Characteristics of Gases

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Characteristics of Gases

•Gases expand to fill a container

•Gases form homogeneous mixtures with other gases

•Gases can be easily compressed

•Readily measured properties of gas are its temperature,volume, and pressure

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1. Gases consist if large number of molecules that are in continuous, randomcontinuous, random motion

2. The volume of all molecules of the gas is negligible compared to the total volumenegligible compared to the total volumein which the gas is contained.

3. Attractive and repulsive forces between gas molecules are negligibleare negligible

4. Energy can be transferred between molecules during collisions, but the averagebut the averagekinetic energy of the molecules does not changekinetic energy of the molecules does not change with time, as long as the temperatureis constant. In other words collisions are perfectly elasticperfectly elastic.

5. The average kinetic energy of the molecules is proportional to absolute temperatureaverage kinetic energy of the molecules is proportional to absolute temperature.At any given temperature all gases have the same average kinetic energy

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P = F

APressure may be defined as a force exerted upon any surface:

Mass = 1.06 kg

Since F= mass x acceleration

Area = 7.85 x 10-3 m2

F = 1.06 kg x 9.81 m/sec2 = 10.4 N where N is an SI unit called a Newton with units of kg-m/s2

P = F

A=

10.4 N

7.85 x 10-3 m2

Gravitational force

= 1.32 x 103 N/m2 or pascals (Pa)

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P=F/A = (10,000 kg)(9.81 m/s2)/1 m2 = 1 x 105 Pa = 101.3 kPa

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Atmospheric pressure can be measured using a barometerAtmospheric pressure can be measured using a barometer

760 mm Hg760 torr101.3 kPa1 atm

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A manometer measures the pressures of encloses gasesA manometer measures the pressures of encloses gases

A closed-tube manometer measures pressures below

atmospheric pressure

An open-tube manometer measures pressures near

atmospheric pressure

For a given pressure difference, the difference in heights of the liquid levels in the two arms of the manometer is inversely proportional to the density of the liquid

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A manometer measures the pressures of encloses gasesA manometer measures the pressures of encloses gases

A manometer is filled with dibutylphthalate (D = 1.05 g/ml) If the conditions are such that h = 12.2 cm at 0.964 atm, what is the pressure of the enclosed gas in mm Hg (D =13.6 g/ml).

Patm = 0.964 atm 760 mm Hg

1 atm

x = 733 mm Hg

Ph = 12.2 cm 1.05 g

ml

x1 cm

x 10 mm = 9.4 mm Hg1 ml

13.6 gx

Because the pressure inside the manometer is less than atmospheric pressure, Pgas = P atm + Ph2

Pgas = 733 + 9.4 = 742 mm Hg

04/19/23 8The volume of a fixed quantity of gas maintained at constant temperature is inversely proportional to the pressure

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Boyle’s Law: V 1/P, where (n, T constant) or PV = k

04/19/23 10The volume of a fixed quantity of gas maintained at constant pressure is inversely proportional to the temperature

Charles’s Law: V T, where (n, P constant) or V/T = k

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P1 = P2 volume and moles

T1 T2 are constant (n, V)

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The pressure of nitrogen gas in a 12.0 L tank at 27°C is 2300 lb/in2. Whatvolume would the gas in the tank have at 1 atm pressure (14.7 lb/in2) if thetemperature remains unchanged?

Boyle’s Law: V 1/P, where (n, T constant) or PV = kP1V1 = P2V2

(2300 lb/in2)(12.0 l) = (14.7 lb/in2)(V2) V2 = 1880 L

The gas pressure in an aerosol can is 1.5 atm at 25° C. Assuming that the gasinside obeys the ideal-gas equation what would be the pressure if the can were heated to 450°C?

Gay-Lussac’s Law: P T, where (n,V constant) or P/T = kP1 P2

T1 T2

1.5 atm = P2

298 K = 723 K

=

=

P2 = 3.6 atm

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Gay-Lussac’s Law: P T, where (n,V constant) or P/T = k


Boyle’s Law: V 1/P, where (n, T constant) or PV = k

V1 V2

T1 T2

P1V1 = P2V2

P1 P2

T1 T2

the Combined gas Law is:the Combined gas Law is:P1V1 P2V2

T1 T2

=

=

=

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A quantity of helium gas occupies a volume of 16.5 L at 78°C and 45.6 atm.What is its volume at STP?

The Combined gas Law is:The Combined gas Law is:

P1V1 P2V2

T1 T2

=

(45.6 atm)(16.5L) (1 atm)(V2)

351 K 273 K=

V2 = 585 L

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Volume 1LPressure 1 atmTemperature 0°CMass of gas 1.783 g# of particles 2.66 x 1022



Ar N2H2

The volume of a fixed quantity of gas maintained at constant pressure and temperature is directly proportional to the number of moles of gas

Avogadro’s Law: V n, where (R,T constant)

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Avogadro’s Law: V n, where (R,T constant) or V = kn


Boyle’s Law: V 1/P, where (n, T constant) or PV = k Since these Are TrueSince these Are True then:then:

V nT

P

V = RnT

P

PV = nRT

R = 0.0821 L-atm/K molR = 0.0821 L-atm/K mol = 62.36 L mmHg/K mol= 62.36 L mmHg/K mol

= 62.36 L torr/K mol= 62.36 L torr/K mol = 8.31 L kPa/K mol= 8.31 L kPa/K mol

= 8.31 J/K mol= 8.31 J/K mol

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What volume would 3.5g of Methane occupy at STP?

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Molar Mass and Gas DensitiesMolar Mass and Gas Densities•The ideal gas law may be used to determine the density or the molecularmass of a gas

n PSince PV = nRT, then:

V=

RT

Let M = the molar mass which is grams per one mole of a substance.

n M

V= P M

RT

If we multiply both sides of the expression by M, then the top left-hand sideof the expression becomes: or g/L which is density (d)moles

literx grams

mole

d = P MRT

M = dRTP

or

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Determine the molar mass of a gas if 2.889 g of gas completely fills a 936 ml container.

M = dRTP

2.889g

.936 L(0.0821 l-atm/mol-K) (304K)

735

760

1 atm

M = 79 7 g/mol

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(just in case you need it)

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Volumes of Gases in Chemical ReactionsVolumes of Gases in Chemical Reactions

The industrial synthesis of nitric acid involves the reaction of nitrogen dioxide gas with water:

3NO2(g) + H2O(l) 2HNO3(aq) + NO(g)

1. How grams of nitric acid can be prepared using 540 L of NO2 at a pressure of 5.00 atm and a temperature of 295 K?

nNO2 = PV

RT

(5.00 atm)( 540 liters)

(0.0821 l-atm/mol-K)(295 K)=

2 mol HNO3 46g

3 mol NO2 mol= 3418 g HNO3

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Gas Mixtures and Partial pressureGas Mixtures and Partial pressure•The ideal gas law may be used to determine the partial pressure of a mixture

of gases

Since PV = nRT, then:

This would be true of a single gas or mixture of gases

V

P1 = n1RT

V

P2 = n2RT

V

P3 = n3RT

According to Dalton’s law of partial pressures, the total pressure of a mixture of gas equals the sum ofAccording to Dalton’s law of partial pressures, the total pressure of a mixture of gas equals the sum of the pressures that each gas would exert if it were present alone.the pressures that each gas would exert if it were present alone.

Therefore Pt = (P1 + P2 + P3 +...)V = nRT

VPt = RT(n1+ n2 + n3+ …)

or

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What pressure, in atm, is exerted by a mixture of 2.00 g of H2 and 8.00 g ofN2 at 273 K in a 10 L vessel?

1 mol

2.02 g

2.00 g H2 = 0.990 mol H2

1 mol

28.0 g

8.00 g N2 = 0.286 mol N2

Pt (10.0 L) = (0.0821 l-atm/mol-K)(273 K)(0.990 mol + 0.286 mol)

Pt = 2.86 atm

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If a 5L container with N2 at 3atm is attached to a 3 L container with He at 3.3 atm what

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Partial Pressures and Mole FractionsPartial Pressures and Mole Fractions•The mole fraction, X, of any component of a mixture is simple the ratio of

the total number of moles in the mixture

VP1 =

n1RT

Total moles of mixtureMole fraction of component 1 = X1 = Moles of component 1

We can relate the mole fractions to partial pressures becauseWe can relate the mole fractions to partial pressures because

Therefore

Pt

V

ntRT=

n1

nt

= X1

P1 = X1Pt

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A study of the effects of certain gases on plant growth requires a synthetic atmosphere composed of 1.5 mol percent CO2, 18.0 mol percent O2, and 80.5 mol percent Ar. (a) Calculate the partial pressure of O2 in the mixture if the total pressure of the atmosphere is to be 745 mm Hg. (b) If this atmosphere is to be held in a 120 L space at 295 K, how many moles of O2 are needed?

P02 = 0.18 (745 mm Hg) = 134 mm Hg

•Converting mole percent to pressure in mm Hg

•Converting mm Hg pressure atmospheres

P02 = 134 mm Hg

760 mm Hg

1 atm= 0.176 atm

Since P1 = X1Pt

nO2 = PV

RT

(0.176 atm)( 120L)

(0.0821 l-atm/mol-K)(295 K)= = 0.872 mol

•Solving for moles of O2 using the ideal gas equation

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Collecting Gases Over WaterCollecting Gases Over Water

P total = P gas + PH2O

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Collecting Gases Over WaterCollecting Gases Over WaterSuppose the 2.00 L of oxygen is collected over water. The temperature of the waterand gas is 26°C and the pressure is 750 mm Hg. How many moles of O2 are collected?What volume would the O2 gas collected occupy when dry, at the same temperature and pressure?

Note: ( the vapor pressure of water at 26 °C is 25 mm Hg., Appendix B)

(a) Keep in mind that water vapor may be dealt with in Dalton’s partial pressure law just like any(a) Keep in mind that water vapor may be dealt with in Dalton’s partial pressure law just like any other gas.other gas.

P total = P gas + PH2O

PO2 = P total - PH2O

PPO2O2 = 750 = 750 - 25 = 725 mm Hg - 25 = 725 mm Hg

nO2 = PV

RT

(725 mm Hg/760 Hg)(0.200 L)

(0.0821 l-atm/mol-K)(299 K)= = 7.77 x 10= 7.77 x 10-3-3 moles O moles O22

(b) P(b) P11VV11 = P = P22VV22 (725 mm Hg)(0.200 L)1 = (750 mm Hg)(V2)

VV22= 0.193 L= 0.193 L

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At higher temperatures a greater fraction of molecules is movingAt higher temperatures a greater fraction of molecules is movingat greater speeds. This results in a higher average kinetics energyat greater speeds. This results in a higher average kinetics energy

Root-mean-square (rms) speedIndicates the speed of a particlepossessing average kinetic energy ( )such that = ½ m2

How does KMT explain PV=k, P/T = kHow does KMT explain PV=k, P/T = k

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= √3RT M

If two molecules which differ in mass have the same kinetic energy, then becauseEk = ½ m 2 is true, the velocities of the particles must be different; lighter particles must move with a higher average speed.

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Molecules Effusion and DiffusionMolecules Effusion and Diffusion

Calculate the rms speed () of a N2 molecules at 25 °C

= 3(8.31 kg-m2/s2-K-mol)( 298K)

Keep in mind that R needs to fit the specific circumstances you are dealing with.Since we are dealing with particles possessing mass and traveling at some velocity,R = 8.31 J/K-mol and J = kg-m2/s2, therefore R = 8.31 kg-m2/s2-K-mol

28 g1000g

1 kg

= 5.15 x 102 m/s

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Molecules Effusion and DiffusionMolecules Effusion and Diffusion

EffusionEffusion

DiffusionDiffusion

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Graham’s Law of EffusionGraham’s Law of Effusion

The effusion rate of a gas is inversely proportional to the square root of its molar massThe effusion rate of a gas is inversely proportional to the square root of its molar mass

r1

r2

=1

3RT/

2

=M1

3RT/M2

=M2

M1

=M2

M1

r1

r2

or

T1 = √ M1

T2 M2

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=M2

M1

Graham’s Law of EffusionGraham’s Law of Effusion

If an unknown gas effuses at a rate that is only 0.468 times that of O2 at the same temperature,what is the molar mass of the unknown gas?

rx

rO2

=32 g/mol

M1

0.468

M1= 146 g/mol

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Diffusion and Mean Free PathDiffusion and Mean Free Path

Diffusion speeds are slower than molecular speeds

The average distance traveled by a molecule between collisions is calledthe mean free pathmean free path. The higher the density of the gas the smaller the mean free path (the shorter the distance between collisions

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10 atm

PV=nRT deviates at high pressurePV=nRT deviates at high pressure

Deviations From Ideal BehaviorDeviations From Ideal Behavior

n =

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Deviations From Ideal BehaviorDeviations From Ideal Behavior

PV=nRT deviates at low temperaturePV=nRT deviates at low temperature

Note: as the temperature increases, the PV/RT behavior of gases approaches ideal behavior

Ideal Gas

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Deviations From Ideal BehaviorDeviations From Ideal BehaviorThe Van der Waals EquationThe Van der Waals Equation

P + n2aV2

(V-nb) = nRT

(V-nb) is a correction for the finite volume(V-nb) is a correction for the finite volumeof the gas molecules The Van der Waal of the gas molecules The Van der Waal constant, b, varies with each gas and is the constant, b, varies with each gas and is the actual volume occupied by the gas molecules actual volume occupied by the gas molecules and increases with an increase in the mass and increases with an increase in the mass or complexity of the moleculeor complexity of the molecule

The correction for pressure, nThe correction for pressure, n22a/Va/V22, takes, takesinto account the intermolecular attractions into account the intermolecular attractions between molecules. between molecules.

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If 1.000 mol of an ideal gas were confined to 22.41 L at 0.0°C, it would exert a pressureof 1.000 atm. Use the Van der Waals equation and the constants in Table 10.3 to estimate the pressure exert by 1.000 mol of Cl2(g) in 22.4 L at 0.0 °C

Deviations From Ideal BehaviorDeviations From Ideal BehaviorThe Van der Waals EquationThe Van der Waals Equation

P + n2aV2

(V-nb) = nRT

(1.000 mol)(0.0821 l-atm/mol-K)(273.2 K)

Solving for pressure: P = nRTV-nb

- n2aV2

22.4 L -(1.000 mol)(0.0562 l/mol)P = -

(1.000 mol)(6.49 L2-atm)/mol2

(22.4 L)2

P = 0.990 atm

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Characteristics of Gases