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Gas Laws Robert Boyle Jacques Charles Amadeo Avogadro

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Presentation on theme: "Gas Laws Robert Boyle Jacques Charles Amadeo Avogadro"— Presentation transcript:

1 Gas Laws Robert Boyle Jacques Charles Amadeo Avogadro
Joseph Louis Gay-Lussac

2 The Combined Gas Law The combined gas law expresses the relationship between pressure, volume and temperature of a fixed amount of gas.

3 Boyle’s Law Pressure is inversely proportional to volume when temperature is held constant.

4 Charles’s Law The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin. (P = constant) Temperature MUST be in KELVIN!

5 Avogadro’s Law For a gas at constant temperature and pressure (STP), the volume is directly proportional to the number of moles of gas (at low pressures). V = an a = proportionality constant V = volume of the gas n = number of moles of gas

6 PV = nRT Ideal Gas Law P = pressure in atm V = volume in liters
n = moles R = proportionality constant = L atm/ mol·K T = temperature in Kelvin Holds closely at P < 1 atm

7 Example What is the pressure in atm of a mol sample of He gas at a temp of 293 K if its volume is 0.505 L? You can find the density of a gas using the ideal gas law! Let’s see how.

8 Density and the Ideal Gas Law
Combining the formula for density with the Ideal Gas law, substituting and rearranging algebraically: M = Molar Mass P = Pressure R = Gas Constant T = Temperature in Kelvins

9 Gas Density … so at STP…

10 Gas Stoichiometry #1 If reactants and products are at the same conditions of temperature and pressure, then mole ratios of gases are also volume ratios. 3 H2(g) N2(g)  NH3(g) 3 moles H mole N  moles NH3 3 liters H liter N  liters NH3

11 Gas Stoichiometry #2 3 H2(g) + N2(g)  2NH3(g)
How many liters of ammonia can be produced when 12 liters of hydrogen react with an excess of nitrogen? 3 H2(g) + N2(g)  2NH3(g) 12 L H2 2 L NH3 = L NH3 8.0 3 L H2

12 Gas Stoichiometry #3 How many liters of oxygen gas, at STP, can be collected from the complete decomposition of 50.0 grams of potassium chlorate? 2 KClO3(s)  2 KCl(s) + 3 O2(g) 50.0 g KClO3 1 mol KClO3 3 mol O2 22.4 L O2 g KClO3 2 mol KClO3 1 mol O2 = 13.7 L O2

13 Gas Stoichiometry #4 How many liters of oxygen gas, at 37.0C and atmospheres, can be collected from the complete decomposition of 50.0 grams of potassium chlorate? 2 KClO3(s)  2 KCl(s) + 3 O2(g) 50.0 g KClO3 1 mol KClO3 3 mol O2 0.612 = mol O2 g KClO3 2 mol KClO3 = 16.7 L

14 Dalton’s Law of Partial Pressures
For a mixture of gases in a container, PTotal = P1 + P2 + P This is particularly useful in calculating the pressure of gases collected over water.


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