Chapter 5: Gases

Density of Gas

We can calculate the density of a gas by rearranging the ideal gas law so that we have n/V on one side of the equation and them multiplying by the molar mass, so that we obtain units of density (g/L).

Real Gases Formula

[P + a(n/V)^2] x [V - nb] = nRT Correction for intermolecular forces: [P + a(n/V)^2] Correction for particle volume: [V - nb]

Pressure conversions

1atm = 101,325 Pa 1atm = 760 torr = 760 mmHg 1atm = 14.7 psi 1atm = 29.92 inHg

Density of Gas Formula

D = PM/RT At standard T and P, D = molar mass (g/mol)/ molar volume (L/mol) = molar mass (g/mol)/ 22.4 L/mol

Unknown: Pressure (P)

Equation: P= nRT/V Known variables: Amount, temp., volume

Unknown: Temperature (T)

Equation: T = PV/nR Known variables: Pressure, volume, amount

Unknown: Volume (V)

Equation: V = nRT/P Known variables: Amount, temp., pressure

Unknown: Amount (n)

Equation: n= PV/RT Known variables: Pressure, volume, temp.

Real Gases

Gases behave ideally when both of the following 2 conditions are met: (1) the volume of the gas particles is negligible compared to the space between them (2) the forces between the gas particles are not significant

Gases in Chemical Reactions

The number of moles of a gas are found using and rearranging the ideal gas law: n = PV/RT

Mixtures of Gases

The pressure of a single gas (a) in a mixture of gasses is called its partial pressure (Pa). The pressure of a single gas in a mixture is calculated using the ideal gas law and the number of moles of the respective gas.

Simple Gas Laws

The simple gas laws discover the relationship of pressure, temperature, volume and amount of gas.

Dalton's law of partial pressures

The sum of the partial pressures of all the gases in the mixture equals the total pressure P total = Pa + Pb + Pc = n(Total)RT/V

Standard Temperature and Pressure

Standard temperature = 273.15 K = 0°C Standard pressure = 1 atm The volume occupied by one mole of an ideal gas at STP is called molar volume. One mole of any gas occupies the same volume at STP = 22.4L

Charles' Law Example Problem 2: The temperature inside my refrigerator is about 4o Celsius. If I place a balloon which is initially at 22o C and a volume of 0.5 liters in my fridge, what will the volume of the balloon be, after it is fully cooled by my refrigerator?

Step 1: List the known and unknown variables: V1 = 0.50L T1 = 22°C V2 = ? T2 = 4.0°C Step 2: Make sure that the variables are in the appropriate units In this case, T needs to be converted to Kelvins T1 = 22°C + 273.15 K = 295.15 K T2 = 4°C + 273.15 K = 277.15 K Step 3: Identify the simple gas law that would be appropriate to solve the problem, given the information. Charles' Law: V1/T1 = V2/T2 Step 4: Solve: 0.50 L/ 295.15 K = V2/ 277.15 K V2 = 0.5 L/ 295.15 K x 277.15 K = 0.47 L

Guy-Lassac's Law Example Problem 4. A rigid plastic container holds 1.00 L methane gas at 0.9 atm pressure when the temperature is 22.0°C. What pressure would the gas exert if the temperature is raised to 44.6°C?

Step 1: List the known and unknown variables: P1 = 0.9 atm T1 = 22.0°C P2 = ? T2 = 44.6°C Step 2: Make sure that the variables are in the appropriate units In this case, T needs to be converted to Kelvins T1 = 22°C + 273.15 K = 295.15 K T2 = 44.6°C + 273.15 K = 317.75 K Step 3: Identify the simple gas law that would be appropriate to solve the problem, given the information. Guy-Lassac's Law: P1/T1 = P2/T2 Step 4: Solve: 0.9 atm/ 295.15 K x P2/ 317.75 K 0.9 atm/ 295.15 K x (317.75) = 1 atm

Boyle's Law Example Problem 1: A balloon contains 7.2 L of He. When the pressure is reduced to 2.00 atm the balloon expands to occupy a volume of 25.1 L. What was the initial pressure exerted on the balloon?

Step 1: List the known and unknown variables: P1 = ? V1 = 7.2 L P2 = 2.00atm V2 = 25.1L Step 2: Make sure that the variables are in the appropriate units. Step 3: Identify the simple gas law that would be appropriate to solve the problem, given the information. Boyle's law: (P1)(V1)=(P2)(V2) Step 4: Solve: P1 = (7.2 L) = (2.00 atm)(25.1 L) P1 = (2.00 atm)(25.1 L)/(7.2 L) = 7.0 atm

Avogadro's Law Example Problem 3: A 25.5 L balloon holding 3.50 moles of carbon dioxide leaks. If we are able to determine that 1.90 moles of carbon dioxide escaped before the container could be sealed, what is the new volume of the container?

Step 1: List the known and unknown variables: V1 = 25.5L n1 = 3.50 mol (initially) V2 = ? n2 = 1.90 mol (leaked), = 1.60 mol (remaining in ballon) Step 2: Make sure that the variables are in the appropriate units. Step 3: Identify the simple gas law that would be appropriate to solve the problem, given the information. Avogadro's Law: V1/n1 = V2/n2 Step 4: Solve: 25.5 L/ 3.50 mol = V2/ 1.60 mol V2 = 25.5 L/ 3.50 mol x (1.60 mol) = 11.7 L

Ideal Gas Law Example Problem: The pressure exerted by 2.8 moles of argon gas at a temperature of 85oC is 420 torr. What is the volume of this sample?

Step 1: List the known variables: n= 2.8 moles T= 85°C P = 420 torr. Step 2: List the unknown variable: V=? Step 3: Check that the known variables are expressed in the appropriate units. If not, convert them. T= 85°C - need to convert to Kelvin . K= °C + 273.15 = 85°C + 273.15 = 358.15K Step 4: Substitute into the equation for volume and solve, using the units as guides and canceling them out. V = nRT/P (2.8 mol)(0.08026 atm·L/mol·K)(358.15 K) / 0.553 atm = 148.8 L

Units of Pressure

The SI unit of pressure is pascal (Pa). 1Pa = 1N/m^2 Common units of pressure include atm, mmHg, torr

Ideal Gas Law

The ideal gas law is an equation that relates the volume, temperature, pressure and amount of gas particles to a constant.

Boyle's Law

*Law Definition* For a given amount of gas at constant temperature, the volume of a gas varies inversely with pressure *Constant Properties* n, T = const *Law Formula* (P1)(V1)=(P2)(V2)

Guy-Lassac's Law

*Law Definition* The pressure of a given amount of gas is directly proportional to the Kelvin temperature if the volume is kept constant. *Constant Properties* n, V = const *Law Formula* P1/T1 = P2/T2

Charles' Law

*Law Definition* The volume of a fixed amount of gas is directly proportional to its Kelvin temperature if the pressure is kept constant. *Constant Properties* n, P = const *Law Formula* V1/T1 = V2/T2

Avogadro's Law

*Law Definition* The volume of a gas is directly proportional to the number of moles of gas present in the sample at constant temperature and pressure. *Constant Properties* P, T = const *Law Formula* V1/n1 = V2/n2

Problem 10. A gas mixture in a 1.55L container at 298K contains 10.0 g Ne and 10.0 g Ar. Calculate the partial pressures (in atm) of Ne and Ar, and the total pressure of the mixture. Calculate the mole fraction of each gas.

Given: V = 1.55L, T = 298K, 10.0g Ne, 10.0g Ar Find: PNe, PAr, PTotal, XNe, XAr Strategy: PNe, PAr --> PTotal (PNe + PAr) XNe = Mol Ne/Mol Ne + Mol Ar Mol (Ne, Ar) --> PTotal <-- Total Mol --> XNe, XAr PNe = XNe x PTotal Solution: 10.0g Ne x 1 mol Ne/20.18g Ne = 0.49554 mol Ne 10.0g Ar x 1 mol Ar/ 39.95g Ar = 0.25031 mol Ar PTotal = (0.49554 mol Ne) + (0.25031 mol Ar) = 0.74585 mol PTotal = (0.74585 mol)(0.08206l*atm/K*mol)(298K)/1.55L = 11.8P XNe = 0.49554 mol / 0.74585 mol = 0.664 XAr = 0.25031 mol / 0.74585 mol = 0.336 PNe = (0.664)(11.8 atm) = 7.84 atm PAr = (0.336)(11.8 atm ) = 3.96 atm PTotal = 7.84 atm + 3.96 atm = 11.8 atm PNe = 7.84 atm PAr = 3.96 atm PTotal = 11.8 atm XNe = 0.664 XAr = 0.336

Dalton's law of partial pressures Example: At depths of about 350 ft, deep-sea divers are subject to a pressure of approximately 10 atm. A typical gas cylinder used for such depths contains 51.2 g of O2 and 326.4 g of He and has a volume of 10.0 L. What is the partial pressure of each gas at 20.00°C, and what is the total pressure in the cylinder at this temperature?

Given: T = 20.00°C --> 293.15 K M (O2) = 51.2 g M (He) = 326.4 g V = 10.0 L Find: P(He), P(O2), P(total) Strategy: 1) Calculate the number of moles of He and O2 present. 2) Use the ideal gas law to calculate the partial pressure of each gas. Then add together the partial pressures to obtain the total pressure of the gaseous mixture. Solution: n(He) = 326.4 g/ 4.003 g/mol = 81.54 mol n(O2) = 51.2 g/ 16.00 g/mol = 1.60 mol P(He) = n(He)RT/V P(He) = (81.54 mol)(0.08026 atm·L/mol·K)(293.15 K)/ 10.0 L = 196.2 atm P(O2) = n(O2)RT/V P(O2) = (1.60 mol)(0.08026 atm·L/mol·K)(293.15 K)/ 10.0 L = 3.85 atm P(Total) = P(He) + P(O2) 196.2 atm + 3.85 atm = 200.1 atm

Density of Gas Example: Calculate the density of xenon gas at 45 °C and a pressure of 742 mmHg.

Given: T = 45°C --> 318.15 K P = 742 mmHg --> 0.976316 atm Find: D (Xe) Strategy: Use the equation for density and molar mass of Xe from the periodic table Molar Mass = 131.29 g/mol Solution: D = PM/RT (0.976136 atm)(131.29 g/mol)/ (0.08026 atm·L/mol·K)(318.15 K) = 4.91 g/L

Gases in Chemical Reactions Example: In the following reaction, 4.58 L of O2 is formed at 745 mmHg and 308K. How many grams of Ag2O decomposed? 2Ag2O(s) --> 4Ag(s) + O2(g)

Given: V= 4.58 L O2 T = 308 K P = 745 mmHg --> 0.9803 atm Find: M Ag2O Strategy: Calculate the moles of Ag2O from the number of moles of oxygen gas (using the stoichiometric ratio). The moles of O2 are obtained using the ideal gas law. Solution: n(O2) = PV/ RT (0.9803 atm)(4.58 L)/ (0.08026atm·L/mol·K)(308 K) = 0.1776 mol O2 0.1776 mol O2 x 2 mol Ag2O/ 1 mol O2 x 231.74 g Ag2O/ 1 mol Ag2O = 82.3 g Ag2O

Problem 1. The pressure in Denver, CO averages about 0.832 atm. Convert this pressure to mmHg and psi.

Given: 0.832 atm Find: mmHg, psi Strategy: Conversion Solution: 0.832 atm x 760 mmHg/ 1 atm = 632 mmHg 0.832 atm x 14.7 psi/ 1 atm = 12.2 psi

Problem 2. Convert 30.44 inHg to psi.

Given: 30.44 inHg Find: psi Strategy: Conversion Solution: 30.44 inHg x 1 atm/ 29.92 inHg x 14.7psi/ 1 atm = 15.0 psi

Pressure

It is the result of the collision of the gas particles. The pressure therefore depends on the number of particles in a given volume, as well as on the temperature the gas is at.

Ideal Gas Law Equation

PV = nRT P = pressure (atm) V = volume (L) n = amount (mol) R = 0.08206 atm·L/mol·K T = temperature (K)

Pressure formula

Pressure = force/area

The Kinetic Molecular Theory

• A gas is modeled as a collection of particles, whose behavior is independent of attractive and repulsive forces • A particle moves in a straight line until collision (with wall or other particle) • has 3 main assumptions: o The size of a particle is negligibly small o The average kinetic energy of a particle is proportional to the temperature (K) o The collision of one particle with another (or with the walls) is completely elastic