Properties of Gases. assume the volume and shape of their containers. most compressible of the states of matter

Gases

Properties of Gases assume the volume and shape of their containers most compressible of the states of matter mix evenly and completely with other gases much lower density than other forms of matter

Substances that Exist as Gases

Elements that exist as gases at 25 C and 1 atm. The Noble gases (the Group 8A elements) are monatomic species; the other elements exist as diatomic molecules. Ozone (O 3 ) is also a gas.

Some substances found as gases at 1 atm and 25 ºC Elements H 2, N 2, O 2, O 3, F 2, Cl 2 He, Ne, Ar, Kr, Xe, Rn Compounds HF, HCl, HBr, Hl, CO, CO 2, NH 3, NO, NO 2, N 2 O, SO 2, H 2 S, HCN

Pressure of a Gas

Pressure The force exerted on an object divided by the surface area of the object; P = F A Any gas confined to a container is found to exert a pressure on the container.

SI Units of Pressure customary units 1 standard atm = 760 mm Hg = 760 torr SI units pressure = force/area pressure = Newton/m 2 = Pascal 1 standard atmosphere = 101,325 Pa or 1 atm = 101.3 kpa

vacuum atmospheric pressure mercury A Torricellian Barometer

vacuum 760 mm Hg (1 standard Atmosphere) atmospheric pressure mercury A Torricellian Barometer

The Gas Laws Boyle s Law Charles Law Avogadro s Law The Ideal Gas Law

The Gas Laws and the Scientific Method Observations Laws Theory Hypothesis Experiment Boyle Charles Avogadro Ideal Gas Kineticmolecular theory

Boyle s Law Robert Boyle 1626 1691

Manometer atmospheric pressure P gas = P atm - h Hg

Manometer atmospheric pressure P gas = P atm + h Hg

Boyle s Data Pressure (mm Hg) 724 869 951 998 1230 1893 2250 Volume (arbitrary units) 1.50 1.33 1.22 1.16 0.94 0.61 0.51

1.2 1.0 Pressure (atm) P/2 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 Volume (L) 2V

Pressure (mm Hg) Volume (arbitrary units) PV 724 1.50 1090 Boyle s 869 1.33 1160 Data 951 1.22 1160 998 1.16 1160 1230 0.94 1200 1893 0.61 1200 2250 0.51 1100

Boyle s Law at constant temperature, the volume of a constant amount of gas is inversely proportional to the pressure constant n, constant T PV = k P = k 1 V

pg.422 problem 2 The pressure of a sample of helium in a 1.00 L container is 0.988 atm. What is the new pressure if the sample is placed in a 2.00L container? PV = k

pg.422 problem 2 The pressure of a sample of helium in a 1.00 L container is 0.988 atm. What is the new pressure if the sample is placed in a 2.00L container? P 1 V 1 = k = P 2 V 2

pg.422 problem 2 The pressure of a sample of helium in a 1.00 L container is 0.988 atm. What is the new pressure if the sample is placed in a 2.00L container? P 1 V 1 = P 2 V 2 (1.00 L) (0.988 atm) = (2.00 L) (P 2 ) (1.00 L) (0.988 atm) (2.00 L) = P 2 0.494 atm = P 2

Charles s Law Jacques Charles 1746 1823

Charles s Law at constant pressure, the volume of a constant amount of gas is directly proportional to the absolute temperature constant n, constant P V = kt V T = k

50 He 40 CH 4 V(mL) 30 273.15 º C H 2 20 10 N 2 O 0 300 200 100 0 100 200 300 400 t(ºc)

Celsius scale 5727 C Temperature at the surface of the sun Kelvin scale 6000 K 1064 C 100 C 37 C 0 C Melting point of gold Boiling point of water Body temperature Melting point of ice 1337 K 373.15 K 310 K 273.15 K 196 C Boiling point of N 2 77K 268.95 C 273.15 C Boiling point of He Lowest temperature (absolute zero) 4.2 K 0 K

pg.425 problem 6 A gas at 89º C occupies a volume of 0.67 L. At what Celsius temperature will the volume increase to 1.12 L? V 1 = T 1 k = V 2 T 2

pg.425 problem 6 A gas at 89ºC occupies a volume of 0.67 L. At what Celsius temperature will the volume increase to 1.12 L? V 1 V 2 = T 1 T 2 0.67 L (8989ºC + 273 K) = 1.12 L T 2 The temperature must be expressed in Kelvin

pg.425 problem 6 A gas at 89ºC occupies a volume of 0.67 L. At what Celsius temperature will the volume increase to 1.12 L? V 1 V 2 = T 1 T 2 0.67 L (89 + 273 K) = 1.12 L T 2 605 K = T 2 605 K - 273 T 2 = 1.12 L (89 + 273K) 0.67 L 332 ºC = T 2

Gay-Lussac s Law at constant volume, the pressure of a constant amount of gas is directly proportional to the absolute temperature constant n, constant V P = kt P T = k

pg.427 problem 9 A gas in a sealed container has a pressure of 125 KPa at a temperature of 30.0º C. If the pressure in a container is increased to 201 KPa, what is the new temperature? P 1 = T 1 k = P 2 T 2

pg.427 problem 9 A gas in a sealed container has a pressure of 125 kpa at a temperature of 30.0º C. If the pressure in a container is increased to 201 kpa, what is the new temperature? 125 kpa = (30.0 + 273 K) T 2 210 KPa T 2 210 kpa (30.0 + 273K) = 125 kpa P 1 P 2 = T 1 T 2 509 K = T 2 509 K - 273 = 236ºC = T 2

The Combined Gas Law Boyle s, Charles s and Gay-Lussac s laws combined into a single law. constant n, P 1 V 1 P 2 V 2 = T 1 T 2

pg.427 problem 9 A helium-filled balloon has a volume of 2.1 L at 0.998 atm and 36ºC. If it is released and rises to an elevation at which the pressure is 0.900 atm and the temperature is 28ºC, what is the new volume of the balloon? P 1 V 1 P 2 V 2 = T 2 T 1 (V 2 ) (0.998 atm) (2.1 L) = (0.900 atm) (36.0+ 273 K) (28.0 + 273 K)

pg.427 problem 9 A helium-filled balloon has a volume of 2.1 L at 0.998 atm and 36ºC. If it is released and rises to an elevation at which the pressure is 0.900 atm and the temperature is 28ºC, what is the new volume of the balloon? P 1 V 1 P 2 V 2 = T 2 T 1 (0.998 atm) (2.1 L) (28.0º C + 273) (36.0º C + 273) (0.900 atm) = (V 2 )

pg.427 problem 9 A helium-filled balloon has a volume of 2.1 L at 0.998 atm and 36ºC. If it is released and rises to an elevation at which the pressure is 0.900 atm and the temperature is 28ºC, what is the new volume of the balloon? P 1 V 1 P 2 V 2 = T 2 T 1 (0.998 atm) (2.1 L) (301 K) (309 K) (0.900 atm) = (V 2 ) 2.27 L = V 2

Avogadro s Law Amadeo Avogadro (1776 1856)

Avogadro s Law at constant temperature and pressure, the volume of a gas is directly proportional to the number moles constant T, constant P V = kn equal volumes of different gases contain equal numbers of molecules

The Ideal-Gas Equation

The Ideal Gas Law PV = nrt constant n, constant T (Boyle s Law) constant n, constant P (Charles Law) constant P, constant T (Avogadro s Law)

The Ideal-Gas Law PV = nrt P is pressure in atmospheres V is volume in liters n is number of moles T is absolute temperature in Kelvins R is called the gas constant

An ideal gas is a hypothetical gas whose pressure-volume-temperature behavior can be completely accounted for by the ideal-gas equation. large volumes high temperatures

Standard temperature and pressure PV = nrt standard conditions defined P = 1 atm T = 0ºC (273 K) STP

Standard Molar Volume The volume occupied by 1 mole of a gas at STP is 22.414 L

The Gas Constant R The volume occupied by 1 mole of a gas at STP is 22.414 L PV = nrt R = R = PV nt (1 atm) (22.414 L) (1 mol) (273.15 K) R = 0.082057 (L atm/mol K)

V = 30.6 L Example What is the volume occupied by 49.8 g of HCl at STP? 1mol 49.8 g HCl x = 1.366 mol 36.46 g P = 1 atm HCl T = 273 K nrt V = P (1.366 mol) (0.0821 L atm/mol K) (273K) 1 atm

Example What is the volume occupied by 49.8 g of HCl at STP? (alternative solution) 49.8 g n = = 1.366 mol 36.46 g/mol P = 1 atm T = 273 K V = x 1.366 mol 22.4 L 1 mol V = 30.6 L

Example A compound has the empirical formula BH 3. At 27 º C, 74.3 ml of the gas exerted a pressure of 1.12 atm. If the mass of the gas was 0.0934 g, what is its molecular formula? PV = nrt n = PV RT

Example A compound has the empirical formula BH 3. At 27 ºC, 74.3 ml of the gas exerted a pressure of 1.12 atm. If the mass of the gas was 0.0934 g, what is its molecular formula? n = (1.12 atm) (.0743 L) (0.0821 L atm/mol K) n = 0.00338 mol (27 ºC + 273) n = PV RT

Example (cont.) A compound has the empirical formula BH 3. At 27 º C, 74.3 ml of the gas exerted a pressure of 1.12 atm. If the mass of the gas was 0.0934 g, what is its molecular formula? n = 0.00338 mol 0.0934 g = 27.6 g/mol 0.00338 mol BH 3 = 13.8 g / empirical formula B 2 H 6

The ideal gas law is often used to calculate the changes that will occur when the conditions of a gas are changed PV = nrt PV = nrt If nrt are constant If nrp are constant 1 P = ( nrt) nr V = V P ( ) PV = ( nrt ) = P 2 V V 2 2 = ( ) = V T nr P T T 2

The ideal gas law is often used to calculate the changes that will occur when the conditions of a gas are changed PV = nrt If PRT are constant RT V = ( ) V = n P RT ( ) P n = V 2 n 2 P T PV = nrt If nrv are constant P = nr ( ) V nr =( ) V = T P 2 T 2

Example A sample of oxygen gas initially at 0.97 atm is cooled from 21 º C to -68 º C at constant volume. What is its final pressure. PV = nrt P T nr =( ) V = P 2 T 2 P 1 P = 2 T 1 T 2

Example A sample of oxygen gas initially at 0.97 atm is cooled from 21 º C to -68 º C at constant volume. What is its final pressure. P 1 P = 2 T 1 T 2 (0.97 atm) 292 K = P 2 205 K P 2 = 0.68 atm

Density Calculations

Example What is the density of UF 6 gas at 62 º C and 779mmHg? n = P 779mmHg V RT n = 760mmHg/atm V (0.0821 L atm/mol K) (335 K) = 0.0373 mol 352.03g x L 1 mol = 13.1g/L A common unit for gasses

Example n Cyanogen, empirical formula CN, is a gas with a density of 2.335 g/l at 0ºC and 1 atm. What is its molecular formula? P V n = (1atm) (1 L) = RT (0.0821 L atm/mol K) (273K) 2.335g = 0.0446 mol = 52 g/mol Cyanogen 26g /mol (CN) C 2 N 2 = 2

Plot of P versus V is a hyperbola PV = k P = (1/V ) k is of the form y = mx + b, which is the equation for a straight line

P P 0.6 atm 0.3 atm 2L 4L V (a) 1 V (b)

Density Calculations density = mass V n = mass Molar mass = g g/mol n V = P RT mass Molar mass(v) = P RT Therefore: density = P RT (Molar mass)