Ch. 5 Gases

11th ed. 53-61 odd, 67-73 odd, 81-87 odd, 93-9410th ed. 51-59 odd, 63-69 odd, 77-83 odd, 89, 90

Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

HW: 19-25 odd, 31-49 odd

Elements that exist as gases at 250C and 1 atm

•Gases adopt the shape of their containers (fluidity).

•Gases adopt the volume of their containers (Diffusion/compressibility).

•Gases will mix evenly and completely when in the same volume

•Gases have much lower densities than liquids and solids.

Physical Characteristics of Gases

WF6 gas: 13 g/L11x heavier than air, but still >75x less dense than water

Properties of GasesDIFFUSION - Uniform spreading of gas molecules

EFFUSION - Movement of gas through small hole

FLUIDITY - Ability to flow and take shape of their container (liquids and gases)

5

Units of Pressure1 pascal (Pa) = 1 N/m2

1 atm = 760 mmHg = 760 torr

1 atm = 101,325 Pa

Pressure = Force

Area

Decreased Area,Increased Pressure

Increased Area,Decreased Pressure

Barometervacuum

Hg(l)

Sea level 1 atm

4 miles 0.5 atm

10 miles 0.2 atm

Atmospheric pressure: Dependant on elevation, temperature, weather

Pressure of a gasPressure = the collision of gas particles with a surface;

force per unit area

As number of collisions increase, pressure increases

As force of collisions increase, pressure increases

Manometers Used to Measure Gas Pressures

closed-tube open-tube

For below 1 atm pressures For above atm pressures

Pressure conversion Example 1

What is the pressure in atmospheres if the barometer reading is 688 mmHg (torr)?

We need to know:

Kinetic Molecular Theory Summary

1. Gas particles separated from each other by large distances

2. Gas particles are in constant motion in random directions, and they frequently collide.

3. Negligible intermolecular forces

4. Kinetic energy is proportional to temperature (gases at the same temperature will have the same average KE)

Be able to cite each and know each of their implications

Kinetic Molecular Theory of Gases

1. A gas is composed of molecules that are separated from each other by large distances far greater than their own dimensions. The molecules can be considered to possess mass but have insignificant volume.

Very low density (molecule per volume)

Kinetic Molecular Theory of Gases2. Gas molecules are in constant motion in random

directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic (don’t lose energy).

Kinetic Molecular Theory of Gases

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

Molecules too far apart to effect each other

“Negligible intermolecular forces”

(liquid)

These bonds only occur in liquids and solids

Inter = “between”

As temperature goes up, KE goes up

As KE goes up, molecule velocity goes up.Therefore, as Temp ↑, molecule velocity ↑

Kinetic Molecular Theory of Gases4. The average Kinetic energy (KE) of the molecules

is proportional to the temperature of the gas

KE = ½ mv2 ; v2 is average square velocity

*Any two gases at the same temperature will have the same average KE, regardless of size.

Any two gases at the same temperature will have the same average KE, regardless of size.

KE = ½ mv2

We can compare a two gas mixture with relative speedat an arbitrary KE value of 5 at given temperature.

Helium gas (4 g/mol) vs Chlorine gas (Cl2; 71 g/mol)

He: 5 = ½ 4*v2

Cl2: 5 = ½ 71*v2

v = 1.6

v = 0.38

HeCl2

= 1.60.38

= 4.2

At this temperature, Helium is moving 4.2 times faster than chlorine gas

88

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.

NH3

17 g/molHCl36 g/mol

NH4Cl(s)

v1

v2

M2

M1

=

random molecular path

Because two gases have the same KE at the same temp (KE = ½*m*v2). We can relate them to determine an unknown gas’s mass using a standard gas’s diffusion rate. Graham’s law of diffusion

Gas effusion is the process by which gas under pressure escapes from one compartment of a container to another by passing through a small opening.

v1

v2

M2

M1

=t2

t1

=

Smaller effuses faster

time unitssec or min

rate unitsm/s

(d/t1)

(d/t2)=

Distance/time

Example: Effusion of a gasA flammable hydrocarbon (CxHy) gas is found to effuse through a porous barrier in 1.50 min.

It takes an equal volume of bromine vapor 4.73 min to effuse through the same barrier (with same conditions).

Calculate the molar mass of the unknown gas, and suggest what this gas might be.

Gas effusion. Gas molecules move from a high-pressureregion (left) to a low-pressureone through a pinhole.

M2

M1

=t2

t1

*Remember Bromine is diatomic: Br2(g)

Example: Effusion solution

Solution From the molar mass of Br2, we write

Where is the molar mass of the unknown gas

Because the molar mass of carbon is 12.01 g and that of hydrogen is 1.008 g, the gas is methane (CH4).

Apparatus for Studying Molecular Speed Distribution and average molecular speed

After numerous hits, the molecular deposition will eventually become visible. Density of each region is measured

82

The distribution of speedsfor nitrogen gas moleculesat 3 different temperatures

The distribution of speedsof three different gasesat the same temperature

Speed of sound (340 m/s)

Fastest flight speed (990 m/s)

vrms MM

~800 °F

-280 °F

80 °F

80 °F

Average gas molecule speed (vrms)Total kinetic energy of a mole of gas equals 3/2RT

NA•(1/2mv2) = 3/2•R•T

MM (molar mass) = NAm

1 Joule = 1 kg m2/s2

R = 8.314 J/K · mol

R = Gas constant (we’ll explain later)

Because our constant (R) uses Joules, which uses kg,we must express Molar mass in kg. We must also use Kelvin for temperature.

vrms MM

Example: Speed of a gas

Calculate and compare the root-mean-square speeds of helium atoms and nitrogen molecules in m/s at 25°C.

R = 8.314 J/K · mol 1 Joule = 1 kg m2/s2

The molar mass of He is 4.003 g/mol, or 0.004 kg/mol.

vrms MMvrms MM

The temperature is 25°C , but needs to be expressed as 298 K

Considering 1 J = 1 kg m2/s2, the rest of the units cancel out

At 25°C, helium travels on average ~3000 mph

vrms MM

Example: Speed of a gas Solution

The procedure is the same for N2, the molar mass of which is 28.02 g/mol, or 2.802 × 10−2 kg/mol so that we write

Escape velocity is the speed where an object’s KE is equal to the gravitational potential energy. The speed needed to “break-free” of Earth’s gravity is ~ 11,000 m/s

Earth’s atmosphere has low abundance of H2 & He because the molecules are light and travel fast enough to escape the Earth’s pull.

Example: Speed of a gas Solution

Crash Course: Passing Gaseswww.youtube.com/watch?v=TLRZAFU_9Kg

Three physical properties can describe a sample of gas

• Volume

• Pressure

• Temperature

They are each interconnected with each other – if one changes, the others must change with it.

There are several scientific gas laws that define the behavior of gases with these parameters

Apparatus for Studying the Relationship BetweenPressure and Volume of a Gas

Doubling Pressure

Tripling Pressure

As P increases

V decreases

P x V = constant (k1)

P1 x V1 = P2 x V2

Boyle’s Law: Pressure-Volume relationship

Constant temperatureConstant amount of gas

“Inversely proportional”

P ∝ 1/V

1VP = k1 *

“ ” ∝ = proportional

Boyle’s Law (Pressure-Volume)As Volume decreases, Collisions become more frequent in smaller space.

More collisions = more pressure

As Volume ↓; Pressure ↑ orAs Pressure ↓; Volume ↑

P1 x V1 = P2 x V2

Boyle’s Law Practice:

• If we have 5 L of Nitrogen gas (N2) at 2 atm, what is the volume if we apply 4 times the pressure?

• Helium gas is found in a container with a volume of 3L at 4 atm. If the container double in size, what is the new pressure?

P1 x V1 = P2 x V2

2 atm x 5 L = 8 atm x V2 V2 = 1.25 L

4 atm x 3 L = P2 x 6 L P2 = 2 atm

Units must be equal on both sides

As T increases V increases

Variation in Gas Volume with Temperature at Constant Pressure

KMT explains this: as molecules travel faster (at higher T), Volume would

have to increase to maintain the same

Pressure

Variation of Volume with Temp.

V ∝ T

K = 0C + 273

Charles’s Law

Temperature must bein Kelvin

proportional

(Constant Pressure)

Also, P T∝(Constant V)

VT

= k2

V ∝ TV1/T1 = V2 /T2

T (K) = t (0C) + 273.15

First Determination of Absolute zero

Lord Kelvin realized each line extrapolated to the same point

• To a theoretically lowest attainable staring temperature

• He identified -273 0C as absolute zero

Data stops here from condensation

As Temp ↑; Volume ↑

Charles’s Law Practice:

• If we have 0.8 mL of Oxygen gas (O2) at 30 °C, what is the

volume if we heat the gas to 90 °C?

• Air is found at 1 atm 25 °C in a fixed volume. If the pressure increases 5-fold, what is the new temperature?

= V2 = 0.96 mL

T2 = 1,365 K ~ 2,000 ° F

Units must be equal on both sides, must use Kelvins

V2

363 K0.8 mL303 K

= 5 atm T2

1 atm273 K

Veritasium: Fire Syringewww.youtube.com/watch?v=4qe1Ueifekg

Avogadro’s Law (Moles of gas)

V number of moles (n)

V = constant x n

V1 / n1 = V2 / n2

Constant temperatureConstant pressure

V ∝ n

4 volumes → 2 volumes

First proposed by Avogadro, 1811: Gas volume does not depend on molecule size, only # particles

Mass is conserved, but volume is NOT

Avogadro’s Law

Combustion of hydrocarbons (like octane) drive the pistons in combustion engines from the expansion of gases (CO2 and H2O vapor)

*Potato guns also work based on this. Typically combustion of alcohols in hairspray.

Candle wax demo

Paraffin wax: a mixture of long hydrocarbons

Remember: Combustion of a hydrocarbon gives CO2 and H2O

2C20H42(s) + 61O2(g) → 40CO2(g) + 42H2O(l)

Conversion of available oxygen gas to CO2 produces less moles of gas, so there’s a decrease in pressure.

You can watch it done here if missed in class:https://www.youtube.com/watch?v=0WGOpSpuDYQ

Less pressure inside, causes atmospheric pressure to push the water up (just like mercury in a barometer)

Ideal Gas Equation

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

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

Boyle’s law: P (at constant n and T)

1V

V

V = constant x = RnTP R is the gas constant

PV = nRTIdeal Gas Equation: assumes negligible intermolecular forces between particles

nTP

nTP

Combining them all we see:

+

The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). (SATP 25 °C & 1 atm)`

PV = nRT

R = PVnT

=(1 atm)(22.414L)

(1 mol)(273.15 K)

Gas constantR = 0.082057 (L • atm) / (mol • K)

Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. (regardless of molecular identity)

Must use these unitsRemember, K = C° + 273; 1 atm = 760 mmHg

Example: Ideal Gas #1Sulfur hexafluoride (SF6) is a colorless and odorless gas. Due to its lack of chemical reactivity, it is used as an insulator in electronic equipment.

Calculate the pressure (in atm) exerted by 265.9 grams of the gas in a steel vessel of volume 5.43 L at 69.5°C.

*Convert grams to moles with MM(265.9 g) / (146.1 g/mol) = 1.82 mol

*Add 273 to °C to use Kelvin

PV = nRT

P = nRTV

= 9.42 atmCrash Course: The Ideal Gas Lawwww.youtube.com/watch?v=BxUS1K7xu30

Calculate the volume (in L) occupied by 7.40 g of NH3 at STP

*This relation can only be used at STP conditions

Or we could use the ideal gas equation where 7.40 g NH3 = 0.435 moles of NH3, and then applying V = nRT/P.

V = (0.435 mol)(0.082057)(273.15 K)1 atm

= 9.74 L

Combined Ideal Gas Equation

Can be used when gas sample conditions change

PV = nRT

R =

(Before change)

R =

(After change)

Sometimes called the Modifed gas law

The R’s are the same so we can set them equal

Can be used in place of each prior gas law used alone

Example: Combined Gas Law #1

Argon is an inert gas used in light bulbs to stop the vaporization of the tungsten filament.

A certain light bulb containing argon at 1.20 atm and 18°C is heated to 85°C at constant volume.

Calculate its final pressure (in atm).

Electric light bulbs are usually filled with argon.

n1 = n2 because bulb is sealed V1 = V2 because bulb volume does not expand

which is Charles’ law

Next we write

Initial Conditions Final Conditions P1 = 1.20 atm P2 = ?

T1 = (18 + 273) K = 291 K T2 = (85 + 273) K = 358 K

The final pressure is given by

Check At constant volume, the pressure of a given amount of gas is directly proportional to its absolute temperature. Therefore the increase in pressure is reasonable.

Example: Combined Gas Law #1 Solution

A small bubble rises from the bottom of a lake, where the temperature and pressure are 8°C and 6.4 atm, to the water’s surface, where the temperature is 25°C and the pressure is 1.0 atm.

Calculate the final volume (in mL) of the bubble if its initial volume was 2.1 mL.

We can remove n because it’s constant in this problem

Example: Combined Gas Law #2

The given information is summarized:

Initial Conditions Final Conditions P1 = 6.4 atm P2 = 1.0 atm

V1 = 2.1 mL V2 = ?

T1 = (8 + 273) K = 281 K T2 = (25 + 273) K = 298 K

Rearranging

Example: Combined Gas Law #2 Solution

An inflated He balloon at sea level (1.0 atm) with a V = 7.1 L (basketball) is allowed to rise to a height of 8.8 km (Mt. Everest), where the pressure is about 0.33 atm. The change in temperature drops from 20°C to -30°C.

What is the final volume of the balloon?

n1 = n2 because balloon is sealed

Practice: Combined Gas Law #3

Dalton’s Law of Partial Pressures

V and T are constant

P1 P2

Individual gas pressures are cumulative regardless of chemical identity (negligible intermolecular forces)

Ptotal = P1 + P2

=+

Dalton’s Law

V and T are constant

Consider a case in which two gases, A and B, are in a container of volume V.

PA = nART

V

PB = nBRT

V

nA is the number of moles of A

nB is the number of moles of B

PT = PA + PB XA = nA

nA + nB

XB = nB

nA + nB

PA = XA PT PB = XB PT

Pi = Xi PTmole fraction (Xi ) =

ni

nT

A mixture of gases contains 4.46 moles of neon (Ne), 0.74 mole of argon (Ar), and 2.15 moles of xenon (Xe).

Calculate the partial pressures of the gases if the total pressure is 2.00 atm at a certain temperature.

Example: Dalton’s Law

XNe= 4.46 mol Ne

7.35 mol Total= 0.607

Check Make sure that the sum of the partial pressures is equal to the given total pressure; that is, (1.21 + 0.20 + 0.586) atm = 2.00 atm.

Example: Dalton’s Law Solution

More Practice: Ptotal = 0.78 atm of 1.2 mol CO2 & 3.4 mol O2

Density (d) Calculations

d = mV

=PMRT

n is moles of gasm is the mass of the gas in gramsM is the molar mass of the gas

Molar Mass (M) of a Gaseous Substance

dRTP

M =d is the density of the gas in g/L

mn

& M = P RT

n V

=PV = nRT(grams)(mole)

mM

so, n =

Finding Molar Mass by Gas Density

Calculate the density of carbon dioxide (CO2) in grams per liter (g/L) at 0.990 atm and 55°C.

We will use T = 273 + 55 = 328 K and 44.01 g/mol for the molar mass of CO2

PMRT

d =

PMRT

d =

Example: Molar mass to Density

Gas Stoichiometry

Similar to gravimetric analysis of solid samples

Example: Gas Stoichiometry #1

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

The N2 gas produced quickly inflates the bag.

Calculate the N2 volume generated at 80°C and 823 mmHg by the decomposition of 60.0 g of NaN3.

Airbag Deploying in Slow Mo - The Slow Mo Guys www.youtube.com/watch?v=KRcajZHc6Yk

2 mol NaN3 3 mol N2

Example: Gas Stoichiometry #2

Calculate the volume of O2 (in liters) required for the complete combustion of 7.64 L of acetylene (C2H2) measured at STP.

The reaction of calcium carbide (CaC2) with water produces acetylene (C2H2), a flammable gas.

5 mol O2 2 mol C2H2

5 Liters O2 2 Liters C2H2

Avogadro’s Law: gas volume is independent of its identity:

71

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

Collecting a Gas over Water

PT = PO2 + PH2O

72

Vapor of Water and Temperature

Example 5.15

Oxygen gas generated by the decomposition of KClO3 is collected as shown

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

The volume of oxygen collected at 24°C and atmospheric pressure of 762 mmHg is 128 mL.

Calculate the mass (in grams) of oxygen gas obtained.

The pressure of the water vapor at 24°C is 22 mmHg.

Example 5.15

Therefore,

From the ideal gas equation we can determine moles of O2:

To use R, we must convert to from mmHg to atm and C to K

n = PV RT

= (0.974 atm)(0.128 L)(0.0821)(297 K)

= 0.00511 moles O2

= 0.00511 moles O2 = 0.164 grams O2

Chemistry in Action:

Scuba Diving and the Gas Laws

V

Depth (ft) Pressure (atm)

0 1

33 2

66 3

Ascending too fast can cause the “bends” - Decompression sickness (gas bubbles enlarging in the blood stream)

D P

(Boyle’s Law)

Non-Ideal Gas: Effect of intermolecular forces on the pressure exerted by a gas.

“Attractive forces “lessen” the force exerted on the walls

1) At High Pressure, density increases, and intermolecular forces are no longer negligible.

2) Molecules slow down at Low Temperature and lowers KE to overcome attractive forces

Deviations from Ideal Behavior

1 mole of ideal gasPV = nRT

n = PVRT = 1.0

Repulsive Forces

Attractive Forces

Most gases act “ideally” below ~ 5 atm

Van der Waals Gas equationFor non-ideal gas

P + (V – nb) = nRTan2

V 2( )}

correctedpressure

}

correctedvolume

*Experimentally determined

a correlates to molecule attraction

b “roughly” correlates to molecule size

Example 5.18

Given that 3.50 moles of NH3 occupy 5.20 L at 47°C, calculate the pressure of the gas (in atm) using

(a) the ideal gas equation

(b) the van der Waals equation

Example 5.18

(b) It is convenient to first calculate the correction terms in Equation (5.18) separately. From Table 5.4, we have

a = 4.17 atm · L2/mol2

b = 0.0371 L/mol

so that the correction terms for pressure and volume are

Example 5.18 Solution

Finally, substituting these values in the van der Waals equation:

The value is 1.5 atm lower than when using the ideal gas equation.

This makes sense as we expect the pressure to be reduced by added intermolecular forces.

Crash Course: Real Gaseswww.youtube.com/watch?v=GIPrsWuSkQc