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Vanessa Prasad-PermaulValencia Community College

CHM 1045

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Elements that exist as gases at 250C and 1 atmosphere

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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.

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a) Gas is a large collection of particles moving at random throughout a volume

b) Collisions of randomly moving particles with the walls of the container exert a force per unit area that we perceive as gas pressure

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HOW IS PRESSURE DEFINED?

The force the gas exerts on a given area of the container in which it is contained. The SI unit for pressure is the Pascal, Pa.

Pressure = ForceArea

• If you’ve ever inflated a tire, you’ve probably made a pressure measurement in pounds (force) per square inch (area).

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Units of Pressure

1 pascal (Pa) = 1 kg/m·s2

1 atm = 760 mmHg = 760 torr

1 atm = 101,325 Pa

1 bar = 105 Pa

1 atm = 14.69 lb/in2

Barometer

760mm

mercury

Height is proportional to the barometric pressure

Hg is used instead of H2O :• more dense • better visibility• accuracy

EXERCISE 5.1A GAS CONTAINER HAD A MEASURED PRESSURE OF57kPa. CALCULATE THE PRESSURE IN UNITS OF ATMAND mmHg. First, convert to atm (57 kPa = 57 x 103 Pa). 57 x 103 Pa x 1 atm = 0.562 =

0.56 atm 1.01325 x 105 Pa

Next, convert to mmHg. 57 x 103 Pa x 760 mmHg = 427.5 = 4.3 x

102 mmHg 1.01325 x 105 Pa

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EMPIRICAL GAS LAWS

You can predict the behavior of gases based on the following properties:

• Pressure

• Volume

• Amount (moles)

• Temperature

* If two of these physical properties are held constant, it is possible to show a simple relationship between the

other two…

9As P (h) increases V decreases

Hg Hg

Boyle’s experiment: A manometer to study the relationship between

pressure (P) & Volume (V) of a gas

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BOYLE’S LAW

the volume of a sample of gas at a given temperature varies inversely with the applied pressure

So….

For a given amount of gas (n) &

@ constant temperature (T) :

If pressure (P) increases, the volume (V) of the gas decreases

P1 * V1 = P2 * V2

• Pressure–Volume Law (Boyle’s Law):

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P a 1/VP * V = constant

P1 * V1 = P2 * V2

Boyle’s Law

Constant temperatureAny given amount of gas

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EXERCISE 5.2A volume of carbon dioxide gas equivalent to 20.0 L was collected @ 23oC and 1.00atm pressure. What would be the volume of gas at constant temperature and 0.830atm?

P1 * V1 = P2 * V2

Application of Boyle’s law gives

V2 = V1 x P1 = 20.0 L x 1.00atm = 24.096 = 24.1 L

P2 0.830atm

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Charles’ Law: relating volume and temperature

• Temperature–Volume Law (Charles’ Law):

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Charles’ Law

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For a given amount of gasat constant pressure

V a T

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

Charles’ Law

Temperature must be in Kelvin

V / T = constant

Variation of gas volume with temperature

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CHARLES’ LAW

the volume occupied by any sample of gas at a constant pressure is directly proportional to the absolute temperature

So….

For a given amount of gas (n) & @ constant pressure (P) :

If temperature (T) increases, the volume (V) of the gas increases

V1 V2

T1 = T2

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EXERCISE 5.3A chemical reaction is expected to produce 4.3dm3 of oxygen at 19oC and 101kPa. What will the volume be at constant pressure and 25oC?

First, convert the temperatures to the Kelvin scale.Ti = (19 + 273) = 292 KTf = (25 + 273) = 298 K

Following is the data table.Vi = 4.38 dm3 Pi = 101 kPa Ti = 292 KVf = ? Pf = 101 kPa Tf = 298 K

Apply Charles’s law to obtainVf = Vi x Tf = 4.38 dm3 x 298K = 4.470 = 4.47 dm3

Ti 292K

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COMBINED GAS LAW

Relating Volume, Temperature and Pressure

Taking Boyle’s Law and Charles’ Law:

The volume occupied by a given amount of gas is proportional to the absolute temperature divided by the pressure:

V = constant x T or PV = constant (for a given amount of gas)

P T

P1V1 = P2V2

T1 T2

We can combine Boyle’s and charles’ law to come up with the combined gas law

Use Kelvins for temp, any pressure, any volume

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1

11 T

VP

T

VP

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Combined Gas Law

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EXERCISE 5.4A balloon contains 5.41dm3 of helium at 24oC and 10.5kPa. Suppose the gas in the balloon is heated to 35oC and the pressure is now 102.8kPa, what is the volume of the gas?

First, convert the temperatures to kelvins.Ti = (24 + 273) = 297 KTf = (35 + 273) = 308 K

Following is the data table.Vi = 5.41 dm3 Pi = 101.5 kPa Ti = 297 KVf = ? Pf = 102.8 kPa Tf = 308 K

Apply both Boyle’s law and Charles’s law combined to getVf = Vi x Pi x Tf =5.41 dm3 x 101.5kPa x 308K = 5.539 = 5.54 dm3

Pf Ti 102.8kPa 297K

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V number of moles (n)V = constant x nV1/n1 = V2/n2

Constant temperatureConstant pressure

equal volumes of any two gases at the same temperature & pressure contain the same number of molecules

• The Volume–Amount Law (Avogadro’s Law):

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Avogadro’s Law

nV

11

1 knV

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Avogadro’s Law

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Avogadro’s Number = one mole of any gas contains the same number of molecules 6.023 x 1023

Must occupy the same volume at a given temperature and pressureThe conditions 0 0C and 1 atm are called

standard temperature and pressure (STP).

Experiments show that at STP, 1 mole of an ideal gas occupies 22.4 L.

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

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

Boyle’s law: V (at constant n and T)1P

V nT

P

V = constant x = RnTP

nTP

R is the molar gas constant

PV = nRT

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PV = nRT

R = PV

nT=

(1 atm)(22.414L)

(1 mol)(273.15 K)

R = 0.082057 L • atm / (mol • K)R = 8.3145 J / (mol • K)R = 1.9872 cal / (mol • K)

*The units of pressure times volume are the units of energy joules (J) or calories (cal)

IDEAL GAS EQUATION

Ideal gases obey an equation incorporating the laws of Charles, Boyle, and Avogadro.

1 mole of an ideal gas occupies 22.414 L at STP

STP conditions are 273.15 K and 1 atm

pressure

The gas constant R = 0.08206 L·atm·K–1·mol–1

P has to be in atm

V has to be in L

T has to be in K

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TRnVP

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EXERCISE 5.5Show that the moles of gas are proportional to the pressure for constant volume and temperature

Use the ideal gas law, PV = nRT, and solve for n:n = PV RT

n = V x P RT

Note that everything in parentheses is constant. Therefore, you can write

n = constant x POr, expressing this as a proportion,

n P

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EXERCISE 5.6What is the pressure in a 50.0L gas cylinder that contains 3.03kg of oxygen at 23oC?T = 23oC + 273K = 296KV = 50.0LR = 0.08206 L . atm/K . moln = 3.03kg x 1000g x 1mol = 94.688 mol O2

1 kg 31.998gP = ?PV= nRT P = nRT = 0.0347mol x 0.08206 L . atm x 296K V K . mol = 46.0 atm 50.0L

• Density and Molar Mass Calculations:

• You can calculate the density or molar mass (MM) of a gas. The density of a gas is usually very low under atmospheric conditions.

TR

MMP

V

MMnd

volume

mass

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The Ideal Gas Law

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EXERCISE 5.7Calculate the density of helium (g/L) at 21oC and 752mmHg. The density of air under these conditions is 1.188g/L. What is the difference in mass between 1 liter of air and 1 liter of helium?

Variable Value

P752 mmHg x 1 atm = 0.98947 atm 760mmHg

V 1 L (exact number)T (21 + 273) = 294 Kn ?

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Using the ideal gas law, solve for n, the moles of helium.

n = PV = 0.98947 atm x 1L = RT 0.08206 L . atm/ K . mol x 294K

Now convert mol He to grams.0.04101 mol He x 4.00g He = 0.16404g He = 0.164g/L 1.00 mol He 1 LTherefore, the density of He at 21°C and 752 mmHg is 0.164 g/L.

The difference in mass between one liter of air and one liter of He:

mass air − mass He = 1.188 g − 0.16404 g = 1.02396 = 1.024 g difference

0.4101 mol

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EXERCISE 5.8A sample of a gaseous substance at 25oC and 0.862 atm has a density of 2.26 g/L. What is the molecular mass of the substance?

Variable Value P 0.862 atm V 1 L (exact number) T (25 + 273) = 298 K n ?

From the ideal gas law, PV = nRT, you obtain

n = PV = 0.862 atm x 1 L = 0.03525 mol RT 0.08206 L . atm/K . Mol x 298K

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Dividing the mass of the gas by moles gives you the mass per mole (the molar mass).

Molar mass = grams of gas = 2.26g = 64.114 g/mol moles of gas 0.03525mol

Therefore, the molecular mass is 64.1 amu.

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Dalton’s Law of Partial Pressures

V and T are

constant

P1 P2 Ptotal = P1 + P2

• For a two-component system, the moles of components A and B can be represented by the mole fractions (XA

and XB).

1

BA

BA

BB

BA

AA

XX

nn

nX

nn

nXMole fraction is related

to the total pressure by:

totii PXP

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Dalton’s Law of Partial Pressures

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EXERCISE 5.10 A 10.0L flask contains 1.031g O2 and 0.572g CO2 at 18oC. What are the partial pressures of each gas? What is the total pressure? What is the mole fraction of oxygen in this mixture?

Each gas obeys the ideal gas law.

1.031 g O2 x 1 mol = 0.0322188 mol O2

32.00g

P = nRT = 0.0322mol x 0.0821L.atm/K.mol x 291K = 0.076936 atm V 10.0L

0.572 g CO2 x 1 mol = 0.012997 mol CO2

44.01g

P = nRT = 0.0130mol x 0.0821L.atm/K.mol x 291K = 0.031036 atm V 10.0L

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The total pressure is equal to the sum of the partial pressures:

PT = PO2 + PCO2

= 0.076936atm + 0.031036atm =

0.10797 = 0.1080 atm

The mole fraction of oxygen in the mixture is

Mole fraction O2 = PO2 = 0.076936atm = 0.7122 =

0.712 PT 0.1080atm

0.712 x 100% = 71.2 mole % of O2 in this gas mixture

This theory presents physical properties of gases in terms of the motion of individual molecules.

• Average Kinetic Energy Kelvin Temperature

• Gas molecules are points separated by a great distance

• Particle volume is negligible compared to gas volume

• Gas molecules are in constant random motion

• Gas collisions are perfectly elastic

• Gas molecules experience no attraction or repulsion 40

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Average Kinetic Energy (KE) is given by:

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KE 1

2mu2

MM

RT

mN

RTu

A

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U = average speed of a gas particle

R = 8.314 J/K mol

m = mass in kg

MM = molar mass, in kg/mol

NA = 6.022 x 1023

The Root–Mean–Square Speed: is a measure of the average molecular speed.

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MM

RTu

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Taking square root of both sides gives the equation

MM

RTurms

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

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• Maxwell speed distribution curves.

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Kinetic Molecular Theory

• Diffusion is the

mixing of different

gases by random

molecular motion

and collision.

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Graham’s Law

• Effusion is when

gas molecules

escape without

collision, through a

tiny hole into a

vacuum.

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Graham’s Law

• Graham’s Law: Rate of effusion is proportional to its rms speed, urms.

• For two gases at same temperature and pressure:

MM

RTRate rms

3 u

1

2

1

2

2

1

MM

MM

MM

MM

Rate

Rate

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Graham’s Law

•At higher pressures, particles are much

closer together and attractive forces become

more important than at lower pressures.

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Behavior of Real Gases

•The volume taken up by gas particles is actually less important at lower pressures than at higher pressure. As a result, the volume at high pressure will be greater than the ideal value.

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Behavior of Real Gases

• Corrections for non-ideality require van der Waals equation.

nRTbnVV

naP –

2

2

IntermolecularAttractions

ExcludedVolume

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Behavior of Real Gases

A sample of argon gas has a volume of 14.5 L at

1.56 atm of pressure. What would the pressure be

if the gas was compressed to 10.5 L? (at constant

temperature and moles of gas)

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Example 1: Boyle’s Law

A sample of CO2(g) at 35C has a volume of 8.56

x10-4 L. What would the resulting volume be if

we increased the temperature to 85C? (atconstant moles and pressure)

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6.53 moles of O2(g) has a volume of 146 L. If we

decreased the number of moles of oxygen to 3.94

moles what would be the resulting volume?(constant pressure and temperature)

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Oxygen gas is normally sold in 49.0 L steel containers at a pressure of 150.0 atm. What volume would the gas occupy if the pressure was reduced to 1.02 atm and the temperature raised from 20oC to 35oC?

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An inflated balloon with a volume of 0.55

L at sea level, where the pressure is

1.0 atm, is allowed to rise to a height

of 6.5 km, where the pressure is about

0.40 atm. Assuming that the

temperature remains constant, what is

the final volume of the balloon?

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Sulfur hexafluoride (SF6) is a colorless,

odorless, very unreactive gas.

Calculate the pressure (in atm) exerted

by 1.82 moles of the gas in a steel

vessel of volume 5.43 L at 69.5°C.

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What is the volume (in liters) occupied by 7.40 g of CO2 at STP?

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What is the molar mass of a gas with a

density of 1.342 g/L at STP?

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What is the density of uranium hexafluoride, UF6, (MM = 352 g/mol) under conditions of STP?

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The density of a gaseous compound is 3.38 g/L at 40°C and 1.97 atm. What is its molar mass?

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Exactly 2.0 moles of Ne and 3.0 moles of Ar were placed in a 40.0 L container at 25oC. What are the partial pressures of each gas and the total pressure?

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A sample of natural gas contains 6.25 moles of methane (CH4), 0.500 moles

of ethane (C2H6), and 0.100 moles of

propane (C3H8). If the total pressure of

the gas is 1.50 atm, what are the partial pressures of the gases?

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What is the mole fraction of each component in a mixture of 12.45 g of

H2, 60.67 g of N2, and 2.38 g of NH3?

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On a humid day in summer, the mole

fraction of gaseous H2O (water vapor)

in the air at 25°C can be as high as 0.0287. Assuming a total pressure of 0.977 atm, what is the partial pressure

(in atm) of H2O in the air?

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In gas stoichiometry, for a constant temperature and pressure, volume is proportional to moles.

Example: Assuming no change in temperature and pressure, calculate the volume of O2 (in liters) required for the

complete combustion of 14.9 L of butane (C4H10):

2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O(l)

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All of the mole fractions of elements in a given compound must add up to?

1. 1002. 13. 504. 2

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Hydrogen gas, H2, can be prepared by letting zinc metal react with aqueous HCl. How many liters of H2 can be prepared at 742 mm Hg and 15oC if 25.5 g of zinc (MM = 65.4 g/mol) was allowed to react?

Zn(s) + 2 HCl(aq) H2(g) + ZnCl2(aq)

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Under the same conditions, an unknown gas diffuses 0.644 times as fast as

sulfur hexafluoride, SF6 (MM = 146

g/mol). What is the identity of the unknown gas if it is also a hexafluoride?

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What are the relative rates of diffusion of the three naturally occurring isotopes of neon: 20Ne, 21Ne, and 22Ne?

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• Deviations result from assumptions about ideal gases.

1. Molecules in gaseous state do not exert any force, either attractive or repulsive, on

one another.

2. Volume of the molecules is negligibly small compared with that of the container.

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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. (a = 4.17, b = 0.0371)

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Assume that you have 0.500 mol of N2 in a volume of 0.600 L at 300 K. Calculate the pressure in atmospheres using both the ideal gas law and the van der Waals equation.

For N2, a = 1.35 L2·atm mol–2, and b = 0.0387 L/mol.

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