CHAPTER 13 – GASES PRESSURE – Force per unit area Due to the constant bombardment of the inside walls of the container by the gas molecules 6A-1 (of 34)

Atmospheric pressure is measured with a BAROMETER STANDARD PRESSURE: 760.0 mm Hg 760.0 torr 1.000 atm 6A-2

Close-Ended Manometer Gas pressure is measured with a MANOMETER Close-Ended Manometer 125 mm more than 0 125 mm Open-Ended Manometer 125 mm more than 765 mm 890. mm 6A-3

BOYLE’S LAW – The volume of a gas is inversely proportional to its pressure, provided the temperature and quantity of gas do not change V  1 ___ p V = k ___ p pV = k p1V1 = k p2V2 = k p1V1 = p2V2 6A-4

40. 0 mL of oxygen gas are collected with a pressure of 250. mm Hg 40.0 mL of oxygen gas are collected with a pressure of 250. mm Hg. Calculate the volume of the gas if the pressure changes to 300. mm Hg. p1V1 = p2V2 p1 = V1 = 250. mm 40.0 mL p2 = V2 = 300. mm ? mL p1V1 = V2 ______ p2 = (250. mm Hg)(40.0 mL) _____________________________ (300. mm Hg) = 33.3 mL 6A-5

TEMPERATURE – A measure of the average kinetic energy of molecules

The lowest possible temperature is the temperature where molecular motion stops ABSOLUTE SCALE – One with 0 as the lowest possible value KELVIN SCALE (K) – A temperature scale where 0 is the lowest possible temperature K = Cº + 273.2 ABSOLUTE ZERO – The lowest possible temperature 6A-7

Change to K: 42 ºC -22.0 ºC 42 ºC + 273.2 = 315 K -22.5 ºC + 273.2 = 250.7 K Change to ºC: 150. K 300.5 K 150. K - 273.2 = -123 ºC 300.5 K - 273.2 = 27.3 ºC Standard temperature: 0.0 ºC 273.2 K 6A-8

T must be measured in Kelvin so that when T = 0, V = 0 CHARLES’ LAW – The volume of a gas is directly proportional to its temperature, provided the pressure and quantity of gas do not change V  T V = kT V = k ___ T V1 = k ___ T1 V2 = k ___ T2 V1 = V2 ___ ___ T1 T2 T must be measured in Kelvin so that when T = 0, V = 0 6A-9

15. 0 mL of hydrogen gas are collected at 10. ºC 15.0 mL of hydrogen gas are collected at 10.ºC. Calculate the volume of the gas if the temperature changes to 40.ºC. V1 = V2 ___ ___ T1 T2 V1 = T1 = 15. 0 mL 283 K V2 = T2 = ? mL 313 K V1T2 = V2 ______ T1 = (15.0 mL)(313 K) _____________________ (283 K) = 16.6 mL 6A-10

Combining Charles’ and Boyle’s Laws: V1 = V2 ___ ___ T1 T2 p1V1 = p2V2 p1V1 = p2V2 ______ ______ T1 T2 6A-11

30. 0 mL of a gas are collected at 22ºC and 750. torr 30.0 mL of a gas are collected at 22ºC and 750. torr. Calculate the volume of the gas at standard temperature and pressure. 273.2 K, 760.0 torr p1V1 = p2V2 ______ ______ T1 T2 p1 = V1 = T1 = 750. torr 30.0 mL 295 K p2 = V2 = T2 = 760.0 torr ? mL 273.2 K p1V1T2 = V2 _________ p2T1 = (750. torr)(30.0 mL)(273.2 K) ____________________________________ (760.0 torr)(295 K) = 27.4 mL 6A-12

AVOGADRO’S LAW – The volume of a gas is directly proportional to the quantity of gas, provided the pressure and temperature of gas do not change V  n V = kn 6A-13

V  n Avogadro’s Law V  nT ____ p or pV  nT V  1 Boyle’s Law ___ p V  T Charles’ Law 6A-14

IDEAL GAS LAW EQUATION pV = nRT p = pressure V = volume n = quantity T = temperature R = Universal Gas Constant (atm) (L) (mol) (K) (0.08206 Latm/molK) 6A-15

pV = RTn 6A-16

Calculate the Celsius temperature of 0. 100 moles of a gas in a 700 Calculate the Celsius temperature of 0.100 moles of a gas in a 700. mL container at a pressure of 2,500. torr. pV = nRT pV = T ____ nR 2,500. torr x 1 atm _____________ 760.0 torr = 3.289 atm 5 (3.2895 atm)(0.700 L) _________________________________________ (0.100 mol)(0.08206 Latm/molK) = 281 K 281 K – 273.2 = 8ºC 6A-17

Calculate the pressure of 2. 00 moles of nitrogen gas in a 5 Calculate the pressure of 2.00 moles of nitrogen gas in a 5.00 liter container at 20.ºC. pV = nRT p = nRT _____ V 20.ºC + 273.2 = 293 K (2.00 mol)(0.08206 Latm/molK)(293 K) ________________________________________________ (5.00 L) = 9.62 atm 6A-18

PARTIAL PRESSURES DALTON’S LAW OF PARTIAL PRESSURES – The total pressure of a mixture of gases is the sum of the pressures exerted by each gas pN2 = 591 torr pO2 = 161 torr pAr = 8 torr ___________ 760. torr PARTIAL PRESSURE – The pressure a gas would exert if it were alone in a container 6A-19

COLLECTING GASES IN LAB To determine the quantity of gas collected in lab, you must measure the (1) gas volume (2) gas temperature (3) gas pressure pV = n ____ RT EUDIOMETER – A gas collecting tube 6A-20

Atmospheric pressure: 765 mm Hg Gas pressures? 1st – Hg level THE SAME in and out ∴ pressure inside eudiometer = pressure outside ∴ pressure of gas = atmospheric pressure p = 765 mm Hg 6A-21

Atmospheric pressure: 765 mm Hg Gas pressures? 2nd – Hg level HIGHER inside ∴ pressure inside eudiometer < pressure outside ∴ pressure of gas < atmospheric pressure p = 765 mm Hg – 13 mm Hg = 752 mm Hg 6A-22

Atmospheric pressure: 765 mm Hg Gas pressures? 3rd – Hg level LOWER inside ∴ pressure inside eudiometer > pressure outside ∴ pressure of gas > atmospheric pressure p = 765 mm Hg + 6 mm Hg = 771 mm Hg 6A-23

COLLECTING GASES BY WATER DISPLACEMENT Collecting O2 gas When a gas is collected by water displacement, some water evaporates into the eudiometer The eudiometer contains both oxygen gas and water vapor If the total pressure is measured, the partial pressure of the water vapor must be subtracted to get the partial pressure of the oxygen gas The partial pressure of the water vapor (called WATER VAPOR PRESSURE) depends only on temperature, and can be looked up 6A-24

Atmospheric pressure: 765 mm Hg Temperature: 25ºC Gas pressures? 1st – H2O level THE SAME in and out ∴ p of O2 (g) + p of H2O (g) = atmospheric pressure ∴ p of O2 (g) = atmospheric pressure – p of H2O (g) Water Vapor Pressure at 25ºC = 24 mm Hg p = 765 mm Hg – 24 mm Hg = 741 mm Hg 6A-25

Atmospheric pressure: 765 mm Hg Temperature: 25ºC Gas pressures? 2nd – H2O level HIGHER inside ∴ p of O2 (g) + p of H2O (g) < atmospheric pressure Water level differences are converted to mercury level differences by dividing by the ratio of mercury’s density to water’s density: 13.6 27 mm H2O ÷ 13.6 = 2.0 mm Hg p = 765 mm Hg – 2.0 mm Hg – 24 mm Hg = 739 mm Hg 6A-26

Atmospheric Pressure = AP Water Vapor Pressure = WVP 1st – AP 2nd – AP – h 3rd – AP – WVP 4th – AP – (h÷13.6) – WVP 6A-27

VOLUME CALCULATIONS IN CHEMICAL REACTIONS Calculate the mass of magnesium metal that would react with sulfuric acid to produce 250. mL of hydrogen gas at 23ºC and 1.05 atm. Mg + H2SO4 → MgSO4 + H2 x g 250. mL 1 mol 1 mol 0.0108 mol 0.0108 mol pV = n ____ RT = (1.05 atm)(0.250 L) ____________________________________ (0.08206 Latm/molK)(296 K) = 0.0108 mol H2 1 0.01081 mol H2 x 1 mol Mg _____________ 1 mol H2 x 24.31 g Mg ______________ 1 mol Mg = 0.263 g Mg 6A-28

Calculate the mass of carbon that must be burned to produce 4 Calculate the mass of carbon that must be burned to produce 4.50 L of carbon dioxide gas at 29ºC and 0.987 atm. C + O2 → CO2 x g 4.50 L 1 mol 1 mol 0.179 mol 0.179 mol pV = n ____ RT = (0.987 atm)(4.50 L) ____________________________________ (0.08206 Latm/molK)(302 K) = 0.179 mol CO2 2 0.1792 mol CO2 x 1 mol C _____________ 1 mol CO2 x 12.01 g C ____________ 1 mol C = 2.15 g C 6A-29

Calculate the volume of hydrogen gas produced at STP by the reaction of 3.29 g of zinc metal with hydrochloric acid. Zn + HCl → 2 ZnCl2 + H2 3.29 g x L 1 mol 1 mol 0.0503 mol 0.0503 mol 3.29 g Zn x 1 mol Zn _____________ 65.38 g Zn x 1 mol H2 ____________ 1 mol Zn = 0.0503 mol H2 2 V = nRT _____ p = (0.05032 mol)(0.08206 Latm/molK)(273 K) _____________________________________________________ (1.00 atm) = 1.13 L H2 6A-30

Calculate the volume of oxygen gas at 25ºC and 745 torr needed to burn 5.00 g of propane, C3H8. C3H8 + O2 → 5 3 CO2 + H2O 4 5.00 g x L 1 mol 5 mol 0.113 mol 0.567 mol 5.00 g C3H8 x 1 mol C3H8 ________________ 44.11 g C3H8 x 5 mol O2 ______________ 1 mol C3H8 = 0.566 mol O2 8 V = nRT _____ p = (0.5668 mol)(0.08206 Latm/molK)(298 K) ____________________________________________________ (0.9803 atm) = 14.1 L O2 745 torr x 1 atm ____________ 760.0 torr = 0.980 atm 3 6A-31

Calculate the volume of air at 25ºC and 745 torr needed to burn the 5 Calculate the volume of air at 25ºC and 745 torr needed to burn the 5.00 g of propane, if air is 20.9% oxygen by volume. air = 20.9% O2 100 L air = 20.9 L O2 14.14 L O2 x 100 L air ___________ 20.9 L O2 = 67.7 L air 6A-32