Chapter 11 molecular composition of gases
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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
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
1-1 Gay-Lussac’s Law of Combining Volumes

  • hydrogen + oxygen  water vapor

  • hydrogen + chlorine  hydrogen chloride


11 1 volume mass relationships of gases1
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
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 law1
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
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 gases1
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
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
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 law1
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
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 law1
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 law2
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 law3
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 law4
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
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
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 diffusion1
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.


11 4 effusion and diffusion2

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 diffusion3
11-4 Effusion and Diffusion energy.

  • 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 diffusion4
11-4 Effusion and Diffusion energy.

  • Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure.


11 4 effusion and diffusion5
11-4 Effusion and Diffusion energy.

  • 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.