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Chapter 11 – Molecular Composition of Gases. 11-1 Volume-Mass Relationships of Gases. Joseph Gay-Lussac, French chemist in the 1800s, found that at constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers.
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11-1 Volume-Mass Relationships of Gases • Joseph Gay-Lussac, French chemist in the 1800s, found that at constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers. • This is called Gay-Lussac’s law of combining volumes of gases. • Today we know that these volume relationships are given by the coefficients of a balanced chemical equation and are equivalent to the mole ratios of gaseous reactant and products.
1-1 Gay-Lussac’s Law of Combining Volumes • hydrogen + oxygen water vapor • hydrogen + chlorine hydrogen chloride
11-1 Volume-Mass Relationships of Gases • In 1811, Amedeo Avogadro proposed that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. • This is known as Avogadro’s law.
11-1 Avogadro’s Law • H2(g) + Cl2(g) 2HCl(g) • One volume of hydrogen combines with one volume of chlorine to form two volumes of hydrogen chloride. • One molecule of hydrogen combines with one molecule of chlorine to form two molecules of hydrogen chloride.
11-1 Avogadro’s Law • Gas volume (V) is directly proportional to the number of particles (n) at constant temperature and pressure. V = kn or k = V/n
11-1 Molar Volume of Gases • One mole of any gas will occupy the same volume as any other gas at the same temperature and pressure, regardless of the mass of the particle. • The volume occupied by one mole of any gas at STP is called the standard molar volume of a gas and is equal to 22.4 L/mol.
11-1 Molar Volume of Gases • A sample of hydrogen gas occupies 14.1 L at STP. How many moles of the gas are present? • At STP, a sample of neon gas occupies 0.55 L. How many moles of neon gas does this represent?
11-2 The Ideal Gas Law • To describe a gas sample, four quantities are needed: pressure, volume, temperature, and number of moles • V α 1/P (Boyle’s Law) • V α T (Charles’ Law) • V α n (Avogadro’s Law) • V α 1/P x T x n • V = R x 1/P x T x n • PV = nRT • R is the gas constant. P is pressure in atm or kPa V is volume in L (dm3) N is moles T is temp in Kelvin
11-2 The Value of R • Remember, one mole of any gas at STP has a volume of 22.4 L. We can use this definition to determine the value of R. • If pressure is in atm… • If pressure is in kPa… • Remember, 1 L = 1 dm3
11-2 The Ideal Gas Law • The ideal gas law is the mathematical relationship among pressure, volume, temperature and the number of moles of a gas. • PV = nRT P – pressure (atm or kPa) V – volume (L) n – moles R – gas constant (0.0821 L.atm/mol.K or 8.31 L.kPa/mol.K) T – temperature (K)
11-2 Ideal Gas Law • What volume will be occupied by 0.21 moles of oxygen gas at 25°C and 1.05 atm of pressure?
11-2 Ideal Gas Law • A sample of carbon dioxide gas has a mass of 1.20 g at 25°C and 1.05 atm. What volume does this gas occupy?
11-2 The Ideal Gas Law • Variations: n = m/MM so PV = mRT/MM and MM = mRT/PV D = m/V so D = MMP/RT
11-2 The Ideal Gas Law • What is the molar mass of a gas if 0.427 g of the gas occupies a volume of 125 mL at 20.0°C and 0.980 atm?
11-2 The Ideal Gas Law • What is the density of argon gas, Ar, at a pressure of 551 torr and a temperature of 25°C?
11-3 Stoichiometry of Gases • Volume-Volume calculations – just use the mole ratio! C3H8 + 5O2 3CO2 + 4H2O • What volume, in L, of oxygen, is required for the complete combustion of 0.350 L of propane?
11-4 Effusion and Diffusion • Diffusion – the gradual mixing of two gases due to their spontaneous, random motion. • Effusion – the process by which the molecules of a gas confined in a container randomly pass through a tiny opening in the container.
11-4 Effusion and Diffusion The rates of effusion and diffusion depend on the relative velocities of the gas molecules. • The velocity of a gas varies inversely with its mass. Lighter molecules move faster than heavier molecules at equivalent temperatures.
Remember, temperature is a measure of average kinetic energy. Particles of two gas samples (A and B) at the same temperature have the same average kinetic energy. KE =1/2 mv2 Graham’s Law of Effusion compares rates of effusion and diffusion for gases. The relationship can be derived easily. 11-4 Effusion and Diffusion
11-4 Effusion and Diffusion • The rate of effusion or diffusion for gases depends on the average velocities of the particles. • Graham’s Law of Effusion – For two gases, A and B, at the same temperature, the following relationship exists…
11-4 Effusion and Diffusion • Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure.
11-4 Effusion and Diffusion • If a molecule of neon gas travels at an average of 400 m/s at a given temperature, estimate the average speed of a molecule of butane gas, C4H10, at the same temperature.