chapter 11 – molecular composition of gases. 11-1 volume-mass relationships of gases joseph...
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Chapter 11 – Molecular Composition of GasesChapter 11 – Molecular Composition of Gases
11-1 Volume-Mass Relationships of Gases11-1 Volume-Mass Relationships of Gases
Joseph Gay-Lussac, French Joseph Gay-Lussac, French chemist in the 1800s, found that chemist in the 1800s, found that at constant temperature and at constant temperature and pressure, the volumes of pressure, the volumes of gaseous reactants and products gaseous reactants and products can be expressed as ratios of can be expressed as ratios of small whole numbers. small whole numbers.
This is called Gay-Lussac’s law This is called Gay-Lussac’s law of combining volumes of gases. of combining volumes of gases.
Today we know that these Today we know that these volume relationships are given volume relationships are given by the coefficients of a balanced by the coefficients of a balanced chemical equation and are chemical equation and are equivalent to the mole ratios of equivalent to the mole ratios of gaseous reactant and products. gaseous reactant and products.
1-1 Gay-Lussac’s Law of Combining 1-1 Gay-Lussac’s Law of Combining VolumesVolumes
hydrogen + oxygen hydrogen + oxygen water vapor water vapor
hydrogen + chlorine hydrogen + chlorine hydrogen chloride hydrogen chloride
11-1 Volume-Mass Relationships of Gases11-1 Volume-Mass Relationships of Gases
In 1811, Amedeo In 1811, Amedeo Avogadro proposed Avogadro proposed that equal volumes of that equal volumes of gases at the same gases at the same temperature and temperature and pressure contain equal pressure contain equal numbers of molecules. numbers of molecules.
This is known as This is known as Avogadro’s law. Avogadro’s law.
11-1 Avogadro’s Law11-1 Avogadro’s Law
HH22((gg) + Cl) + Cl22((gg) ) 2HCl( 2HCl(gg))
One volume of hydrogen combines with one One volume of hydrogen combines with one volume of chlorine to form two volumes of volume of chlorine to form two volumes of hydrogen chloride. hydrogen chloride.
One molecule of hydrogen combines with One molecule of hydrogen combines with one molecule of chlorine to form two one molecule of chlorine to form two molecules of hydrogen chloride. molecules of hydrogen chloride.
11-1 Avogadro’s Law11-1 Avogadro’s Law
Gas volume (V) is directly proportional to the Gas volume (V) is directly proportional to the number of particles (n) at constant number of particles (n) at constant temperature and pressure. temperature and pressure.
V = knV = kn
oror
k = V/nk = V/n
11-1 Molar Volume of Gases11-1 Molar Volume of Gases
One mole of any gas will occupy the same One mole of any gas will occupy the same volume as any other gas at the same volume as any other gas at the same temperature and pressure, regardless of the temperature and pressure, regardless of the mass of the particle. mass of the particle.
The volume occupied by one mole of any The volume occupied by one mole of any gas at STP is called the gas at STP is called the standard molar standard molar volumevolume of a gas and is equal to of a gas and is equal to 22.4 L/mol22.4 L/mol..
11-1 Molar Volume of Gases11-1 Molar Volume of Gases
A sample of hydrogen gas occupies 14.1 L A sample of hydrogen gas occupies 14.1 L at STP. How many moles of the gas are at STP. How many moles of the gas are present?present?
At STP, a sample of neon gas occupies 0.55 At STP, a sample of neon gas occupies 0.55 L. How many moles of neon gas does this L. How many moles of neon gas does this represent?represent?
11-2 The Ideal Gas Law11-2 The Ideal Gas Law
To describe a gas sample, four quantities are needed: To describe a gas sample, four quantities are needed: pressure, volume, temperature, and number of molespressure, volume, temperature, and number of moles
V V αα 1/P (Boyle’s Law) 1/P (Boyle’s Law) V V αα T (Charles’ Law) T (Charles’ Law) V V αα n (Avogadro’s Law) n (Avogadro’s Law) V V αα 1/P x T x n 1/P x T x n V = R x 1/P x T x nV = R x 1/P x T x n
PV = nRTPV = nRT R is the gas constant. 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 R11-2 The Value of R
Remember, one mole of any gas at STP has a Remember, one mole of any gas at STP has a volume of 22.4 L. We can use this definition to volume of 22.4 L. We can use this definition to determine the value of R. determine the value of R.
If pressure is in atm…If pressure is in atm…
If pressure is in kPa…If pressure is in kPa…
Remember, 1 L = 1 dmRemember, 1 L = 1 dm33
11-2 The Ideal Gas Law11-2 The Ideal Gas Law
The ideal gas law is the mathematical The ideal gas law is the mathematical relationship among pressure, volume, relationship among pressure, volume, temperature and the number of moles of a temperature and the number of moles of a gas. gas.
PV = nRTPV = nRTP – pressure (atm or kPa)V – volume (L)n – molesR – gas constant (0.0821 L.atm/mol.K or 8.31 L.kPa/mol.K)T – temperature (K)
11-2 Ideal Gas Law11-2 Ideal Gas Law
What volume will be occupied by 0.21 moles What volume will be occupied by 0.21 moles of oxygen gas at 25of oxygen gas at 25°C and 1.05 atm of °C and 1.05 atm of pressure?pressure?
11-2 Ideal Gas Law11-2 Ideal Gas Law
A sample of carbon dioxide gas has a mass A sample of carbon dioxide gas has a mass of 1.20 g at 25of 1.20 g at 25°°C and 1.05 atm. What C and 1.05 atm. What volume does this gas occupy?volume does this gas occupy?
11-2 The Ideal Gas Law11-2 The Ideal Gas Law
Variations:Variations:n = m/MMn = m/MM
so PV = mRT/MMso PV = mRT/MM
and MM = mRT/PVand MM = mRT/PV
D = m/VD = m/V
so D = MMP/RTso D = MMP/RT
11-2 The Ideal Gas Law11-2 The Ideal Gas Law
What is the molar mass of a gas if 0.427 g What is the molar mass of a gas if 0.427 g of the gas occupies a volume of 125 mL at of the gas occupies a volume of 125 mL at 20.020.0°C and 0.980 atm?°C and 0.980 atm?
11-2 The Ideal Gas Law11-2 The Ideal Gas Law
What is the density of argon gas, Ar, at a What is the density of argon gas, Ar, at a pressure of 551 torr and a temperature of pressure of 551 torr and a temperature of 2525°C?°C?
11-3 Stoichiometry of Gases11-3 Stoichiometry of Gases
Volume-Volume calculations – just use the Volume-Volume calculations – just use the mole ratio!mole ratio!
CC33HH88 + 5O + 5O22 3CO 3CO22 + 4H + 4H22OO
What volume, in L, of oxygen, is required for What volume, in L, of oxygen, is required for the complete combustion of 0.350 L of the complete combustion of 0.350 L of propane?propane?
11-4 Effusion and Diffusion11-4 Effusion and Diffusion
Diffusion – the gradual Diffusion – the gradual mixing of two gases due to mixing of two gases due to their spontaneous, random their spontaneous, random motion.motion.
Effusion – the process by Effusion – the process by which the molecules of a which the molecules of a gas confined in a container gas confined in a container randomly pass through a randomly pass through a tiny opening in the tiny opening in the container.container.
11-4 Effusion and Diffusion11-4 Effusion and Diffusion
The rates of effusion and diffusion depend on The rates of effusion and diffusion depend on the relative velocities of the gas molecules. the relative velocities of the gas molecules.
The velocity of a gas varies inversely with its The velocity of a gas varies inversely with its mass. Lighter molecules move faster than mass. Lighter molecules move faster than heavier molecules at equivalent heavier molecules at equivalent temperatures. temperatures.
11-4 Effusion and Diffusion11-4 Effusion and Diffusion
Remember, temperature is Remember, temperature is a measure of average a measure of average kinetic energy. kinetic energy.
Particles of two gas Particles of two gas samples (A and B) at the samples (A and B) at the same temperature have same temperature have the same average kinetic the same average kinetic energy.energy.
KE =1/2 mvKE =1/2 mv22
Graham’s Law of Effusion Graham’s Law of Effusion compares rates of effusion compares rates of effusion and diffusion for gases.and diffusion for gases.
The relationship can be The relationship can be derived easily.derived easily.
11-4 Effusion and Diffusion11-4 Effusion and Diffusion
The rate of effusion or The rate of effusion or diffusion for gases diffusion for gases depends on the average depends on the average velocities of the particles. velocities of the particles.
Graham’s Law of Effusion Graham’s Law of Effusion – For two gases, A and B, – For two gases, A and B, at the same temperature, at the same temperature, the following relationship the following relationship exists…exists…
A
B
B
A
MM
MM
rate
rate
A
B
B
A
MM
MM
v
v
11-4 Effusion and Diffusion11-4 Effusion and Diffusion
Compare the rates of effusion of hydrogen Compare the rates of effusion of hydrogen and oxygen at the same temperature and and oxygen at the same temperature and pressure. pressure.
11-4 Effusion and Diffusion11-4 Effusion and Diffusion
If a molecule of neon gas travels at an If a molecule of neon gas travels at an average of 400 m/s at a given temperature, average of 400 m/s at a given temperature, estimate the average speed of a molecule of estimate the average speed of a molecule of butane gas, Cbutane gas, C44HH1010, at the same temperature. , at the same temperature.
