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AP Chemistry Chapter 5 Ch. 5 Gases Agenda 23 September 2017 Per 3 & 5 ○ 5.1 ○ 5.2 The Laws ○ 5.3 The Law ○ 5.4 Gas ○ 5.5 ’s Law of Partial ○ 5.6 The Kinetic Molecular Theory of Gases ○ 5.7 Effusion and ○ 5.8 Real Gases ○ 5.9 Characteristics of Several Real Gases ○ 5.10 Chemistry in the Atmosphere 5.1 Units of Pressure

760 mm Hg = 760 Torr = 1 atm

Pressure = = Newton = (Pa) Area m2 5.2 of Boyle, Charles and Avogadro PV = k Slope = k V = k P

y = k(1/P) + 0 5.2 Gas Laws of Boyle - use the actual data from Boyle’s Experiment Table 5.1 And Desmos to plot vs. Pressure And then Pressure vs 1/V. And PV vs. V And PV vs. P Boyle’s Law 3 centuries later? Boyle’s law only holds precisely at very low pressures.Measurements at higher pressures reveal PV is not constant, but varies as pressure varies. Deviations slight at pressures close to 1 atm we assume gases obey Boyle’s law in our calcs. A gas that strictly obeys Boyle’s Law is called an Ideal gas. Extrapolate (extend the line beyond the experimental points) back to zero pressure to find “ideal” value for k, 22.41 L atm Look Extrapolatefamiliar? (extend the line beyond the experimental points) back to zero pressure to find “ideal” value for k, 22.41 L atm The volume of a gas at constant pressure increases linearly with the Charles’ Law of the gas.

All the gases extrapolate to zero volume at the same temperature. The slopes are different because the samples contain different numbers of moles of gas. V = bT

0K is called Question. A sample of gas is heated from 50℃ to 100℃. What happens to the volume? Assume pressure remains constant. Question. A sample of gas is heated from 50℃ to 100℃. What happens to the volume? Assume pressure remains constant. V = bT V/T = constant

V1 = V2

T1 T2 Question. A sample of gas is heated from 50℃ to 100℃. What happens to the volume? Assume pressure remains constant. V = bT ℃ V/T = constant T1 = 50 + 273 = 323K ℃ V1 = V2 T2 = 100 + 273 = 373K

T1 T2 Question. A sample of gas is heated from 50℃ to 100℃. What happens to the volume? Assume pressure remains constant. V = bT ℃ V/T = constant T1 = 50 + 323 = K ℃ V1 = V2 T2 = 100 + 273 = 373K

323K 373K Question. A sample of gas is heated from 50℃ to 100℃. What happens to the volume? Assume pressure remains constant. V = bT V/T = constant

373KV1 = V2 1.15 V1 = V2

323K Avogadro’s Law

V = an Where V is the volume of a gas, n is the number of moles of gas particles, and a is a proportionality constant. For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas. 5.3 The PV = nRT for a gas.

R is the combined proportionality constant called the universal R is 0.08206 Latm/Kmol n = number of moles of gas

Empirical equation, based on experimental data 5.3 The Ideal Gas Law PV = nRT Equation of state for a gas. A gas that obeys this equation is said to behave “ideally.” - Expresses behavior that real gases approach at low pressure and high . - Ideal gas is a hypothetical substance - Only minimal errors result from assuming ideal behavior for most gases for situations we encounter Question 6 moles of hydrogen gas in a container. 3 moles of gas in a separate container. The two gases are removed from their containers and placed into a third container where they react.

If all three containers have the same volume, and all the gases are at the same temperature, which container has the lowest pressure? Sketch the situation: Use PV = nRT P = n (RT/V)

P=6(Constant) P=3(Constant) P=6(Constant) Question If all three containers have the same volume, and all the gases are at the same temperature, which container has the lowest pressure?

Container with oxygen in it before the reaction has the lowest pressure. 5.4 Gas Stoichiometry 1 of an ideal gas at 0℃ (273.2K) and 1atm. PV = nRT V = nRT/P = (1.000mol)(0.08206 L atm)(273.2K) Kmol V = 22.42 L is the

Remember 0℃ (273.2K) and 1atm is called STP, standard temperature and pressure. 5.4 Gas Stoichiometry What would the volume of 1 mole of an ideal gas be at a normal room temp. (22℃) and 1atm? PV = nRT 5.4 Gas Stoichiometry What would the volume of 1 mole of an ideal gas be at a normal room temp. (22℃) and 1atm? PV = nRT V = nRT/P = (1.000mol)(0.08206 L atm)(273.2 + 22)K Kmol V = 24.22 L Molar of a Gas PV = nRT n = number of moles of gas = grams of gas n = m/ molar mass

P = m RT = RT Molar mass V Molar Mass Molar Mass of a Gas P = m RT = density RT Molar mass V Molar Mass

Or

Molar mass = density RT P Question If 100.0g of Oxygen gas at STP is put into a balloon, what would the volume of the balloon be? Question If 100.0g of Oxygen gas at STP is put into a balloon, what would the volume of the balloon be? O2 Molar mass = 2(16.00) = 32.00 g/mol

PV = mass RT Molar mass V = 100g(0.08206Latm mol-1 K-1) (273.2K) = 32.00 gmol-1 1 atm Question If 100.0g of Oxygen gas at STP is put into a balloon, what would the volume of the balloon be?

O2 Molar mass = 2(16.00) = 32.00 g/mol

PV = mass RT Molar mass V =100(0.08206 L)(273.2) = 70.05L 32.00 Question A certain mass of neon gas is contained in a rigid steel container. The same mass of gas is added to this container. What will happen to the pressure in the container, assuming that the temperature stays constant? Question A certain mass of neon gas is contained in a rigid steel container. The same mass of Argon gas is added to this container. What will happen to the pressure in the container, assuming that the temperature stays constant? Say there is 1 mole of Neon, around 20 g. If we add 20g of Argon 20g x 1mol = 0.5 mol 39.95g New P = nRT/V = 1.5 (RT/V) 5.5 Dalton’s Law of Partial Pressures

For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each of the gases would exert if it were alone.

P total = P1 + P2 + P3 + …

P = nRT/V the total number of moles of particles is what is important not identity of gas 5.5 Dalton’s Law of Partial Pressures

P = nRT/V the total number of moles of particles is what is important not identity of gas

Note this is telling us that for ideal gases, the volume of the individual gas is not important in deciding the pressure exerted, and between particles do not influence the pressure either. Question 67 (p.221) Consider the flasks in the following diagram. What are the final partial pressures of H2 and N2 after the stopcock between the two flasks is opened? (Assume the final volume is 3.00L.) What is the total pressure (in torr)?

0.200 atm 475 torr 5.6 Kinetic Molecular Theory of Gases

Postulates 1. The particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero). 5.6 Kinetic Molecular Theory of Gases

Postulates 2. The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas. 5.6 Kinetic Molecular Theory of Gases

Postulates 3. The particles are assumed to exert no forces on each other; they are assumed neither to attract not to repel each other. 5.6 Kinetic Molecular Theory of Gases

Postulates 4. The average kinetic of a collection of gas particles is assumed to be directly proportional to the temperature of the gas. 5.6 Kinetic Molecular Theory of Gases

How well do the predictions of the KMT (model) fit the experimental observations?

Pressure and Volume P = (nRT) 1/V KMT says if volume decreased, particles will collide with container more often, P increases. 5.6 Kinetic Molecular Theory of Gases How well do the predictions of the KMT (model) fit the experimental observations?

Pressure and Temperature P = (nR/V) T KMT says if temp. increased, particles will collide with container more often as they speed up, P increases. 5.6 Kinetic Molecular Theory of Gases How well do the predictions of the KMT (model) fit the experimental observations? Volume and Temperature V = (nR/P) T KMT says if temp. increased, particles will collide with container more often as they speed up, and so only way to keep P constant is to increase volume. 5.6 Kinetic Molecular Theory of Gases How well do the predictions of the KMT (model) fit the experimental observations? Volume and number of moles V = (RT/P) n KMT says if number of moles of gas increases, particles will collide with container more often, and so only way to keep P constant is to increase volume. 5.6 Kinetic Molecular Theory of Gases How well do the predictions of the KMT (model) fit the experimental observations? Mixtures of gases Total pressure exerted by a mixture of gases is the sum of the pressures of the individual gases - as expected because KMT assumes all gas particles are independent of each other and the the of individual particles are unimportant. Maxwell-Boltz mann Distributions

As a result of all collisions that occur between gas particles - stop them having a straight path from A to B Maxwell-Boltz mann Distributions

As temp. Increases, both the average velocity and the spread of velocities increases The peak of the curve reflects the most probable velocity (velocity found most often as we sample the movement of the various particles in the gas). 5.7 Effusion and Diffusion Diffusion - mixing of gases. Rate of diffusion is rate of mixing of gases.

Effusion - movement of gas through a tiny orifice into an evacuated chamber Graham’s Law 5.7 Effusion and Diffusion Effusion - movement of gas through a tiny orifice into an evacuated chamber Graham’s Law

Rate of effusion for gas 1 = √ M2

Rate of effusion for gas 2 √ M1

Follows from KMT (using root mean sq. velocity) Diffusion https://www.youtube.com/watch?v=Rf9j0ztzcs4

Distance traveled by ammonia = urms NH3

Distance traveled by HCl urms HCl

= M HCl = 36.5 = 1.5

MNH3 17 Observed ratio is less than 1.5 -quantitative analysis of diffusion requires more analysis. 5.8 Real gases Ideal gas behavior can best be thought of as the behavior approached by real gases under certain conditions. Many gases come close at low pressures and/or high temperatures. Why do you think this is what we observe? 5.8 Real gases In a the atoms or molecules have finite volumes - so the volume available to a given particle in a real gas is less than the volume of the container because the gas particles themselves take up some of the space. 5.8 Real gases Van der Waals, took volume of gas particles into account using a correction factor nb (n was number of moles, b constant found by experiment. P’ = nRT V - nb 5.8 Real gases Van der Waals, took attractions among particles into account

Pobs = (P’ - correction factor) = nRT - correction factor V - nb 5.8 Real gases Van der Waals, took attractions among particles into account

Pobs = (P’ - correction factor) = nRT - correction factor V - nb

Pobs = nRT - a n 2 V-nb V 5.8 Real gases 2 nRT = [ Pobs + a (n/v) ] (V - nb)

Experimental studies indicate that the changes van der Waals made in the basic assumptions of the KMT correct the major flaws in the model.

5.8 Real gases Van der Waals Equation 2 nRT = [ Pobs + a (n/v) ] (V - nb)

Close to ideal greater deviations Low pressure vs. High Pressure High T vs. Low T 5.8 Real gases Van der Waals Equation 2 nRT = [ Pobs + a (n/v) ] (V - nb)

Close to ideal greater deviations Small molecules vs. large molecules Non-polar molecules vs. polar molecules Question Which of the following gases is expected to behave most ideally at a given temperature and pressure?

H2O CO2 N2 Xe

NH3 CH4 Cl2 He 5.10 Chemistry in the Atmosphere

N2 0.78084 Sea level

O2 mole fraction 0.20948

What will the at sea level and 22℃ be? 5.10 Chemistry in the Atmosphere

N2 mole fraction 0.78084 Sea level

O2 mole fraction 0.20948

Estimate the density of air at sea level and 22℃. 5.10 Chemistry in the Atmosphere

N2 mole fraction 0.78084 Sea level

O2 mole fraction 0.20948

Estimate the density of air at sea level and 22℃ V = 1mol 0.08206 Latmmol-1K-1 (273+22)K = 1atm 5.10 Chemistry in the Atmosphere

N2 mole fraction 0.78084 Sea level

O2 mole fraction 0.20948

Estimate the density of air at sea level and 22℃ V = 1mol 0.08206 Latmmol-1K-1 (273+22)K =24.21 L 1atm Mass = 5.10 Chemistry in the Atmosphere Density of air at sea level and 22℃ V = 1mol 0.08206 Latmmol-1K-1 (273+22)K =24.21 L 1atm Mass = (0.78 x 28g) + ((0.21 x 32g) = 5.10 Chemistry in the Atmosphere Density of air at sea level and 22℃ V = 1mol 0.08206 Latm mol-1K-1 (273+22)K =24.21 L 1atm Mass = (0.78 x 28g) + ((0.21 x 32g) = 29g

Density = mass /volume = 29g/ 24 L = 1.2 g/L