Ch 10 Gases

Gases are made up of particles that may be single atoms (He, Ne) or molecules (O2, CH4). Although gases may have different chemical properties, they all have very similar physical properties – which are what we study in this chapter.

All gases:have massare easily compressedexpand to fill their containersdiffuse through each other easily (form

homogeneous mixtures)

exert pressure on the walls of their container

Most common gases have low molecular weights and are Composed of only non-metals.

Vapor – word used to describe substances that are liquids or

solids under normal conditions, but can exist in thegaseous state.

Kinetic – Molecular Theory – explains behaviors of gases by describing behavior of its particles. Parts of KM theory:

a. A gas consists of small particles that have mass

b. Particles in a gas are separated from each other by relatively large distances (compression results from decreasing the distance between particles)

c. Particles of a gas are in constant, rapid, random motion. (explains how they fill container)

d. Collisions of gas particles with the walls of the container are perfectly elastic – meaning they do not lose any kinetic energy in the collision. The average kinetic energy of all particles stays the same. (assuming constant temp)

e. Gas particles that collide with the walls of the container or an object exert pressure (force per unit area). The magnitude of the pressure

depends on the how often and with what force the particles strike the walls.

So….Assuming a constant volume, increasing the number of gas particles will increase pressure; increasing temp (KE) will increase pressure

f. The average KE of a gas depends ONLY on its temp. Gas particles speed up with increased temp and slow down with decreased temp.

( Kinetic Energy = ½ mv2) Doubling the temperature will double the

KE Halving the temp. will halve the KE

g. Gas particles have essentially no attraction or repulsion to each other.

Atmospheric pressure: pressure exerted by the gases in the

atmosphere.1. caused by gravity’s attraction to the

mass of gas molecules in the atmosphere (although the

2. tiny size of the gas molecules allows their thermal energy to override the gravitational force, and allows them to not pile up on the Earth’s surface – like us!)

3. the thicker (taller) the atmosphere, the more mass, causing a greater pressure

4. atmospheric pressure varies with altitude: lower altitude = higher pressure

greater altitude = lower pressure

Earth’s atmosphere is mainly O2 & N2Water vapor in the air is lighter than O2 or N2, so the higher the proportion of water in the atmosphere the less mass of the air, resulting in a lower pressure due to a reduced force of gravity. (low pressure systems = wet weather)

Various units of pressure

pascal (Pa) - the SI unit for pressure1 Pa = 1 N/m2 Newton/meter

atmosphere (atm)1 atm = avg pressure at sea level1 atm = 101325 Pa1 atm = 101.3 kPa1 atm = 760 mmHg = 760 torr1 atm = 14.70 lb/in2

1 bar = 100,000 Pa = 0.9869 atm

Barometers: instrument used to measure the atmospheric pressure Composed of a column of mercury in a dish of

mercury the height of the mercury in the tube is

determined by the downward force of the atmosphere on the Hg in the dish.

The units are mmHg

Pressure conversion problems:These are basic conversion problems:

1. Write given quantity as a fraction over 12. Select the conversion factor from your reference

sheet that has the two units in the problem.3. Multiply your given quantity by the conversion

factor

Sample problem: The column of Hg in a barometer is 745mmHg. What is the pressure in Pa.

745mmHg x 101325 Pa =99300 Pa 1 760 mmHg

Enclosed gases: Pressure inside a closed container of gas can be different

from the surrounding atmosphere.If you open the container, gas will flow in or out until the

pressure in the container = pressure of the atmosphere.

Manometer: instrument used to measure the pressure in a closed container.

It consists of a U shaped tube of mercury open to the atmosphere on one end and attached

to the container on the other end.

The difference in height of the Hg column on both sides equals the difference in the pressure of the atmosphere compared to the gas in the container.

If the level of Hg is the same on both sides of the U, then the pressure of the container =

pressure of the atmosphere.

If the pressure of the container is greater than the atmospheric pressure, it will push down on the Hg and making the level of lower

on the side attached to the container, in which case the

Pressure in = Atmosphere pressure + differenceContainer in ht Hg

Note you must work in mmHg If the pressure in the container is less than the

atmospheric pressure the Hg on the container side will be higher and

Pressure in = Atmosphere pressure - difference

Container in ht Hg

Sample problem: You have a closed container attached to a manometer. The Hg in the open ended side is higher, and the difference in height is 27mm. The atmospheric pressure is 755mmHg. What is the pressure in the container in atm?

755 + 27 = 782 mmHg

(Add the difference b/c the pressure in the container is more than in the atmosphere) 782 mmHg x 1 atm = 1.03 atm 1 760 mmHg

Group practice: A sample of gas is placed in a container attached to a manometer. The atmospheric pressure is 764.7 mmHg. The height of the Hg on the open side of the manometer is 136.4 mm. The height on the side connected to the canister is 103.8 mm. What is the pressure of the gas in atmospheres? Kilopascals?

To predict the behavior of gases we need to know 4 variables

1. amount of gas (moles)2. volume3. temperature, K (C + 273 = K)4. pressure

1. Amount of gas, (n) in moles

n = mass, g or x g x 1 mole molar mass, g/mol 1 molar mass,g

2. Volume (V) because a gas completely fills its container the

volume of container = vol of gas Volume is measured in liters, mL, dL

recall 1 L = 1000 mL = 1000cm3 or cc

3. Temperature (T) on Kelvin scale any oC must be converted to K

K = oC +273

4. Pressure (P) the outward force on walls of the container due

to collisions of the gas particles Pressure = Force

Area

Gas Laws – mathematical models of observed relationships among the four gas variables (T, P, V,

n)

Note that STP of gases is 0oC (273K) and 1 atm, compared to standard conditions for enthalpy that were 25oC and 1 atm.

Boyle’s Law : Given constant temperature and a fixed quantity of gas, pressure and volume are inversely proportional. P * V = k1 where k1 is a constant

(the value of k1 depends on the temp and the quantity of gas)

If pressure increases, volume decreasesIf pressure deceases , volume increases

If you have a sample of gas and measure the pressure and volume (P1 and V1) then change the system to have a different pressure and volume (P2 and V2), since P1V1 = k1 and P2V2 = k1 then

P1V1 = P2V2

Sample problem: A 25L helium tank is filled with H2 gas at 30.0 atm. If this is dispensed into balloons at 1.04 atm, what volume of balloons will it fill?

30 x 25 = 1.04 V2

V2 = 720 L It will fill 720L of balloons.

Group Practice:a. The gauge on a 12L tank of compressed oxygen

reads 3800 mmHg. How many liters would this same gas occupy at a pressure of 0.75 atm at constant temperature?

Charles’s Law: at constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature (Kelvin) or V = k2

T(the value of k2 depends on the pressure and quantity of gas)

We can consider two sets of volume & temperatures:

V1 = V2 or re-written V1T2 = V2T1T1 T2

V1, V2 = volumeT1, T2 = temperature in Kelvin

If you’re interested…derivation of these equations: Since volume is directly proportional to temp,KV = k2T which can be rearranged to V = k2

TAnd a sample of gas with a volume (V1) and temp (T1) can be altered to have a different volume (V2) and temp (T2), and since V1 = k2 = V2 T1 T2 you can rearrange the constants to get your useful Charles’s law equation V1T2 = V2T1

Sample Problem: What will be the volume of a gas sample at 355K if its volume at 273K is 8.57L? Assume the pressure of the gas is unchanged.

V1T2 = V2T1V1 ( 273) = (8.57) ( 355)

V1 = 11.1 L

Group problem: #2. The pressure in a natural gas tank is maintained at 2.20 atm. On a day when the temperature is -15° C, the volume of the gas is 3.25 x103m2. What is the volume of the same quantity of gas when the temperature is 31oC?

Avogadro’s Law: equal volumes of gasses at the same temperature and pressure will contain an equal

number of particles.

Since all gases show the same physical behavior, a gas with a larger volume must consist of a greater number of particles (assuming constant T and P)

V = k3n V = volumek3 = Avogadro’s constant (22.4L)n = moles

Recall that the volume of 1 mole of gas at STP is 22.4 L (so the value of k3 must be 22.4 L/mol)Quick summary of relationships:Boyle’s Law V is proportional to 1/P (constant n, T)Charles’s Law V is proportional to T (constant n, P)Avogadro’s Law V is proportional to n (constant P,T)

Ideal Gas Law; a combination of previous laws

PV = nRT P = pressureV = volumen = molesT = temperature , KelvinR = Universal Gas Constant

The Universal Gas Constant, R, has three values depending on the units.

= 0.08206 atm L / mol K= 8.314 Pa m3 / mol K

= 8.314 J / mol KYou choose the unit that matches the other units in your

problem.Additional values include 1.987 cal/mol K and 62.36 L Torr/

mol K

The ideal gas law describes the physical behavior of an ideal gas An ideal gas is a model gas that behaves

according to the kinetic molecular theory an ideal gas is NOT a real gas but many real gases behave like an ideal gas

under ordinary conditions We are making the assumption that

o the molecules of gas do NOT interact

o occupy essentially no volume in comparison to volume of the container

real gases do NOT act like ideal gases under low temperature and high pressure

Sample problem: How many moles of a gas at 100oC does it take to fill a 1.00L flask to a pressure of 1.5 atm?

n = ?T = 100o C V = 1.00LP = 1.5 atm Equation: PV = nRT

Convert temp to K : 100oC + 273 = 373K

Choose your R value that matches your other units in the problem: 0.0821 atm L / mol KPlug values into equation and solve for n

(1.50) (1) = (n (0.0821) ( 373)n = 0.0490 moles

Group problems: #1 A sample of CO2 gas is collected in a 250mL flask. The gas has a pressure of 1.3 atm at a temperature of 31oC. how many moles of CO2 were produced?

#2: If a tennis ball has a volume of 144 cm3 and contains 0.33 g of N2 gas, what is the pressure of the ball at 24oC?

Combined Gas Law – allows you to solve problems that involve changes in more than one variable. You assume n remains constant for this law.

P 1V1 = P2V2 T1 T2

All volume and pressure units must be the sameTemp. must be in K.

Example: A gas at 110 kPa and 30oC fills a flexible container with an initial volume of 2L. If the temperature is raised to 80.0oC, and the pressure increases to 440 kPa, what is the new volume?

P1 = 110 kPaT1 = 30 C 30 + 273 = 303 KV1 = 2 LP2 = 440 kPaT2= 80 C 80 + 273 = 353 KV2 = ?

V2 = 0.58L

Group practice:#1. An inflated balloon has a volume of 6.0L at sea level (1 atm). It is allowed to ascend until the pressure is 0.45 atm. During ascent the temperature falls from 22°C to -21oC. Calculate the volume of the balloon at its final altitude.

#2. A 0.50 mol sample of oxygen gas is confined at 0°C and 1.0 atm in a cylinder with a movable piston. The piston compresses the gas so the final volume is ½ its original volume and the final pressure is 2.2 atm. What is the final temp of the gas in Celsius?

Gas Densities and Molar Mass

The density of a gas is directly related to its pressure and molar mass, and inversely related to its temperature.

d = PΜ d = density of a gas RT P = pressure

Μ = molar massR = universal gas constantT = temperature, K

the greater the molar mass or pressure, the denser the gas

The higher the temperature, the less dense the gas

The equation can be re-arranged to solve for molar mass :

M = dRT P

Practice problems:What is the density of carbon tetrachloride vapor at 714 torr and 125°C?P = 714 torr * 1 atm/760 torr = 0.939 atmT = 125°C + 273 = 398KR = 0.08206 L atm/mol KM = molar mass of CCl4 = 153.8 g/mold = PM = (0.939)(153.8) = 4.42 g/L RT (0.08206)(398)

A large evacuated flask has an initial mass of 134.567g. When the flask is filled with a gas of unknown molar mass , to a pressure of 735 torr at 31°C, its mass is 137.456 g. When the flask is evacuated again and filled with water at 31°C, its mass is 1067.9g. The density of water is 0.997 g/mL) Assuming the ideal gas law applies, calculate the molar mass , M, of the gas.

1. Find the mass of the gas by subtracting the mass of the empty flask from the mass of the flask plus gas.137.456 – 134.567 = 2.889 g

2. The gas volume equals the water volume – but you must convert g water to mL water.1067.9g – 134.567g = 933.3 g water

9333 g water * 1 mL = 936 mL = 0.936L0.997 g

3. We know the mass and volume of the gas, so we can calculate density = m/V = 2.889g/.936L = 3.09 g/L

4. Convert 735Torr * 1 atm = 0.09671 atm 760 torr

5. Convert temp to K31 + 273 = 304 K

6. Use the equation for MM = dRT = (3.09)(0.08206)(304) = 79.7 g/mol

P 0.09671

Group practice:#1 The molar mass of the atmosphere at the surface of Titan, Saturn’s largest moon, is 28.6 g/mol. The surface temp. is 95K, and the pressure is 1.6 atm. Assuming

ideal gas conditions, what is the density of Titan’s atmosphere?

#2. Calculate the average molar mass of dry air if it has the density of 1.17 g/L at 21°C and 740.0 torr.

Additional types of problems may be solved using stoichiometry and the molar ratios represented by the coefficients in a balanced chemical equation, combined with the gas laws.

Example problem: Car air bags are inflated by nitrogen gas produced by the rapid decomposition of sodium azide in the following reaction: 2NaN3 (s) 2 Na(s) + 3 N2 (g)

If an air bag has a volume of 36L and is to be filled with N2 at 1.15 atm and 26.0°C, how many grams of NaN3 must be decomposed?

Sequence to solve:Gas laws solve for mol N2 , convert to mole NaN3 using stoichiometry, then convert to g NaN3 using 1 mole/molar mass.

Group practice: In the first steps of the industrial process for making nitric acid, ammonia reacts with oxygen gas in thre presence of a suitable catalyst to form nitric oxide and water vapor: 4 NH4 (g) + 5 O2 (g) 4 NO(g) + 6 H2O(g)

How many L of NH4 at 850°C and 5.00 atm are required to react with 1.00 mol O2 in this reaction?

Daltons Law of Partial Pressures; the sum of the partial pressures of all the components in a gas mixture is equal to the total pressure of the gas mixture.

PT = p1 + p2 + p3 + ….Where PT = total pressure

P1 = partial pressure of gas 1P2 = partial pressure of gas 2

P3 = partial pressure of gas 3

partial pressure: pressure of one type of gas in a mixture

Dalton’s law says that each type of gas in a mixture exerts the same pressure as it would if it were alone (at constant T)

Sample problem: What is the atmospheric pressure if the partial pressures of N2, O2, and Ar are 604.5 mmHg, 162.8mmHg and 0.5 mmHg, respectively?

PT = pN2+ pO2 + pAr = 604.5 + 162.8 + 0.5 = 767.8 mmHg

Group practice: The gases CO2, O2, N2, Ne, and Kr are mixed in a container. All gases have the same partial pressure and the total pressure is 33500 Pa. What is the partial pressure of N2?

We can combine the law of partial pressure with the ideal gas law.

We know PT = p1 + p2 + p3

And we know p1 = n1RT etc for each of the V component gases

Thus PT = nTRT where nT = total moles of all

V gases

Sample problem: A mixture of 6.00g O2 and 9.00g CH4 is placed in a 15.0 L vessel at 0°C. What is the partial pressure of each gas, and what is the total pressure in the vessel?

Group practice: What is the total pressure exerted by a mixture of 2.00g of H2 and 8.00g N2 at 273K in a 10.0L vessel?

Solving problems involving gases produced in a lab experiment, and collected in a bottle over water.

1. Gas produced due to a reaction is collected in water-filled, inverted bottle.

2. The bottle is moved up or down in the pan of water until the water level is the same inside and outside the bottle – this ensures the gas inside is at atmospheric pressure – and the volume of gas can now be read.

3. The total pressure of gas in the bottle is a combination of the gas produced and water vapor (from the liquid water). Use the equation:

PT = Pgas + Pwater Values for Pwater at various

temps are in Appendix Bpg 1058

Practice problem: When a sample of KClO3 is partially decomposed according the the following reaction:2 KClO3(s) 2 KCl (s) + 3 O2(g)

, 0.250L of gas is collected at 26°C and 765 torr total pressure. A. How many moles of O2 are collected? B. How many grams of KClO3 decomposed?

1. We can use the equation po2 = no2RT to solve for no2

V But we need po2 which we can get by subtracting

the vapor pressure of water at 26° C (Appendix B) from the total pressure.765 torr = PO2 + 25 torr PO2 = 740 torr

Convert 740 torr * (1 atm/760 torr) = 0.973 atm

Convert temp to K 26 + 273 = 299K

0.973 = no2 * 0.0821 * 299 nO2 = .00992 mol

0.250L

2. Use stoichiometry to convert moles O2 to moles KClO3, then to g KClO3

0.00992 mol O2 * 2 mol KClO3 * 122.6g KClO3 =0.811 g KClO3

1 3 mol O2 1 mol KClO3

Group practice: When a sample of ammonium nitrate is decomposed according to the following reaction:NH4NO2 (s) N2 (g) + 2 H2O(l) ; 511mL of N2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH4NO3 were decomposed?

Getting back to the Kinetic Molecular Theory….

Recall that at a constant pressure, all the molecules of gas in a container have the same average kinetic energy (and the same average speed), however, different particles are moving at different speeds.

Why? Conservation of momentum (mv) occurs with each collision, although some particles may leave the collision at a higher speed than others.

Distribution of molecular speed (v) for N2(g).

The peak of the graph represents the most probable speed (ump), which is the speed of the largest number of molecules.

The root-mean-square speed (urms) is the speed of a molecule with an average kinetic energy of the sample.

Avg KE = ½m(urms)2

Because mass does not change with temperature, increasing temperature will cause an increase in average KE and an average urms .

Use the Kinetic Molecular Theory to explain why1. An increase in volume at constant

temperature causes pressure to decrease.

2. A temperature increase at constant volume causes pressure to increase.

Practice Problems: A sample of O2 gas initially at STP is compressed to a smaller volume at constant temp. What effect does this have on

a. The avg KE of the molecules?b. Their avg speed?c. The number of collisions they make with the

container walls per unit time?d. The number of collisions they make with a unit area

of container wall per unit time?

Avg KE is determined only by temperature. Since temp remains constant, avg KE is unchanged.

Because avg KE remains the same, avg speed remains the same.

The number of collisions they make with the walls of the container per unit time increases b/c the particles are moving in a smaller volume, but with the same avg speed, causing them to encounter a wall more frequently.

The number of collisions with a unit area of the wall in a unit time will also increase because the total number of collisions has increased, and the area of the wall has decreased.

Group practice:How is the rms speed of N2 molecules in a gas sample changed by

a. an increase in temperature?

b. An increase in volumec. Mixing with a sample of Ar at the same temp?

Speeds of molecules in a mixture of gases1. According to the Kinetic molecular theory – all

particles in a mixture of gases have the same avg KE = ½ m(urms)2 at a given temperature.

2. If the mixture contains molecules of different masses, such as He (M=4) and Xe (M=131), the He molecules must have a greater rms speed than the Xe molecules in order to have the same avg KE.

3. This can be stated mathematically:

urms = M = molar mass of particles

The less massive the gas particles, the higher their rms speed.

Practice problem: Calculate the rms speed of N2 molecules at 25°C.

Convert temp to K: 25 + 273 = 298K Molar mass N2 = 28 g/mol

But looking at R values, we must use R = 8.314 J/mol-K. Recall that 1 J = kg m2/s2

So…..we must put molar mass, M, in terms of kg!

28 g/mol * 1 kg/1000g = 0.028 kg/mol

Plug into equation:

Urms = 3 * 8.314 * 298 = 515 m/s 0.028

Group practice: What is the rms speed of an atom in a sample of He gas at 25°C?

The dependence of molecular speed on mass has two consequences:

1. Effusion: the escape of gas molecules through a small hole

2. Diffusion : the spread of one substance throughout space or throughout a second substance.

3. The effusion rate of a gas is inversely proportional to the square root of its molar mass. Grahams Law of Effusion:

r 1 = M2 r1 & r2 = effusion rates of gases 1 & 2 r2 M1 M1 & M2 = molar mass of gases 1 & 2

The rate of effusion is directly proportional to its rms speed. Applying this to a mixture of gases we can say that the smaller gas molecules will effuse more rapidly than the larger gas molecules in the mixture.

4. Diffusion is also faster for smaller molecules, however, it is more complicated because as a particle of gas moves, it makes many collisions with other molecules.

Real gases and deviations from Ideal Gas Behavior1. Real gases do not follow ideal gas behavior at high

pressures Because attractions between gas particles

becomes significant And b/c the volume of the gas molecules

becomes larger and not insignificant compared to the volume of the container. At high pressures, actual gas volumes are larger than predicted by the Ideal Gas Law.

2. Real gases do not follow ideal gas behavior at low temperatures as the gas approaches its change to a liquid b/c the attractions between real gas particles become significant and they begin to “stick” to each other instead of bounce apart.

The van der Waal’s Equation: Is useful to predict the behavior of real gases at high pressures.

(P + )(V - nb) = nRT

P = pressurea = a constant that is a measure of the attraction between gas moleculesb = a constant; the finite volume of the molecules

The term accounts for the attractive forces. Attractive forces tend to reduce pressure Attractive forces increase as the square of the number

of molecules per unit volume

The term nb accounts for the volume occupied by the gas molecules.

Note that larger, more massive molecules have larger volumes (b) and tend to have greater attractive forces (a).

Practice problem: If 1 mol of an ideal gas were confined to 22.41L at 0°C, it would exert a pressure of 1.0 atm. Use the van der Waals equation to estimate the pressure exerted by 1.00 mol of Cl2 gas in 22.41 L at 0°C.

We can re-arrange the vdW equation to solve for P:

P = n RT - n 2 a V – nb V2

Convert temp to K = 273 K Use R = 0.08206 L atm / mol K to match other

units

P = (1 mol) ( 0.08206) ( 273K) - (1 mol) ( 6.49) 22.41 L – (1.0 mol* 0.0562L) (22.41 L)2

P = 1.003 atm – 0.013 atmP = 0.990 atm

Group practice: A sample of 1.0 mol CO2(g) is in a 3 L container at 0°C. Calculate the pressure of the gas using the ideal gas equation and then again using van der Waals equation.