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Mole concept applied to gases

Mole concept applied to gases. 1.4.4 Apply Avogadro’s law to calculate reacting volumes of gas 1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations 1.4.6 Solve problems between temperature, pressure and volume for a fixed mass of an ideal gas.

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Mole concept applied to gases

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  1. Mole concept applied to gases 1.4.4 Apply Avogadro’s law to calculate reacting volumes of gas 1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations 1.4.6 Solve problems between temperature, pressure and volume for a fixed mass of an ideal gas. 1.4.7 Solve problems relating to the ideal gas equation, PV=nRT 1.4.8 Analyse graphs relating to the ideal gas equation.

  2. Avogadro’s Hypothesis • At a constant temperature and pressure, a given volume of gas always has the same number of particles. • The coefficients of a balanced reaction is the same ratio as the volumes of reactants and products

  3. 2CO (g) + O2 (g) 2CO2(g) For the above example, it is understood that half the volume of oxygen is needed to react with a given volume of carbon monoxide. This can be used to carry out calculations about volume of gaseous product and the volume of any excess reagents.

  4. Example • 10cm3 of ethyne (C2H2) is reacted with 50cm3 of hydrogen to produce ethane (C2H6), calculate the total volume and composition of the remaining gas mixture, assuming constant T and P. 1st get balanced equation: C2H2(g) + 2H2(g)  C2H6(g) 2nd look at the volume ratios: 1 mol ethyne to 2 mol of hydrogen, therefore 1 vol to 2 vol 3rd analyse: If all 10cm3 of ethyne is used, it needs only 20cm3 of hydrogen, therefore hydrogen is in excess by 50cm3-20cm3 = 30 cm3. In the end you’ll have 10 cm3 Ethane and the leftover 30 cm3 hydrogen

  5. Molar volume • The temperature and pressure are specified and used to calculate the volume of one mole of gas. • Standard temperature and pressure (STP) is at sea level 1 atm = 101.3 kPa and 0oC = 273 K this volume is 22.4 dm3 (or 22.4 L) Molar gas volume, Vm. It contains 6.02 x 1023 molecules of gas

  6. Example Calculate how many moles of oxygen molecules are there in 5.00 dm3 at STP n= VSTP = 5.00 = 0.223 mol 22.4 dm3 22.4 dm3

  7. Boyle’s Law (1659) • Boyle noticed that the product of the volume of air times the pressure exerted on it was very nearly a constant, or PV=constant. • If V increases, P decreases proportionately and vice versa. (Inverse proportions) • Temperature must be constant. • Example: A balloon under normal pressure is blown up (1 atm), if we put it under water and exert more pressure on it (2 atm), the volume of the balloon will be smaller (1/2 its original size) • P1V1=P2V2

  8. Charles’ Law (1787) • Gas expands (volume increases) when heated and contracts (volume decreases) when cooled. • The volume of a fixed mass of gas varies directly with the Kelvin temperature provided the pressure is constant. V= constant x T • V1 = V2 T1 T2

  9. Gay-Lussac’s Law • The pressure of a gas increases as its temperature increases. • As a gas is heated, its molecules move more quickly, hitting up against the walls of the container more often, causing increased pressure. • P1 = P2 T1 T2

  10. Laws combined… • P1V1 = P2V2 T1 T2 T must be in Kelvins, but P and V can be any proper unit provided they are consistently used throughout the calculation

  11. Practice • If a given mass of gas occupies a volume of 8.50 L at a pressure of 95.0 kPa and 35 oC, what volume will it occupy at a pressure of 75.0 kPa and a temperature of 150 oC? 1st convert oC to K: 35 + 273 = 308 K 150 + 273 = 423 K 2nd rearrange equation and solve problem: V2 = V1 x P1 x T2= 8.50 x 95.0 x 423 = 14.8 L P2 x T1 75.0 x 308

  12. Temperature • Kelvin temperature is proportional to the average kinetic energy of the gas molecules. • It is a measure of random motion of the gas molecules • More motion = higher temperature

  13. Ideal gas behaviour • Ideal behaviour is when a gas obeys Boyle’s, Charle’s and Gay-Lussac’s laws well • At ordinary temperature and pressures, but there is deviation at low temperature and high pressures

  14. Ideal gas • where all collisions between molecules are perfectly elastic and in which there are no intermolecular attractive forces. • Its like hard spheres bouncing around, but NO interaction.

  15. Ideal gas law • PV = nRT • P= pressure (kPa) • Volume = (dm3) • n= number of moles • R=universal gas constant =8.3145 J mol-1 K-1 • T= temperature (K)

  16. Example 3.376 g of a gas occupies 2.368 dm3 at 17.6 oC and a pressure of 96.73 kPa, determine its molar mass. PV= nRT rearrange equation for n n= PV/RT = (96.73 x 2.368) / (8.314 x 290.6) = 0.09481 mol Molar mass = mass/ mole = 3.376 g / 0.09481 mol = 35.61 g/mol

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