Leaving Certificate Chemistry The Gas Laws Leaving Certificate Chemistry

Boyle’s Law Robert Boyle – Born in Lismore, Co. Waterford to the Earl of Cork Referred to as the founder of modern chemistry 1662 – He studied the relationship between the pressure exerted by a gas and the volume (space) it occupies. Robert Boyle

Boyle’s Law Boyle’s Law: At a constant temperature, the pressure exerted by a gas is inversely proportional to the volume occupied by that gas. Volume (V) Pressure (P)

Charles’s Law Jacques Charles – French inventor, scientist, mathematician, and balloonist He launched the world's first (unmanned) hydrogen-filled balloon in August 1783 He was interested in the relationship between the temperature of a gas and the volume (space) it occupies. Jacques Charles His law became known as Charles’s Law

Charles’s Law Charles’s Law: 2002 Q. 4 (i) Charles’s Law: At a constant pressure, the temperature of a gas is proportional to the volume occupied by that gas. 2004 Q. 4 (d) Volume (V) Temperature (T)

Avogadro’s Law Amedeo Avogadro – Italian Scientist In tribute to him, the number of elementary entities (atoms, molecules, ions or other particles) in 1 mole of a substance, 6.02214179 ×1023, is known as the Avogadro constant. Avogadro’s Law: Equal volumes of all gases, at the same temperature and pressure, contain the same number of particles, or molecules.

Gay-Lussac’s Law Joseph Gay-Lussac – French was a French chemist and physicist Jacques Charles

N2(g) + 3H2(g) -----> 2NH3(g) Gay-Lussac’s Law: N2(g) + 3H2(g) -----> 2NH3(g) 1 litres 3 litres 2 litre If all the reactants and products in a reaction are gases then the ratio of the volumes of the gases can be expressed as simple whole numbers The volume of gases taking part in a chemical reaction show simple whole number ratios to one another when those volumes are measured at the same temperature and pressure.

2H2(g) + O2(g) -----> 2H2O(g) What is the ratio of the volumes of gases in the reaction? If all the reactants and products in a reaction are gases then the ratio of the volumes of the gases can be expressed as simple whole numbers

Converting units Celsuis K ( + 273 ) Kelvin Celsuis (- 273 )

Changing units of measurement Convert the following to the Kelvin: -10oC -272oC 25oC 100oC My notes = normal pressure Nm -2 and this is equal to 1Pa Normal pressure = 100000 Pa r 105 Pa or 100kPa Standard temperature and pressure = 273 K and 1 x 10 5Pa

P1 V1 P2 V2 = T1 T2 The Combined Gas Law The combined gas law is a gas law which combines Charles's law, Boyle's law, and Gay-Lussac's law P1 V1 P2 V2 = T1 T2 Temperature is measured in degrees Kelvin: Add 273 to convert degrees Celsius to degrees Kelvin P1, V1 and T1 are initial pressure, volume and temperature of the gas 20 oC = 20 + 273 = 293 K 100 oC = 100 + 273 = 373 K P2, V2 and T2 are a later pressure, volume and temperature of the gas —50 oC = —50 + 273 = 223 K

P1 V1 T1 = P2 V2 T2 The Combined Gas Law The combined gas law is a gas law which combines Charles's law, Boyle's law, and Gay-Lussac's law P1 V1 T1 = P2 V2 T2 20 oC = 20 + 273 293 100 oC = 100 + 273 373 —50 oC = + 273 223 13

(Standard temperature = 0 oC, standard pressure = 101,325 Pa) If a definite mass of gas occupies 500 cm3 at a pressure of 101,000 Pa and a temperature of 27 oC, what is its volume in cm3 at s.t.p.? (Standard temperature = 0 oC, standard pressure = 101,325 Pa) P1 V1 P2 V2 = T1 T2 101,000 × 500 101,325 × V2 = 300 273 101,000 × 500 × 273 453.5 cm3 = V2 300 × 101,325

P1 V1 P2 V2 Q.200 = T1 T2 P1 = 101000Pa P2 = 101325 Pa V1 = 500cm3 V2 = ? cm3 T1 = 27+ 273= 300K T2 = 0 + 273 = 273K (101000)(500) = (101,325)(V2) 300 273 (101000)(500) (273) = V2 (101,325)( 300) = V2

= Q.201 P1 = 975kPa P2 = 100kPa V1 = 30cm3 V2 = ? cm3 T1 T2 P1 = 975kPa P2 = 100kPa V1 = 30cm3 V2 = ? cm3 T1 = 25+ 273= 298K T2 = 0 + 273 = 273K (975)(30) = (100)(V2) 298 273 (975)(30) (273) = V2 (100)( 298) 267.9614 = V2 Answer =267.9614cm3

P1 V1 P2 V2 Q.202 = T1 T2 P1 = 101kPa P2 = 108 kPa V1 = 1.2m3 V2 = ? m3 T1 = 5+ 273= 278K T2 = 30+ 273 = 303K (101)(1.2) = (108)(V2) 278 303 (101)(1.2) (303) = V2 (108)( 278) 1.2231= V2 Answer = 1.2231 m3

P1 V1 P2 V2 Q.203 = T1 T2 P1 = 50,662.5 Pa P2 = 50,662.5 x2= 101325Pa V1 = 283cm3 V2 = ? cm3 T1 = 25+ 273= 298K T2 =278 x 4 = 1192 K (59,662.5)(298) = (101,325)(V2) 278 1112 (59662.5)(283) (1192) = V2 (101,325)( 298) 566 = V2 Answer = 566cm3

The kinetic theory of gases The kinetic theory of gases attempts to explain properties of gases Assumptions of the kinetic theory: 2009 Q. 10 (a) Gases are made up of particles whose diameters are negligible There are no attractive or repulsive forces between gas particles. Particles are in constant random motion colliding with each other. A perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision. An inelastic collision is one in which part of the kinetic energy is changed to some other form of energy in the collision. Any macroscopic collision between objects will convert some of the kinetic energy into internal energy and other forms of energy, so no large scale impacts are perfectly elastic The average kinetic energy of the particles is proportional to the Kelvin temperature. Gas particle no attraction or repulsion forces diameters are negligible All collisions are perfectly elastic.

Deviations from the ideal gas laws Two of the characteristics of ideal gases: The gas molecules themselves occupy no appreciable volume ( diameters are almost negligible in size) The gas molecules have no attraction or repulsion for each other Real gas molecules, however, do have an appreciable volume and do attract one another – especially under conditions of low temperature and high pressure So no gas is completely ideal, and some are less ideal than others. At high pressures, and low volumes, the intermolecular distances can become quite short, and attractive forces between molecules becomes significant Neighboring molecules exert an attractive force, which will minimize the interaction of molecules with the container walls. And the apparent pressure will be less than ideal (PV/RT will thus be less than ideal). As pressures increase, and volume decreases, the volume of the gas molecules becomes significant in relationship to the container volume In an extreme example, the volume can decrease below the molecular volume, thus PV/RT will be higher than ideal (V is higher) At high temperatures, the kinetic energy of the molecules can overcome the attractive influence and the gasses behave more ideal At higher pressures, and lower volumes, the volume of the molecules influences PV/RT and its value, again, is higher than ideal

Which will deviate most from real gas behaviour? Consider the following gases – H2, HF, NH3 and F2 HF has hydrogen bonding between molecules - strong attractive forces NH3 – has hydrogen bonding between molecules (but electronegativity difference between N and H is less so attractive forces between molecules will be weaker than in 1) F2 – has Van Der Walls forces attracting molecules – very weak intermolecular forces H2 – Van Der Waals forces are even less here(less electrons present )– weakest attractive forces between molecules Least like an ideal gas (deviates the most) Most like an ideal gas

Ideal gases Approximations about the behaviour of gases can be made by using the ideal gas formula: PV = nRT Temperature ( K) Pressure (Pa) Gas constant R = 8.31JK-1mol-1 Volume (m3) Number of moles of the gas

Mandatory experiment Determination of the relative molecular mass of a volatile liquid using the ideal gas equation

Determination of the RMM of a volatile liquid A known amount of a volatile liquid is vaporised at a known temperature, pressure and in a container of known volume PV = nRT By doing this experiment we can find n ( number of moles of the gas) and then use this to find the relative molecular mass of the gas

Converting units Celsuis K ( + 273 ) Kelvin Celsuis (- 273 ) Pa (÷ 1000 ) kPa Pa (x 1000 ) cm3 m3 ( x 10 -6 ) m3 cm3 ( x 10 6 )

Changing units of measurement Convert the following to the Kelvin: -10oC -272oC 25oC 100oC Convert the following to Pascals: 100kPa 1 x 10 3 kPa 2 x 10 – 6 kPa 1.01 x 105 KPa Convert the following to cm3: 1000m3 1 m3 2L 10 x 10 4cm3 My notes = normal pressure Nm -2 and this is equal to 1Pa Normal pressure = 100000 Pa r 105 Pa or 100kPa Standard temperature and pressure = 273 K and 1 x 10 5Pa

Calculations using the ideal gas law How many moles of nitrogen gas are present in 200cm3 of the gas at 35oC and a pressure of 200kPa? (The gas constant is 8.31JK-1 mol-1 ) PV = nRT P = Pressure = 200kPa = 200,000Pa V = Volume = 200cm3 = .0002m3 n = number of moles = ? R = gas constant = 8.31JK-1 mol-1 T = temperature = 35oC = 308K Calculations using the ideal gas law

Calculations using the ideal gas law PV = nRT (200,000)(0 .000002) = n (8.31) (308) (200,000)(0 .000002) = n (8.31) (308) 0.4 = n 2559.48 0.000156281 = n ( the number of moles) Calculations using the ideal gas law

Calculations using the ideal gas law Q209. How many moles of nitrogen gas are present in 100cm3 of the gas at 30oC and a pressure of 200kPa? (The gas constant is 8.31JK-1 mol-1 ) PV = nRT P = Pressure = 200kPa = 200,000Pa V = Volume = 100cm3 = .0001m3 n = number of moles = ? R = gas constant = 8.31JK-1 mol-1 T = temperature = 30oC = 303K Calculations using the ideal gas law

Calculations using the ideal gas law PV = nRT (200,000)(0 .0001) = n (8.31) (303) (200,000)(0 .0001) = n (8.31) (303) 20 = n 2517.93 0.0079 = n ( the number of moles) Calculations using the ideal gas law

Calculations using the ideal gas law Q210. 1.236g of a volatile liquid gave 512cm3 of vapour at 20oC and a pressure of 101325Pa. Calculate the relative molecular mass (The gas constant is 8.31JK-1 mol-1 ) PV = nRT P = Pressure = 101.325kPa = 101325Pa V = Volume = 512cm3 = .000512m3 n = number of moles = ? R = gas constant = 8.31JK-1 mol-1 T = temperature = 20oC = 293K Calculations using the ideal gas law

Calculations using the ideal gas law PV = nRT (101325)(0 .000512) = n (8.31) (293) (101325)(0 .000512) = n (8.31) (293) 51.8784 = n 2424.83 0.0213 = n ( the number of moles) Calculations using the ideal gas law

Calculations using the ideal gas law 261e) Calculate the relative molecular mass of the volatile liquid Moles of X Mass of X in g X Relative Molecular Mass ÷ Relative Molecular Mass 0.0213 moles x RMM = 1.236g RMM = 1.236/ 0.0213 RMM = 58.0282 RMM = 58 Calculations using the ideal gas law

Calculations using the ideal gas law Q211. What is the relative molecular mass to the nearest whole number if 1.00g occupies a volume of 200cm3 of vapour at 25oC and a pressure of 101.3kPa. (The gas constant is 8.31JK-1 mol-1 ) PV = nRT P = Pressure = 101.3kPa = 101300Pa V = Volume = 200cm3 = .0002m3 n = number of moles = ? R = gas constant = 8.31JK-1 mol-1 T = temperature = 25oC = 298K Calculations using the ideal gas law

Calculations using the ideal gas law PV = nRT (101300)(0 .0002) = n (8.31) (298) (101300)(0 .0002) = n (8.31) (298) 20.26 = n 2476.38 0.0082 = n ( the number of moles) Calculations using the ideal gas law

Calculations using the ideal gas law 261e) Calculate the relative molecular mass of the volatile liquid Moles of X Mass of X in g X Relative Molecular Mass ÷ Relative Molecular Mass 0.0082 moles x RMM = 1.0 g RMM = 1.0/ 0.0082 RMM = 121.9512 Calculations using the ideal gas law

Calculations using the ideal gas law Q212. 1.1g of a volatile liquid occupied 3.15 x 10 -4 m3 at 300K and a pressure of 105Pa. Calculate the relative molecular mass (The gas constant is 8.31JK-1 mol-1 ) PV = nRT P = Pressure = 105 Pa V = Volume = 3.15 x 10 -4 m3 n = number of moles = ? R = gas constant = 8.31JK-1 mol-1 T = temperature = 300K Calculations using the ideal gas law

Calculations using the ideal gas law PV = nRT (100000)(3.15 x 10 -4 m3 ) = n (8.31) (300) (100000)(0 .000315) = n (8.31) (300) 31.5 = n 2493 0.0126 = n ( the number of moles) Calculations using the ideal gas law

Calculations using the ideal gas law Moles of X Mass of X in g X Relative Molecular Mass ÷ Relative Molecular Mass 0.0126moles x RMM = 1.1 g RMM = 1.1/ 0.0126 RMM = 87.3016 Calculations using the ideal gas law

Calculations using the ideal gas law Q213. A small quantity of a volatile organic solvent propanone evaporates at room temperature and pressure. Use the ideal gas equation to calculate the volume, in litres of propanone vapour formed when 0.29g of liquid propanone evaporated taking temperature of 20oC and a pressure of 101kPa. (The gas constant is 8.31JK-1 mol-1 ) PV = nRT P = Pressure = 101kPa = 101000Pa V = Volume = ? n = number of moles = ? ( but mass is known) R = gas constant = 8.31JK-1 mol-1 T = temperature = 293 K Calculations using the ideal gas law

Calculations using the ideal gas law Moles of X Mass of X in g X Relative Molecular Mass ÷ Relative Molecular Mass moles x 58 = 0.29 g Moles = 0.29 / 58 Moles = 0.005 Calculations using the ideal gas law

Calculations using the ideal gas law PV = nRT (101000)V = (0.005)(8.31)(293) V = (0.005)(8.31)(293) 101000 V = 0.0001m3 Calculations using the ideal gas law

To go from m3 to litres multiply by 103 Volume has to be in litres... 0.0001m3 x 10 3 = 0.1 L

Calculations using the ideal gas law Q214. Assuming ideal behaviour how many moles of carbon dioxide are present in 720cm3 at 10oC and a pressure of 105 Pa. (The gas constant is 8.31JK-1 mol-1 ) PV = nRT P = Pressure = 100000Pa V = Volume = 720cm3 = .00072m3 n = number of moles = ? R = gas constant = 8.31JK-1 mol-1 T = temperature = 10oC = 283K Calculations using the ideal gas law

Calculations using the ideal gas law PV = nRT (100000)(0 .00072) = n (8.31) (283) (100000)(0 .00072) = n (8.31) (283) 0.0306 = n ( the number of moles) Calculations using the ideal gas law

Calculations using the ideal gas law Q215. Carbon dioxide is stored under pressure in liquid form in a fire extinguisher. 2.5 Kilograms of carbon dioxide are released into the air as a gas on teh discharge of the fire extinguoisher. What volume does this gas occupy at a temperature of 288K and a pressure of 100000Pa. (The gas constant is 8.31JK-1 mol-1 ) PV = nRT P = Pressure = 100000Pa V = Volume = ? n = number of moles = ? ( but mass is known) R = gas constant = 8.31JK-1 mol-1 T = temperature = 288K Calculations using the ideal gas law

Calculations using the ideal gas law Moles of X Mass of X in g X Relative Molecular Mass ÷ Relative Molecular Mass moles x 44 = 2500 g Moles = 2500 / 44 Moles = 56.8182 Calculations using the ideal gas law

Calculations using the ideal gas law PV = nRT (100000)V = (56.8182)(8.31)(288) V = (0.005)(8.31)(293) 101000 V = 1.3598m3 Calculations using the ideal gas law

Check your learning Describe some assumptions of the Kinetic gas theory What is an ideal gas? Give two reasons why real gases deviate from ideal behaviour Under what conditions do gases depart from ideal behaviour? What is the ideal gas law?

Calculations using the ideal gas law Q216(d) In an experiment to measure the relative molecular mass of a volatile liquid was vaporised at 97oC. The volume occupied was found to be 95cm3 and a pressure of 105Pa. Calculate the number of moles and relative molecular mass (The gas constant is 8.31JK-1 mol-1 ) PV = nRT P = Pressure = 1 x 105Pa V = Volume = 95cm3 = 95 x 10 -6 m3 n = number of moles = ? R = gas constant = 8.31JK-1 mol-1 T = temperature = 370K Calculations using the ideal gas law

Calculations using the ideal gas law PV = nRT (100000)(0.000095 ) = n (8.31) (370) (100000)(0 .000095) = n (8.31) (370) 0.0031 = n ( the number of moles) Calculations using the ideal gas law

Calculations using the ideal gas law 261e) Calculate the relative molecular mass of the volatile liquid Moles of X Mass of X in g X Relative Molecular Mass ÷ Relative Molecular Mass 0.0031 moles x RMM = 0.275 g RMM = 0.275/ 0.0031 RMM = 88.7097 To the nearest whole number – RMM = 89 Calculations using the ideal gas law

Calculations using the ideal gas law Q217. In order to determine the relative molecular mass of a volatile liquid a conical flask was filled with vapour of the liquid. The results of the experiment were summarised as follows ( on sheet). Calculate the relative molecular mass. ( To the neasreast whole number) (The gas constant is 8.31JK-1 mol-1 ) PV = nRT P = Pressure = 98000Pa V = Volume = 328cm3 = 0.000328m3 n = number of moles = ? R = gas constant = 8.31JK-1 mol-1 T = temperature = 373K Calculations using the ideal gas law

Calculations using the ideal gas law PV = nRT (98000)(0.000328 ) = n (8.31) (373) (98000)(0 .000328) = n (8.31) (373) 0.0104 = n ( the number of moles) Calculations using the ideal gas law

Calculations using the ideal gas law 261e) Calculate the relative molecular mass of the volatile liquid Moles of X Mass of X in g X Relative Molecular Mass ÷ Relative Molecular Mass 0.0104 moles x RMM = 1.43 g RMM = 1.43/ 0.0104 RMM = 137.5 To the nearest whole number – RMM = 138 Calculations using the ideal gas law

Calculations using the ideal gas law Q218. In order to determine the relative molecular mass of a volatile liquid a conical flask was filled with vapour of the liquid. The results of the experiment were summarised as follows ( on sheet). Calculate the relative molecular mass. ( To the nearest whole number) (The gas constant is 8.31JK-1 mol-1 ) PV = nRT P = Pressure = 100000Pa V = Volume = 0.000315m3 n = number of moles = ? R = gas constant = 8.31JK-1 mol-1 T = temperature = 300K Calculations using the ideal gas law

Calculations using the ideal gas law PV = nRT (100000)(0.000315 ) = n (8.31) (300) (100000)(0 .000315) = n (8.31) (300) 0.0126 = n ( the number of moles) Calculations using the ideal gas law

Calculations using the ideal gas law 261e) Calculate the relative molecular mass of the volatile liquid Moles of X Mass of X in g X Relative Molecular Mass ÷ Relative Molecular Mass 0.0126 moles x RMM = 1.1 g RMM = 1.1/ 0.0126 RMM = 87.3016 To the nearest whole number – RMM = 87 Calculations using the ideal gas law