Chemistry 068, Chapter 11. Gas Properties and Gas Laws This section deals with the relationships...

Chemistry 068, Chapter 11

Gas Properties and Gas Laws

• This section deals with the relationships between pressure, volume, temperature, and amount (moles) of gas phase reactions.

• We look at several different gas laws – gas laws are mathematical summaries of experimental observations about gas behavior.

• We begin by looking at the empirical gas laws (which come from experiment and observation), then the ideal gas law, and finally, mixtures of gases.

Properties of Common Gases

• The word gas is usually used to describe substances that are gases at room temperature and pressure.

• Vapor is used for substances that are in the gas phase but are normally not at room temperature and pressure.

• Of the element which are gases, all are diatomic other than the noble gases.

Properties of Common Gases (Cont’d)

• Gases have several properties that other states of matter do not have.– Gases can be easily compressed.– Gases spontaneously expand to fill a

container.– Gas density varies by pressure.

• You can easily relate gas volume, pressure, temperature, and amount (moles) using a single equation.


Element Color? Odor? Toxic?

H2 None None No

O2 None None No

N2 None None No

Cl2 Green/YellowChoking Yes

F2 Yellow Pungent YesHe None None NoNe None None NoAr None None NoKr None None NoXe None None NoRn None None No


Compound Color? Odor? Toxic?

CO2 None Faint Pungent NoCO None None Yes

NH3 None Pungent Yes

CH4 None None* No

SO2 None Pungent/Choking Yes

H2S None Rotten Eggs YesHCl None Choking Yes

NO2 Red-Brown Irritating Yes

*An odor is added to natural gas in homes for safety reasons. It has no scent without it.

Kinetic Theory of Ideal Gases

• Explains pressure as the bombardment of container walls by gas molecules.

• Differs from Newton’s earlier theory in which pressure was caused by repulsion between molecules pushing against the container.

• Though described as early as the 1700s, it was mainly developed in the nineteenth century by scientists such as James Clerk Maxwell and Ludwig Boltzmann.

Postulates of Kinetic Theory

• Gas molecules move randomly in straight lines in all directions and at various speeds.– Not all at the same speed.

• Attraction and repulsion forces between molecules are very weak or negligible, except during a collision.– Otherwise wouldn’t move is straight lines at steady

speeds.

– All collisions are perfectly elastic - they bounce off of one another unless a reaction occurs.

– Total kinetic energy constant, none is lost during collisions.

Postulates of Kinetic Theory (Cont’d)

• Gases are composed of molecules whose sizes are negligible compared to the average distances between molecules.– Molecules are very far apart.– Molecules must be much smaller than

intermolecular distances.• The average kinetic energy of a molecule is

proportional to the absolute temperature.– Defines temperature as relating to kinetic energy.

Kinetic Theory and Gas Properties

• Kinetic theory explains three main properties of gasses.– Gasses are compressible.

• Since most of a gas is empty space, gas molecules can be forced close together easily.

– Gasses assume the shape and volume of a container.

• Again, since gas molecules are far apart, they easily expand to fill the available area.

– Gasses have low density.• Since most of a gas is empty space, there is very

little mass per unit volume.

Pressure Measurement and Units

• Pressure affects gases far more than any other state of matter.

• P = force/area = F/A• SI unit is the pascal (Pa) and is defined as

1N/m2. Other common units include atmospheres (atm), Torr, milimeters of mercury (mmHg), and psi (pounds per square inch or lbs/in2).

• 1atm=101325Pa=760Torr=760mmHg=14.68psi• Most gas problems are solved using units of

atm.

Pressure Measurement and Units (Cont’d)

• Pressure is measured using barometers or manometers.– Barometers measure atmospheric pressure.– Manometers measure the pressure of gas in a sealed system.

• At sea level, a mercury barometer has a height of 760mm. Hence 1atm = 760mmHg.

• The pressure reading of a barometer or manometer is related by the equation:P = g*d*hWhere P is pressure, g is the acceleration of gravity (9.8m/s2), d is the density of the liquid, and h is the height of the column. If all units are in SI units, the pressure will have units of pascals.

Pressure and Significant Figures

• Special mention must be made of the rules for significant figures of mmHg (and Torr) measurements.– These rules only apply to mmHg and Torr.

• Normal pressures taken in mmHg are assumed to have at least 3 significant digits as they are measured to the nearest mmHg (or Torr).– Pressures significantly above or below 1 atm

use the normal significant figure rules.

Pressure Unit Conversion Problem

• Convert 2.50atm to torr, mmHg, Pa, and psi.

The Empirical Gas Laws

• Together, the next four laws – Boyle’s, Charles’s, Gay-Lussac’s, and Avogadro’s are called the empirical gas laws.

• They are empirical – that is, based upon experiment and observation, rather than exclusively on theory.

• They can be used separately, or combined to give the combined gas law (all but Avogadro’s combined) or the ideal gas law (all four combined).

The Empirical Gas Laws (Cont’d)

• The empirical gas laws were developed during the 17th through 19th centuries.

• They were the precursors to the modern ideal gas law.

• They were developed by holding constant all but two terms so that the relationship between those two terms could be determined.

Qualitative Explanation of the Gas Laws – Pressure

• Pressure is due to gas phase collisions.• Therefore molecules per volume and average

speed per collision (due to temperature) both increase frequency of collisions.

• Average speed (again, due to temperature) determines the force per collision.

• More collisions means more pressure.• More force per collision likewise means more

pressure.

Boyle’s Law (Pressure – Volume)

• Studied quantitatively by Robert Boyle in 1661, Boyle’s law describes the relationship between volume and pressure.

• At constant temperature and amount (moles), volume is inversely proportional to pressure.

V α 1/P or VP = constant• This can be reaaranged to give:

V1P1 = V2P2

Qualitative Explanation of the Gas Laws – Boyle’s Law

• In Boyle’s law, the temperature and number of moles is kept constant, and VP = constant.

• If you change volume, you change the molecules per volume and thus collision frequency.

• Pressure must therefore change accordingly.• A smaller volume means more molecules per

volume and thus a greater collision frequency, leading to a greater pressure.

Boyle’s Law Problems

• A partially filled helium balloon has a volume of 0.50L at 1.0 atm. What will its volume be high in the air where the pressure is only 0.10 atm?

Charles's Law (Temperature – Volume)

• In 1787, Jacques Alexandre Charles made some of the earliest quantitative measurements of gases at different temperatures. His work was continued by John Dalton (in 1801) and Jospeh Louis Gay-Lussac (in 1802).

• Their work showed that there is a linear relationship between volume and temperature.

• This can be expressed as:V α T or V/T = constant• This can be rearranged to give:V1/T1 = V2/T2

Qualitative Explanation of the Gas Laws – Charles’s Law

• In Charles’s law, the pressure and number of moles is kept constant, and V/T = constant

• If you change the temperature, you change the average speed, which changes the frequency and force of collisions, both of which would change pressure.

• To keep a constant pressure you must change the molecules per volume. Since the number of moles does not change, you must have a change in volume.

Charles's Law Problems

• The gas in a hot air filled balloon has a volume of 100.0L at 400K. What is its volume when it cools to room temperature (293K)?

Gay – Lussac’s Law (Temperature – Pressure)

• Jospeh Louis Gay-Lussac’s (1802) continuation of the work began by Charles culminated in a relationship between temperature and pressure.

• He showed that the pressure of a fixed amount of gas at fixed volume was directly proportional to its absolute temperature.

Qualitative Explanation of the Gas Laws – Gay-Lussac’s Law

• In Gay – Lussac’s law, the volume and number of moles is kept constant, and P/T = constant

• If you change the temperature, you change the average speed, which changes the frequency and force of collisions, both of which would change pressure.

• Since volume and amount are not changing, for P/T to remain constant, pressure must increase as temperature increases. This can be expressed as:

P α T or P/T = constant• This can be rearranged to give:P1/T1 = P2/T2

Gay – Lussac’s Law Problems

• A gas cylinder can withstand 10.0atm of pressure without bursting. At 300K, it has a pressure of 5.0atm. Will it burst if it is heated to 1000K?

Avogadro’s Law (Volume – Moles)

• In 1811, Amedeo Avogadro interprited earlier work by Jospeh Louis Gay-Lussac to arrive at a relationship between volume and amount (moles or molecules).

• He interpreted Gay-Lussac’s work as meaning that equal volumes at equal pressure and temperature were made up of equal numbers of molecules.

• This can be expressed as:V α n or V/n = constant• This can be rearranged to give:

V1/n1 = V2/n2

Qualitative Explanation of the Gas Laws – Avogadro’s Law

• In Avagadro’s law, the pressure and temperature is kept constant, and V/n = constant

• Constant temperature means that average speed is constant.

• If the number of moles is changed you change the molecules per volume and thus change pressure.

• To keep pressure constant you must change volume to restore the original number of molecules per volume.

Avogadro’s Law Problems

• A balloon holding 2.00 mole of argon has a volume of 1.00L under a particular pressure and temperature. What would the volume of 4.50 moles be under the same conditions?

The Combined Gas Law

• States that pressure times volume divided by temperature is a constant for fixed amount (moles).

• Combines Boyle’s, Charles’s, and Gay – Lussac’s laws into a single equation.

• It can be expressed as:

PV/T = constant

• This can be rearranged to give:

P1V1/T1 = P2V2/T2

Combined Gas Law Problems

• A sample of gas has a volume of 1.00L at 1.0atm and 270K. What is its volume at 0.75atm and 300K?

Combined Gas Law Problems (Cont’d)

• A vapor sample has a volume of 200.0mL at 1.2atm and 25oC. It is then compressed to 100.0mL at 6.0atm. What is its new temperature in oC?

Combined Gas Law Problems (Cont’d)

• A sample of gas occupies 10.0L at 6.0atm and 300K. What will the pressure be if it is compressed to 5.0L at 90K?

The Ideal Gas Law

• Combines the empirical laws into one equation.

• Defines and combines the constants to be the molar gas constant, R.

• Fairly accurate even for non – ideal (real) gases.

What is an Ideal Gas?

• Low pressure.

• High temperature.

• Molecules have no attraction or repulsion towards one another – there are no intermolecular forces.

• Gas molecules have negligible volume.

The Equation

• PV = nRT

• P = pressure, usually in atm.

• V = volume, usually in L.

• n = number of mols.

• R = molar gas constant.

• T = temperature, must be in Kelvin.

Constants and Units

• Many equations and problems involve using the molar gas constant, R.

• R has different numerical values depending on the units you use.

• For ideal gas law problems you will normally use:

R = 0.08205746 L*atm/(K*mol)= 62.364 L*mmHg/(K*mol)

• For kinetics and thermodynamics problems you will normally use:

R = 8.314472 J/(K*mol)• If you are unsure which to use, check the units.

Constants and Units (Cont’d)

• When working gas state problems it is important to make sure to use the correct units and constants.

• Temperature should always be in Kelvin.• Pressure units should match the gas

constant you are using. This usually means using units of atm.

• Likewise, volume units should also match. This usually means using L.

Ideal Gas Law Problems

• What is the volume of 50.0g of CO2 at 0.90atm and 300K?

Ideal Gas Law Problems (Cont’d)

• A 40.0g sample of oxygen occupies 2.0L at 16.0atm. What temperature is the sample at?

Gas Density and Molar Mass

• Recall that for gases, density will vary based upon what conditions the gas is under.

• The molar mass of a gas does not vary.

Gas Density and Molar Mass (Cont’d)

• Thus we can relate the molar mass of a gas as follows:

Molar Mass = M = mass/moles = m/n

or: n = m/M

• We can then substitute this into the ideal gas law and get:

PV = nRT PV = mRT/M

or: M = mRT/PV

Gas Density and Molar Mass (Cont’d)

• Recall that for density we can write:Density = d = mass/volume = m/V

• We can also write the ideal gas law as: PV = mRT/M

• Which we can write as:m/V = PM/RT

• Substituting in for density we get: d = PM/RT

• And also:M = dRT/P

Gas Density and Molar Mass Problems

• What is the density of oxygen gas (O2) at 1.0atm and 300K?

Gas Density and Molar Mass Problems (Cont’d)

• 2.50g of a gas occupy 2.0L at 2.0atm and 250K. What is its molar mass?


• A gas has a density of 1.00 g/L at 30oC and 1.5atm. What is the molar mass of the gas?

Gas Phase Mixtures and Dalton’s Law of Partial Pressures

• When describing gas phase mixtures pressures and mole fractions, rather than volumes, are commonly used to describe relative amounts.

• John Dalton showed that each gas within a mixture contributes to the total pressure of that mixture.– This is called Dalton’s law of partial pressures.– Each gas contributes to the total pressure as much

pressure as it would on its own.– Each individual pressure follows the ideal gas law.– Total pressure is related to total number of moles.

Gas Phase Mixtures and Dalton’s Law of Partial Pressures (Cont’d)

• Thus, the sum of partial pressures equals the total pressure. Each substance follows the ideal gas law.

Ptotal = P1 + P2 + P3 + …

P1V = n1RT

• The partial pressure of a substance is related to the mole fraction of the substance.– The mole fraction of a component in the mixture is the

fraction of the whole which the component represents.

Gas Phase Mixtures and Dalton’s Law of Partial Pressures (Cont’d)

• The mole fraction of A is given by:mole fraction A = XA = nA/ntotal

• The mole fraction of A relates to partial pressure by:XA = nA/ntotal = PA/Ptotal

Or:PA = XAPtotal

Partial Pressure and Mole Fraction Practice

• Calculate the mole fraction of A in each of the following mixtures:– 2.0mol A and 2.0mol B.

– 1.0mol A, 2.0mol B, 1.0mol C.

– 2.0mol A only.

Partial Pressure and Mole Fraction Practice (Cont’d)

• Calculate the total pressure of each of the following mixtures:– 2.0atm A and 1.0atm B.

– 1.0atm A, 2.0atmB, and 1.0atm C.

– 1.0atm A, 2.0atm C (no B).

Partial Pressure and Mole Fraction Practice (Cont’d)

• Calculate the partial pressure of A in each of the following mixtures:– 2.0mol A and 2.0mol B, Ptotal = 4.0atm.

– 1.0mol A and 3.0mol B, Ptotal = 12.0atm.

– 1.0mol A, 2.0mol B, and 3.0mol C, Ptotal = 6.0atm.

Partial Pressure and Mole Fraction Problems

• Calculate the total pressure of a mixture of 1.0mol O2 and 1.0mol N2 in a 2.5L flask at 150K.

Partial Pressure and Mole Fraction Problems (Cont’d)

• A mixture is 65% CO2, 25% O2, and 10% H2 all by volume. What is the partial pressure of each gas if the total pressure is 1.5atm?

Partial Pressure and Mole Fraction Problems (Cont’d)

• A mixture consists of 0.5mol CO2, 1.0mol O2, and 1.5mol H2. What is the partial pressure of each gas if the total pressure is 3.0atm?

Composition of (Dry and Clean) Air

Component XA PA (mmHg)*

N2 0.78084 593.4

O2 0.20948 159.2

Ar 9.34*10-3 7.1

CO2 3.1*10-4 0.24

Ne 2*10-5 0.02

He 1*10-5 0.01

*Total pressure of 760mmHg

Collection of Gas Over Liquid

• One application of partial pressures are problems involving vapor pressure.– Vapor pressure is the maximum partial pressure

exerted by the vapor of a liquid over the liquid. Vapor pressure varies with temperature.

• In these problems, a container holds liquid plus gas above the liquid, plus liquid vapor above the liquid.– Most commonly, the liquid is water.

• The total pressure in the container will be the sum of the pressure of the gas and the vapor pressure of the liquid.

Collection of Gas Over Liquid (Cont’d)

• Thus, Ptotal will be given by:

Ptotal = Pgas + Pliquid

Or:

Pgas = Ptotal – Pliquid

• Note that the vapor pressure of liquids varies with temperature.

Vapor Pressure of Water

T (oC) Pwater (mmHg)

15 12.8 16 13.6 17 14.5 18 15.5 19 16.5 20 17.5 21 18.7 22 19.8 23 21.1 24 22.4 25 23.8

T (oC) Pwater (mmHg)

26 25.2 27 26.7 28 28.3 29 30.0 30 31.8 31 33.7 32 35.7 33 37.7 34 39.9 35 42.2

Collection of Gas Over Liquid Problems

• Hydrogen gas is bubbled into a solution of water at 20oC. A pressure reading reads 760mmHg. What is the pressure of hydrogen?

Collection of Gas Over Liquid Problems (Cont’d)

• Helium gas is bubbled into a solution of water at 30oC. A pressure reading reads 150mmHg. The container has a volume of 0.5L. How many moles of helium were bubbled in?

Gas Phase Stoichiometry

• The ideal gas law can be used to calculate the pressure, volume, etc. due to gas formation that is part of a chemical reaction.

• Reaction stoichiometry will give you number of mols, just as it would for any other reaction.

Gas Volume and Chemical Reactions

• Since mole correspond to volume (all else being constant) you can correspond gas volumes reacting just like you can moles.– So if P, R, and T are constant you can use

volumes instead of moles.

• This is called Gay – Lussac’s law of combining volumes.

• Consider the following reaction:

A(g) + 2B(g) AB2(g)

Gas Volume and Chemical Reactions (Cont’d)


A(g) + 2B(g) AB2(g)

– If we react 1.0L of A with 2.0L of B we should produce 1.0L of AB2.

– Likewise, 0.5L of A reacting with 1.0L of B should produce 0.50L of AB2.

Gas Volume in Limiting Reagent Situations

• Just like moles, volumes can be limiting.– Again, assuming everything else is equal.

• These problems work the same way as other limiting reagent problems.

• Again, consider the following reaction:

A(g) + 2B(g) AB2(g)– If we react 1.5L of A with 2.0L of B we should produce

1.0L of AB2; and have 0.5L of A left over.

– Likewise, 0.5L of A reacting with 1.5L of B should produce 0.50L of AB2; and have 0.5L of B left over.

Gas Chemical Reaction Problems


A(g) + 2B(g) AB2(g)

– What volume of A and B is needed to produce 3.5L of AB2?

– How much, in liters, of A would be needed to completely react 1.5L of B?

Gas Chemical Reaction Problems (Cont’d)


CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

– What volume of is CO2 produced when 4.0L of CH4 is combusted with 6.0L of O2 at 1.0atm and 270K?

– Which reagent, if any, is limiting?

STP

• Standard temperature and pressure (STP).

• STP is a reference set of conditions used to compare gas samples.

• STP is defined as 1atm of pressure and a temperature of 0oC (or 273.15K).

• The molar gas volume (Vm) at STP is 22.413996L/mol; usually shortened to 3 or 4 digits.

The Molar Volume of a Gas

• The molar gas volume is the volume of one mole of gas at a specified temperature and pressure.– Molar volume has units of Liters/mole.– It is necessary to specify the temperature and

pressure as these will both affect the volume.– Often, the molar volume will be taken at STP.

• Unlike molar mass, the molar volume of all gas phase substances is the same under the same conditions.

Molar Volume Problems

• Calculate the molar volume of oxygen under each of the following sets of conditions.– STP.

– 1.00atm, 298K.

– 6.00atm, 675K.

Using Molar Volume to Get Density

• It is possible to use density and the molar volume to determine molar mass.

• This is because the molar mass over the molar volume is equal to the density.

• This is because:

d = m/V, M = m/n, and Molar Volume = V/n

Thus, molar mass/molar volume is equal to:

(m/n)/(V/n) = m/V = d


• Calculate the density of O2 under each of the following conditions:– STP.

– 1.00atm, 298K.

– 6.00atm, 675K.

Molar Volume Chemical Calculation Problems

• What is the volume of 12.0g of nitrogen gas at STP?

Molar Volume Chemical Calculation Problems (Cont’d)

• What is the mass of 2.50L of NO2 at STP?



2KNO3(s) 2KNO2(s) + O2(g)

– What volume of O2 will form from 100.0g of KNO3 reacting at 375K and 3.0atm?

– What mass of KNO3 is needed to produce 5.0L of oxygen gas at STP?



O2(g) + 2H2(g) 2H2O(g)

– What volume of H2O vapor forms from 10.0L of O2 and 12.0L of H2?

– Which reagent, if any, is limiting?

Chemistry 068, Chapter 11. Gas Properties and Gas Laws This section deals with the relationships...

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Transcript of Chemistry 068, Chapter 11. Gas Properties and Gas Laws This section deals with the relationships...