Ch. 5 Gases 11th ed. 53-61 odd, 67-73 odd, 81-87 odd, 93-94 10th ed. 51-59 odd, 63-69 odd, 77-83 odd, 89, 90 Copyright © The McGraw-Hill Companies, Inc.
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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
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