of a gas if 50.0 g occupies a volume of 27.8 dm3 at 298 K and 1.01 x 105 Pa. P = V = n = R = T = PV = nRT

5. Calculate the relative molecular mass of a gas if 50 5. Calculate the relative molecular mass of a gas if 50.0 g occupies a volume of 27.8 dm3 at 298 K and 1.01 x 105 Pa. P = 1.01 x 105 Pa V = 27.8 dm3 = 0.0278 m3 n = ? R = 8.31 J K-1 mol-1 T = 298K PV = nRT ∴

5. Calculate the relative molecular mass of a gas if 50 5. Calculate the relative molecular mass of a gas if 50.0 g occupies a volume of 27.8 dm3 at 298 K and 1.01 x 105 Pa. P = 1.01 x 105 Pa V = 27.8 dm3 = 0.0278 m3 n = ? R = 8.31 J K-1 mol-1 T = 298K PV = nRT ∴ n = PV = RT

5. Calculate the relative molecular mass of a gas if 50 5. Calculate the relative molecular mass of a gas if 50.0 g occupies a volume of 27.8 dm3 at 298 K and 1.01 x 105 Pa. P = 1.01 x 105 Pa V = 27.8 dm3 = 0.0278 m3 n = ? R = 8.31 J K-1 mol-1 T = 298K PV = nRT ∴ n = PV = 1.01 x 105 x 0.0278 RT 8.31 x 298 n = 1.134

5. Calculate the relative molecular mass of a gas if 50 5. Calculate the relative molecular mass of a gas if 50.0 g occupies a volume of 27.8 dm3 at 298 K and 1.01 x 105 Pa. P = 1.01 x 105 Pa V = 27.8 dm3 = 0.0278 m3 n = ? R = 8.31 J K-1 mol-1 T = 298K PV = nRT ∴ n = PV = 1.01 x 105 x 0.0278 RT 8.31 x 298 n = 1.134 ∴ Mr = mass = n

5. Calculate the relative molecular mass of a gas if 50 5. Calculate the relative molecular mass of a gas if 50.0 g occupies a volume of 27.8 dm3 at 298 K and 1.01 x 105 Pa. P = 1.01 x 105 Pa V = 27.8 dm3 = 0.0278 m3 n = ? R = 8.31 J K-1 mol-1 T = 298K PV = nRT ∴ n = PV = 1.01 x 105 x 0.0278 RT 8.31 x 298 n = 1.134 ∴ Mr = mass = 50.0 = 44.1 g mol-1 n 1.134

6. A camping gas bottle with cylinder dimensions of H170 mm D203 mm holds 1.81 kg of butane. Assuming the butane is still a gas, calculate the pressure in the cylinder at 10 oC. P = V = n = R = T = PV = nRT ∴

6. A camping gas bottle with cylinder dimensions of H170 mm D203 mm holds 1.81 kg of butane. Assuming the butane is still a gas, calculate the pressure in the cylinder at 10 oC. P = ? Pa V = n = R = T = PV = nRT ∴

6. A camping gas bottle with cylinder dimensions of H170 mm D203 mm holds 1.81 kg of butane. Assuming the butane is still a gas, calculate the pressure in the cylinder at 10 oC. P = ? Pa V =  x (203/2)2 x 170 mm3 = n = mass / Mr = R = 8.31 J K-1 mol-1 T = (10 + 273) = 283 K PV = nRT ∴

6. A camping gas bottle with cylinder dimensions of H170 mm D203 mm holds 1.81 kg of butane. Assuming the butane is still a gas, calculate the pressure in the cylinder at 10 oC. P = ? Pa V =  x (203/2)2 x 170 mm3 = 5.5 x 106 mm3 = 5.5 x 10-3 m3 n = mass / Mr = 1810 / 58 = 31.2 moles R = 8.31 J K-1 mol-1 T = (10 + 273) = 283 K PV = nRT ∴ P = nRT = V

6. A camping gas bottle with cylinder dimensions of H170 mm D203 mm holds 1.81 kg of butane. Assuming the butane is still a gas, calculate the pressure in the cylinder at 10 oC. P = ? Pa V =  x (203/2)2 x 170 mm3 = 5.5 x 106 mm3 = 5.5 x 10-3 m3 n = mass / Mr = 1810 / 58 = 31.2 moles R = 8.31 J K-1 mol-1 T = (10 + 273) = 283 K PV = nRT ∴ P = nRT = 31.2 x 8.31 x 283 V 5.5 x 10-3 P = 13340000 Pa = 133 kPa

6. A camping gas bottle with cylinder dimensions of H170 mm D203 mm holds 1.81 kg of butane. Assuming the butane is still a gas, calculate the pressure in the cylinder at 10 oC. P = ? Pa V =  x (203/2)2 x 170 mm3 = 5.5 x 106 mm3 = 5.5 x 10-3 m3 n = mass / Mr = 1810 / 58 = 31.2 moles R = 8.31 J K-1 mol-1 T = (10 + 273) = 283 K PV = nRT ∴ P = nRT = 31.2 x 8.31 x 283 V 5.5 x 10-3 P = 13340000 Pa = 133 kPa

7. A 153 kg sample of ammonia gas, NH3, was compressed at 800 K into a cylinder of volume 3.00 m3. a) Calculate the pressure in the cylinder assuming that the ammonia remained as a gas. P = V = n = R = T = PV = nRT ∴

P = ? Pa V = 3.00m3 n = mass / Mr = 153 000 / 17 = 9000 moles 7. A 153 kg sample of ammonia gas, NH3, was compressed at 800 K into a cylinder of volume 3.00 m3. a) Calculate the pressure in the cylinder assuming that the ammonia remained as a gas. P = ? Pa V = 3.00m3 n = mass / Mr = 153 000 / 17 = 9000 moles R = 8.31 J K-1 mol-1 T = 800 K PV = nRT ∴ P = nRT = V

P = ? Pa V = 3.00m3 n = mass / Mr = 153 000 / 17 = 9000 moles 7. A 153 kg sample of ammonia gas, NH3, was compressed at 800 K into a cylinder of volume 3.00 m3. a) Calculate the pressure in the cylinder assuming that the ammonia remained as a gas. P = ? Pa V = 3.00m3 n = mass / Mr = 153 000 / 17 = 9000 moles R = 8.31 J K-1 mol-1 T = 800 K PV = nRT ∴ P = nRT = 9000 x 8.31 x 800 V 3.00 P = 1.99 x 107 Pa = 199 kPa

b) Calculate the pressure in the cylinder when the temperature is raised to 1000 K. PV = nRT ∴ P = nRT = 9000 x 8.31 x 1000 V 3.00 P = 2.49 x 107 Pa = 249 kPa