Chapter 5 Gases

Chapter 5

Gases

The Nobel Gases Exist as Monatomic Gases He, Ne, Ar, Kr, Xe, Rn

Diatomic Gases are H2, N2, O2, F2, Cl2. Oxygen also exists as the gas Ozone = O3

Compound GasesIonic Compounds do not exist as gases at standard

temperature and pressure (STP = 0 oC and 1 atm). This is because the cations and anions are held together by very strong electrostatic forces. To overcome these forces, we would have to raise the temperature well above 1000 oC.

Molecular Compounds vary but their boiling points are much lower. Most still exist as solids or liquids at STP. Some common molecular compounds that are gases at STP are the following:HF, HCl, HBr, HI, CO, CO2, NH3, NO, NO2, N2O,

SO2, H2S, HCNMost of these gases are colorless, except F2 (greenish

yellow), Cl2 (greenish yellow), and NO2 (brown)

Physical Characteristics of Gases

• Gases assume the volume and shape of their containers.

• Gases are the most compressible of the states of matter.

• Gases will mix evenly and completely when confined to the same container.

• Gases have much lower densities than liquids and solids.

The SI Unit of PressureWe need to derive pressure from the 7 SI base units.First, we derive force:Force = mass x accelerationForce = kilogram x m/s2

= 1 newton (N) = the SI unit for forceNext, we derive pressure:Pressure = force / area

= N / m2

= (kg x m/s2)/m2

= 1 pascal (Pa)Pascal is the SI unit for pressure

= 760 mm Hg

Instrument used to measure the pressure of the air in the atmosphere.

Used to measure pressures below atmospheric pressure.

The closed-tube manometer The open-tube manometer

Used for measuring pressures equal to or greater than atmospheric pressure.

Boyle’s Law: P a 1/V P = k1 / V and PV = k1

(k is the proportionality constant k1 = nRT)

At constant temperature

Using Boyle’s LawIn Boyle’s Law: PV=k1 and k1 is the proportionality constant

therefore it will be equal to nRT. For this reason, at constant temperature, a given sample of

gas (with the same number of moles = n) at two different sets of conditions (different volumes and pressures) will follow the following equation:

P1V1= P2V2 Example: An inflated balloon has a volume of 0.55 L at sea

level (1.0 atm) and is allowed to rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that the temperature remains constant, what is the final volume of the balloon?

P1V1= P2V2 => V2 = (P1 / P2) x V1

(1.0 atm / 0.40 atm) x 0.55 L = 1.4 L

At Constant Pressure

V a T

V a T V = k2T

V/T = k2

k2 = nR/P

P a T P = k3 TP/T = k3

k3 = nR/V

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Absolute Zero = -273.15 oC = 0 Kelvin

At Absolute Zero, there is no volume for the gases.

Volume is directly proportional to the Temperature in Kelvin.

Meaning if you doubled the temp, the volume doubles!

Using Charles’s LawIn Charles’s Law: V1 / T1 =k2 and k2 is the proportionality

constant therefore it will be equal to nR / P. For this reason, at constant pressure, a given sample of gas

(with the same number of moles = n) at two different sets of conditions (different volumes and temperatures) will follow the following equation:

V1 / T1 = k2 = V2 / T2

Example: A 452-mL sample of fluorine gas is heated from 22 degrees Celsius to 187 degrees Celsius at constant pressure. What is its final volume?

V2 = V1 x (T2 / T1)= 452 mL x (460 K / 295 K)

= 705 mL

V a n

V = k4n

k4 = RT/P

Gas Law Summary So Far:Boyle’s Law: V a 1/P (at constant n and T)Charles’s Law: V a T (at constant n and P)

Avogadro’s Law: V a n (at constant P and T)We can combine all three expressions to form a single master

equation for the behavior of gases:V a nT / P

= R (nT / P)or PV = nRT

where R is the proportionality constant called the gas constant.

The equation PV = nRT is called the ideal gas equation or ideal gas law. It describes the relationship among the four

variables P, V, T and n.

The Ideal Gas Law PV = nRTAn ideal gas is a hypothetical gas whose pressure-

volume-temperature behavior can be completely accounted for by the ideal gas equation.

The molecules of an ideal gas do not attract or repel one another, and their volume is negligible

compared with the volume of the container. Although there is no such thing as an ideal gas, discrepancies in the behavior of real gases over

reasonable temperature and pressure ranges do not significantly affect calculations. Therefore, we can safely use the ideal gas law equation to solve many

gas problems.

The Ideal Gas Law Constant, RAt Standard Temperature and Pressure (STP) – O oC (273.15 K) and 1 atm pressure, many real gases behave

like an ideal gas.Experiments show that under these conditions, 1 mole of gas occupies 22.414 L, which is somewhat greater

than the volume of a basketball.From PV=nRT we can write:

R = PV / nTR = (1 atm)(22.414 L) (1 mol)(273.15 K)R = 0.082057 L x atm

K x mol

Using the Ideal Gas LawExample: Sulfur hexafluoride (SF6) is a colorless, odorless,

very unreactive gas. Calculate the pressure (in atm) exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 L at 69.5 oC.

P = nRT / VP = (1.82 mol) (0.0821 L x atm/K x mol) (69.5 + 273)K

5.43 LP = 9.42 atm

Try this problem on your own:

Calculate the volume (in liters) occupied by 2.12 moles of nitric oxide (NO) at 6.54 atm and 76 oC.

Molar Volume at STP Gas CalculationsExample: Calculate the volume (in liters) occupied by 7.40 g of NH3 at STP.

Remember that 1 mole of an ideal gas at STP occupies 22.4 Liters.

V = 7.40 g NH3 x 1 mol NH3 x 22.4 L

17.03 g NH3 1mol NH3

V = 9.74 L

When conditions change for P, V, T and n we must use a modified form of the ideal gas equation that includes the final and initial conditions.

R = P1V1 (before change)

n1T1

R = P2V2 (after change)

n2T2

So that…. P1V1 = P2V2

n1T1 n2T2

If n1=n2 then, P1V1 = P2V2

T1 T2

Modified Ideal Gas Law

Using the Modified Ideal Gas LawExample: A small bubble rises from the bottom of a lake, where the

temperature and pressure are 8 oC and 6.4 atm, to the water’s surface, where the temperature and pressure is 25 oC and 1.0 atm. Calculate the final volume (in mL) of the bubble if its initial volume was 2.1 mL.

Note that the moles do not change in this question. Therefore we can use the modified ideal gas law.

P1V1 = P2V2 rearranged => V1P1 T2 = V2

T1 T2 P2T1

Therefore… 2.1 mL x 6.4 atm x 298 K

1.0 atm x 281 K

V2 = 14 mL

Electric Lightbulbs are usually filled with argon gas.

Using the Modified Ideal Gas LawExample: Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain light bulb containing argon at 1.2 atm and 18 oC is heated to 85 oC at constant volume. Calculate its final pressure (in atm). P1V1 = P2V2 rearranged => P1 = P2

T1 T2 T1 T2

rearranged => P2 = P1x T2

T1

P2 = 1.20 atm x 358 K 291 K

P2 = 1.48 atm

Density CalculationsIf we rearrange the ideal gas equation, we can calculate

the density of the gas:n = P

V RTWe can find the number of moles (n) by:

n = mass (g = m) molar mass (g/mol =M )

Therefore, plugging the above in for n in the ideal gas law, we get:

m = P M x V RTWhich rearranges into: d = m = P x M

V R x T

Finding the Molar Mass, M, of a Gas Many times we do not know the molar mass, M, of a

gas. From our previous slide, we know that we can find the density of a gas from:

density = m = P x M V R x T

We can rearrange this equation in order to solve for the molar mass, M, of a gas:

M = d x R x T P

Example: A chemist has synthesized a greenish-yellow gaseous compound of chlorine and oxygen and finds that its density is 7.71 g/L at 36 oC and 2.88 atm. Calculate the molar mass of the compound and determine its molecular formula.

A Combined Gas Sample ProblemChemical analysis of a gaseous compound showed that it contained 33.0 % silicon and 67.0 % fluorine by mass. At 35 oC, 0.210 L of the compound exerted a pressure of 1.70 atm. If the mass of 0.210 L of the compound was 2.38 g, calculate the molecular formula of the compound. (R = 0.0821 L x atm/K x mol)Calculate the empirical formula: SiF3

Calculate the molecular molar mass: M = d x R x T = 169 g/mol

P The empirical molar mass of SiF3 = 85.09 g/mol.

169 / 85 = 2Therefore the molecular formula is Si2F6

Gas StoichiometryWhen you notice that you have a gas involved in a chemical reaction, the rules for stoichiometry change.

You will still convert the amount of reactants to the amount of products using moles, but remember that moles (n) vary depending on PV = nRT for gases.To solve Gas Stoichiometry problems you must read the question and:

1) Identify whether the problem states that the gas is at STP = 0oC and 1 atm. If this is the case remember that 1 mole (n) of any gas will occupy 22.4 Liters.

2) If the problem states that each gas is not at STP, then you will have to determine the moles for each gas given based on PV = nRT for each gas.

Gas Stoichiometry Sample ProblemCalculate the volume of O2 (in liters) at STP required for the complete combustion of 2.64 L of acetylene (C2H2) at STP:

2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

2.64 L C2H2 ? L O2

2.64 L C2H2 1 mol C2H2 = .12 mol C2H2

22.4 L C2H2

.12 mol C2H2 5 mol O2 22.4 L O2 = 6.6 L O2

2 mol C2H2 1 mol O2


Sodium azide (NaN3) is used in some automobile air bags. The impact of a collision triggers the decomposition of NaN3 as follows:

2 NaN3(s) 2Na(s) + 3N2(g)

The nitrogen gas produced quickly inflates the bag between the driver and the windshield. Calculate the volume of N2 generated at 80 oC and 823 mmHg by the decomposition of 60.0 g of NaN3.

2 NaN3(s) 2Na(s) + 3N2(g)

60.0g NaN3 ? Volume N2

60.0 g NaN3 1 mol NaN3 3 mol N2 = 1.38 mol N2

65.02 g NaN3 2 mol NaN3

V = nRT = (1.38 mol) x (0.0821 L x atm) x (80 + 273)K = 36.9 L N2

P (K x mol) x (1.08 atm)

(Note : 760 mmHg = 1 atm and R = 0.0821 L x atm / K x mol)


Dalton’s Law of Partial PressuresThe total pressure of a mixture of gases is just the sum of the

pressures that each gas would exert if it were present alone. Therefore:

The pressure exerted by gas A and B is: PA = nART and PB = nBRT

V VPT = PA + PB = nART + nBRT = RT (nA + nB)

V V VXi is the mole fraction of gas i, within a mixture of gases:

Xi = ni

ntotal

The partial pressure, Pi, of any substance, i, is equal to the mole fraction of the substance multiplied by the total pressure of a mixture of gases: Pi = (Xi)(PT)

Whenever Gas is collected over Water Use Dalton’s Law of Partial Pressure to Figure out the Pressure of the Gas Being

Collected. Always look for this in problems!!!!

2KClO3(s) 2KCl(s) + 3O2(g)

PT = PO2 + PH20

PO2 = PT - PH20


P a 1

V

Scuba Diving and Partial PressuresThe composition of the air that we breathe is 20% oxygen and 80% nitrogen gas by

volume. To much oxygen can actually be harmful to us.At sea level, the pressure of the air around us is 1 atm. This means that the pressure of

the oxygen we breathe is : PO= .20 moles x 1 atm = 0.2 atm = PO

(.20 + .80) moles(Remember that moles (n) a V therefore we can replace the volume with moles.)

Our bodies are therefore used to breathing in 0.2 atm of pressure of oxygen. But, as a diver swims deeper into the ocean, at 100 feet the pressure increases to 4 atm and the oxygen pressure now becomes:

PO = .20 moles x 4 atm = 0.8 atm = PO

(.20 + .80) molesThis pressure of oxygen is much too high, therefore the oxygen needs to be diluted in

order to keep it at a level of 0.2 atm. For this to happen, the oxygen must be diluted to a level of 5%.

Nitrogen gas is not used to dilute the oxygen down because at a pressure of 1 atm, nitrogen gas in the lungs will start to dissolve into the blood. This causes nitrogen narcosis. This is also responsible for the bends. As a diver surfaces, the nitrogen dissolved in the blood transforms into bubbles of nitrogen gas and can stop blood flow and impair the nervous system.

Helium is used instead to dilute oxygen down and a special valve automatically adjusts for the pressure of oxygen. Helium is used because it is an inert gas and it has a low solubility in blood.

The Kinetic Molecular Theory of an Ideal Gas

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). They can be considered as “points”, where they posses mass but have zero volume.

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. Collisions among molecules are perfectly elastic, in other words, energy can not be transferred from one molecule to another as a result of a collision, the total energy of all the molecules in a system remains the same.

3. Gas molecules exert neither attractive nor repulsive forces on one another.

4. The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas. Any two gases at the same temperature will have the same average kinetic energy.

The Kinetic Molecular Theory and It’s Applications to the Gas Laws

Boyle’s Law: V a 1/P (at constant n and T)

Charles’s Law: V a T (at constant n and P) P a T (at constant n and V)

Avogadro’s Law: V a n (at constant P and T)

The Average Kinetic Energy of a Molecule

Remember the term Kinetic Energy (KE)? It means molecules in motion. And do you remember that

temperature is the average kinetic energy?The average KE for one molecule in a gas is represented

by:KE = ½ mu2

Where m = mass of one molecule and u = is speed.The term u2 = mean square speed. It is the average of the

square of the speeds of all of the molecules: u2 = u2

1 + u22 + u2

3 + …. + u2n

NN = the number of molecules.

Maxwell Speed Distribution CurvesThe kinetic theory of gases allows us to investigate molecular

motion in more detail.As long as we hold the temperature of a gas constant, the

average KE and the means square speed will stay constant for the same length of time that the temperature is held constant.

The motion of the molecules are totally random and unpredictable. At a given instant, different molecules within a sample are moving at different speeds.

How many molecules are moving at a particular speed? Maxwell Boltzman analyzed the behavior of gas molecules at different temperatures to answer this question. Maxwell speed distribution curves show gas at different temperatures and graph the number of molecules that move at a particular molecular speed.


Root-Mean-Square SpeedHow fast does a molecule move, on the average, at any

temperature T?One way to estimate molecular speed is to calculate the root-

mean-square (rms) speed (urms), which is an average molecular speed.

The Total KE of a mole of any gas = 3/2 RTBecause Average KE is only for one molecule of a gas:

NA (1/2mu2) = 3/2 RTBecause NAm = M , we can write:

u2 = 3RT / M Taking the square root of both sides gives us:

u2 = urms = 3RT / M


Diffusion• A direct demonstration of random motion is provided by diffusion, the

gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.

• Diffusion always proceeds from areas of higher concentrations to areas of lower concentrations.

• The diffusion process takes a long time to complete even if the molecular speeds are great. This is because molecules experience numerous collisions while moving from one end of the bench to the other.

• Also, because the root mean square speed of a light gas is greater than that of a heavier gas, a lighter gas will diffuse through a certain space more quickly than will a heavier gas.

• Because NH3 is lighter and therefore diffuses heavier than HCl, the solid NH4Cl first appears nearer the HCl bottle on the right.

• Calculating the ratio of the effusion rates of two gases or the distance traveled for two gases:(Distance traveled/Effusion rate) for gas 1: urms = 3RT / M 1 = M 1

(Distance traveled/Effusion rate) for gas 2: urms = 3RT / M 2 M 2


Deviation from Ideal BehaviorIdeal gases are said to:1) Not exert any attractive or repulsive forces upon

one another.2) Have negligibly small volumes compared to that of

the container.To study real gases, the van der Waals equation is

used: (P + an2/V2) (V – nb) = nRT

Where a and b are van der Waals constants that can be looked up in an appendix or table.

Corrected Pressure Corrected Volume

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Chapter 5 Gases