Ms. Lina Ai

Gases
Chapter 5

1

Elements that exist as gases at 250C and 1 atmosphere

2

3

Physical Characteristics of Gases
• • • • Gases assume the volume and shape of their containers. Gases are the most compressible state of matter. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids.

NO2 gas

4

Force Pressure = Area
(force = mass x acceleration)

Units of Pressure 1 pascal (Pa) = 1 N/m2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa Barometer
For measuring atmospheric pressure 5

The force experienced by any area exposed to Earth’s atmosphere is equal to the weight of that column of air above it.

10 miles

0.2 atm

4 miles Sea level

0.5 atm 1 atm

6

Manometers Used to Measure Gas Pressures closed-tube open-tube Patm

h = PHg

Pgas>Patm Pgas = Patm + PHg
Measures pressures below atm. pressure Measures pressures ≥ atm. pressure 7

What is the pressure in atmospheres in a cabin if the barometer reading is 672 mmHg? 672 mmHg x 1 atm/760 mmHg = 0.884 atm
8

closed-tube
The hight, h, of liquid level depends on the density of that liquid

For two liquids A and B h(A)/h(B) = d(A)/d(B) (hxd)B = (hxd)A

Measures pressures below atm. pressure

9

Referring to the following scheme: A liquid (d=0.845 g/ml) has a height of 17.5 mm in an open-tube manometer at 745.0 mmHg of atmospheric pressure. Calculate the gas pressure in torr if the density of mercury is 13.6 g/ml? Given: d(L) = 0.845 g/ml dHg = 13.6 g/ml h(L) = 17.5 mm Patm = 745 mmHg From the scheme Pgas>Patm Pgas = Patm + PHg (hxd)L = (hxd)Hg hHg = 17.5 x 0.854 / 13.6 = 1.10 mmHg Pgas = 745.0 + 1.10 = 746.1 torr

17.5 mm

10

Boyle’s law The pressure of a fixed amount of gas maintained at constant temperature is inversely proportional to the volume of the gas PV = 1 P = pressure (atm) n V = Volume (L or mL) 1 = constant For a sample of gas under two different sets and conditions at constant temperature, we have P1V1 =
1

= P2V2

1

= nRT

P1V1 = P2V2
11

Apparatus for Studying the Relationship Between Pressure and Volume of a Gas

P α 1/V P = K1 x 1/V

As P (h) increases

V decreases

12

Boyle’s Law

P ∝ 1/V P x V = constant P1 x V1 = P2 x V2 Constant temperature Constant amount of gas
13

Summary of Gas Laws Boyle’s Law

14

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?

P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg V1 = 946 mL P2 = P1 x V1 V2 P2 = ? V2 = 154 mL

726 mmHg x 946 mL = = 4460 mmHg 154 mL
15

Charles’s and Gay-Lussac’s law OR Charles’s law The volume of a fixed amount of gas maintained at constant pressure is directly proportional to the absolute temperature of the gas
For a given sample of gas at constant pressure, we can compare two sets of volume-temperature conditions

V1/T1 = V1/T1 = PαT P = 3T

2

= V2/T2 = V2/T2

T = temperature (K) V = volume (L or mL) 2 = nR/P

2

Another form of Charles’s law shows that at constant amount of gas and volume the pressure is proportional to temperature

P/T =

3

3

= constant

Two sets of pressure-volume conditions P1/T1 = P2/T2
3

= nR/V

16

Charles Law

17

Variation in Gas Volume with Temperature at Constant Pressure

As T increases

V increases

18

Variation of Gas Volume with Temperature at Constant Pressure
Notice that the temperature at which the gas has zero volume is zero Kelvin = -273.15 °C

Charles’ & Gay-Lussac’s Law

V∝T V = constant x T V1/T1 = V2 /T2

Temperature must be in Kelvin
T (K) = t (°C) + 273.15
19

A sample of carbon monoxide gas occupies 3.20 L at 125 °C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?

V1 /T1 = V2 /T2 V1 = 3.20 L T1 = 398.15 K V2 = 1.54 L T2 = ?

T1 = 125 (°C) + 273.15 (K) = 398.15 K T2 = V2 x T1 V1 = 1.54 L x 398.15 K 3.20 L = 192 K
20

Avogadro’s law At constant pressure and temperature, the VOLUME of a gas is directly proportional to the absolute temperature of the number of moles of the gas present Vαn V = 4n V/n = = RT/P

4

4

21

Avogadro’s Law

22

Avogadro’s Law
V ∝ number of moles (n) V = constant x n V1 / n1 = V2 / n2
Constant temperature Constant pressure

23

Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?

4NH3 + 5O2 1 mole NH3

4NO + 6H2O 1 mole NO

At constant T and P 1 volume NH3 1 volume NO

In chemical reactions the volume of gases reacted or produced are same as their number of moles, if these volumes are measured at the same temperature And pressure.
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In Summary:
Boyle’s law: V α 1/P Charles’s law: V α T Avogadro’s law: V α n (at constant n and T) (at constant n and P) (at constant P and T)

By combining all three expressions, we will get a master equation describing the behavior of gases: V α nT/P V = R x nT/P R = proportionality constant or gas constant
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PV = nRT

Ideal Gas Equation
1 (at constant n and T) V Charles’ law: V ∝ T (at constant n and P) Boyle’s law: P ∝ Avogadro’s law: V ∝ n (at constant P and T) nT V∝ P nT nT V = constant x =R R is the gas constant P P PV = nRT
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What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 °C = 273.15 K P = 1 atm

PV = nRT nRT V= P V=

1 mol HCl 36.45 g HCl n 49.8 g HCl n = 1.37 mol
L•atm mol•K

1.37 mol x 0.0821

x 273.15 K

1 atm

V = 30.7 L
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The conditions 0 °C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

PV = nRT (1 atm)(22.414L) PV R= = nT (1 mol)(273.15 K) R = 0.082057 L • atm / (mol • K)
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Sulfur hexafluoride is colorless, odorless, very unreactive gas. Calculate the pressure (in atm) exerted by 1.39 moles of the gas in a steel vessel of volume 6.09 L at 55 ° C. T = 55 + 273.15 = 328 K PV = nRT P = nRT/V = 1.39 x 0.0821 x 328/6.09 = 6.15 atm Calculate the volume (in liters) occupied by 5.58 g of NH3 at STP? nNH3 = 5.58 / 17.03 = 0.32765 mol 1mol NH3 22.4 L NH3 0.32765 mol NH3 X X = 0.32765 x 22.4 = 7.34 L
29

A small bubble rises from the bottom of a lake, where the temperature and pressure are 8° and 6.4 atm, to water’s surface, where the temperature is C 25° and the pressure is 1.0 atm. Calculate the final volume (in mL) of the C bubble if its initial volume was 2.1 mL.
Initial P1 = 6.4 atm V1 = 2.1 mL T1 = 8 °C ( 8 + 273.15 K) Final P2 = 1.0 atm V2 = ? mL T2 = 25 °C ( 25 + 273.15 K)

P1V1 = n1RT1 P2V2 = n2RT2 Note that n1 = n2 R is constant Now divide both equations P1V1 = n1RT1 P2V2 = n2RT2 P1V1 / P2V2 = T1/ T2 6.4 x 2.1 / 1.0 x V2 = 281.15 /298.15 13.44/V2 = 0.9429 V2 = 13.44/0.9429 = 14 mL

30

Density (d) Calculations m = PM d= V RT m is the mass of the gas in g M is the molar mass of the gas

Molar Mass (M ) of a Gaseous Substance dRT M= P d is the density of the gas in g/L

31

A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 ° C. What is the molar mass of the gas?

dRT M= P g 2.21 L

m = 4.65 g = 2.21 d= V 2.10 L x 0.0821
L•atm mol•K

g L

x 300.15 K

M=

1 atm

M = 54.5 g/mol

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Gas Stoichiometry
What is the volume of CO2 produced at 37 0C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6(s) + 6O2(g) nC6H12O6 = 5.60/180 = 0.0311 mol 1 mol C6H12O6 0.0311 mol Glu 6 mol CO2 X X = 0.187 mol CO2 L•atm x 310.15 K mol•K 1.00 atm

6CO2(g) + 6H2O(l)

V=

nRT = P

0.187 mol x 0.0821

= 4.76 L

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1.25 g of calcium carbonate (CaCO3) decomposed upon heating to give CO2 and CaO, as show in the equation. The produced carbon dioxide was collected in a glass vessel at 25.0 ⁰C. What is the right volume of the vessel that we should use, knowing that the pressure of CO2 gas at that temperature is 740 torr. CaCO3(s) CO2(g) + CaO(s)

GIVEN: 1 mol of CaCO3 1 mol CO2 USED: 1.25 g CaCO3 25.0 ⁰C PCO2 = 740 torr nCaCO3 = 1.25/100.1 = 1.25x10-2 mol nCaCO3 = nCO2 = 1.25x10-2 mol PV=nRT (P should be in atm unit and R unit is atm.L/mol.K) atm. mol. P = 740 torr x 1 atm/760 torr = 0.974 atm Vvessel=VCO2= nRT/P = 1.25 x 0.00821x 298/0.974 atm = 0.314 L (at least the volume of the container)
34

(Remember: (Remember: volume of 1 mole of gas at STP = 22.4 L, which is called the molar volume of gas) 22.

What volume and mass of H2 at STP will produce when 0.150 g of Al is treated 3H2(g) + 2AlO2with NaOH solution as shown: 2Al(s) + OH-(aq) + 2H2O

GIVEN: 2 mol of Al USED: 0.150 g Al STP

3 mol H2

nAl = 0.150/27.1 = 0.00556 mol 2 mol of Al 3 mol H2 0.00556 mol Al x x = 0.00834 mol H2 At STP: 1 mol H2 22.4 L 0.00834 mol H2 y y = 0.187 L

WtH2 = Mwt x n = 2.0 x 0.00834 = 0.017 g
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Dalton’s Law of Partial Pressures
V and T are constant

P1

P2

Ptotal = P1 + P2
36

A 0.25 dm3 of O2 at 25 ⁰C and 80 Kpa was mixed with 0.10 L of CO2 at 25 ⁰C and 50 Kpa. If the final volume is 0.20 L, calculate the total pressure of the mixture?

37

0.25 L, 25 ⁰C, and 80 Kpa of O2 You have

3

0.10 L, 25 ⁰C, and 50 Kpa of CO2 You have

1

2

38

LET US SEE THE SOLUTION!
A 0.25 dm3 of O2 at 25 ⁰C and 80 Kpa was mixed with 0.10 L of CO2 at 25 ⁰C and 50 Kpa. If the final volume is 0.20 L, calculate the total pressure of the mixture?

USED: 0.25 L, 25 ⁰C, and 80 Kpa of O2 0.10 L, 25 ⁰C, and 50 Kpa of CO2 Vf = 0.20 L Constant temperature and number of moles Pi Vi = nRT …………..gas O2 PfVf = nRT……………gas O2 Pi Vi=PfVf Devide i/f 80x0.25=0.20xPf Pi Vi = nRT …………..gas CO2 PfVf = nRT……………gas CO2 Pi Vi=PfVf Devide i/f 50x0.10=0.20xPf

PT=PO2 + PCO2 CO2 = 100 + 25 =125 Kpa the total pressure of both gases together

39

A 4.00 L of high pressure tank was filled with 50.0 g of O2 and 150 g of N2. Calculate the total pressure inside the tank at 25 ⁰C.

USED: 4.0 L, 25 ⁰C, and 50.0 g of O2 4.0 L, 25 ⁰C, and 150 g of N2 nO2 = 50.0/32.00 = 1.56 mol nN2 = 150/28.00 = 5.36 mol PV=nRT PO2 = 1.56 x 0.0821 x 298.15 / 4.00 = 9.54 atm PN2 = 5.36 x 0.0821 x 298.15 / 4.00 = 32.8 atm PT = PN2 + PO2 = 9.54 + 32.8 = 42.3 atm

40

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

nART PA = V nBRT PB = V PT = PA + PB PA = XA PT

nA is the number of moles of A nB is the number of moles of B

nA XA = nA + nB PB = XB PT

nB XB = nA + nB

Pi = Xi PT

mole fraction (Xi ) =

ni nT
41

A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?

Pi = Xi PT

PT = 1.37 atm = 0.0132

0.116 Xpropane = 8.24 + 0.421 + 0.116

Ppropane = 0.0132 x 1.37 atm = 0.0181 atm

42

Assuming that the air contents are only nitrogen an oxygen gases; calculate the mole fractions and mole fraction percent of each gas if their partial pressure; PO2 = 160 torr and PN2 = 600 torr.

USED: PO2 = 160 torr PN2 = 600 torr PT = PN2 + PO2 = 600 + 160 = 760 torr XA = PA/PT XO2 = PO2/PT = 160/760 = 0.211 XN2 = PN2/PT = 600/760 = 0.789

Attention XT = 0.211+0.789 =1

% Mole fraction of O2 = XO2 /XT x 100%= 0.211/1 x 100% = 21.1% % Mole fraction of N2 = XN2 /XT x 100%= 0.789/1 x 100% = 78.9%
43

What pressure in atm is exerted by a mixture containing 21.0 g of nitrogen and 8.00 g of oxygen gasses at 273 K in 5.0 L container.

USED: 5.0 L, 273 K, and 8.00 g of O2 5.0 L, 273 K, and 21.0 g of N2 nT = nO2 + nN2 = 8/32 + 21/28 = 1.00 mol PTVT=nTRT PT = 1.00 x 0.0821 x 273/5.0 = 4.48 atm

44

Collecting a Gas over Water

2KClO3 (s)

2KCl (s) + 3O2 (g)
45
2 2

PH2O = vapor pressure of H2O depends only on the temperature of water and can be obtained from standard tables, so the pressure of the gas is determined by; Pgas = Patm – PH2O

46

A 310 mL of oxygen was collected over water at 20.0 ⁰C and 738 torr. Estimate the partial pressure of the oxygen. What will be the volume of the “dry” oxygen at STP? (PH2O=17.5 torr at this temperature)

USED: 310 mL, 20.0 ⁰C, 738 torr PH2O = 17.5 torr at same temp. PO2 = Patm – PH2O = 738 – 17.5 = 720 torr USED: 720 torr, 310 mL, 20 ⁰C…………….initial 760 torr, V = ? mL, 273.15 K……..STP CONSTANT NUMBER OF MOLES Derive the equation: P1V1/T1 = P2V2/T2 720 x 310 / 293.15 = 760 x V2 / 273.15 V2 = 274 mL
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Kinetic Molecular Theory of Gases
1. A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. 2. Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic. 3. Gas molecules exert neither attractive nor repulsive forces on one another. 4. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy KE = ½ mu2
48

Kinetic theory of gases and …
• Compressibility of Gases • Boyle’s Law
P ∝ collision rate with wall Collision rate ∝ number density Number density ∝ 1/V P ∝ 1/V

• Charles’ Law
P ∝ collision rate with wall Collision rate ∝ average kinetic energy of gas molecules Average kinetic energy ∝ T P∝T

49

Kinetic theory of gases and …
• Avogadro’s Law
P ∝ collision rate with wall Collision rate ∝ number density Number density ∝ n P∝n

• Dalton’s Law of Partial Pressures
Molecules do not attract or repel one another P exerted by one type of molecule is unaffected by the presence of another gas Ptotal = ΣPi

50

The distribution of speeds for nitrogen gas molecules at three different temperatures

The distribution of speeds of three different gases at the same temperature

urms =

√M

3RT

urms = root-mean-square
(average molecular speed)

At the higher temperatures, more molecules are moving at faster speed On average, at a given temperature, the lighter molecules are moving faster 51

Calculate the root-mean-square speeds of helium atoms and nitrogen molecules in m/s at 25 ⁰C. USED: He and N2 Provided: R = 8.314 J/K.mol M in g/mol 1 J = 1 kg•m2/s2

urms = He urms = N2 urms =

√ √ √

3RT

M
3(8.314 J/K•mol)(298K) 4.003 x 10-3 kg/mol 3(8.314 J/K•mol)(298K) 2.802 x 10-2 kg/mol = 1.36 x 103 m/s = 515 m/s
52

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. r1 r2 molecular path

=

√

M2 M1

NH4Cl

M = M.wt

NH3 17 g/mol

HCl 36 g/mol
53

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

r1 r2

=

t2 t1

=

√

M2 M1

Nickel forms a gaseous compound of the formula Ni(CO)x What is the value of x given that under the same conditions methane (CH4) effuses 3.3 times faster than the compound? r1 2 x M1 = (3.3)2 x 16 = 174.2 r1 = 3.3 x r2 M2 = r2 x = 4.1 ~ 4 54 M1 = 16 g/mol 58.7 + x • 28 = 174.2

( )

Calculate the molecular weight of a gas X, if it effuses 0.876 times as rapidly as nitrogen? USED: rx = 0.876 x rN2 rx rN2 =

√

MN2 Mx

0.876 x rN2 = rN2

√

28.0

Mx
Mx = 36.5 g/mol

55

Which diffuses more rapidly and by what factor, ammonia or hydrogen chloride? Given: MNH3 = 17.03 g/mol MHCl = 36.46 g/mol rNH3 rHCl rNH3 rHCl

=

=

√ √

MHCl MNH3
36.46 rNH3 rHCl 17.03 = 1.463 rNH3 = 1.463 rHCl This means ammonia diffuses 1.463 times more rapidly than HCl under same conditions

56

Deviations from Ideal Behavior 1 mole of ideal gas PV = nRT PV = 1.0 n= RT Repulsive Forces

Attractive Forces

Ideal gas only at low pressure ≤ 5 atm As Pressure density distance intermolecular forces Nonideal Gas
57

Nonideal Gas

As Temperature

Average KE

Effect of intermolecular forces on the pressure exerted by a gas.

The speed of a molecule that is moving toward the container wall (red sphere) is reduced by the attractive forces exerted by its neighbors (gray spheres). Consequently, the impact this molecule makes with the wall is not as great as it would be if no intermolecular forces were present. In general, the measured gas pressure is lower than the pressure the gas would exert if it behaved ideally.
58

Van der Waals equation nonideal gas an2 (V – nb) = nRT ( P + V2 ) corrected pressure

P = real/observed pressure, atm V = volume of the container, L a = constant, (atm•L2/mol2) b = constant, (L/mol)
As a the attraction force

GENERALLY: As

}

corrected volume

} the molecular / atomic size
59

b

Given that 2.75 moles of CO2 occupy 4.70 L at 53 °C, calculate the pressure of the gas (in atm) using (a) the ideal gas equation and (b) the van deer Waals equation. (a=3.59 atm.L2/mol2, b=0.0428 L/mol)

(a) PV=nRT

P = 2.75 x 0.0821 x 326 / 4.70 = 15.7 atm

(b) an2/V2 = 3.59 atm•L2/mol2 x (2.75 mol)2 / (4.70)2 = 1.23 atm nb = 2.75 mol x 0.0427 L/mol = 0.117 mol
(P+1.23 atm)(4.70 L-0.117 L) = (2.75 mol)(0.0821 L•atm/K•mol )(326 K)

P = 14.8 atm

60