Chapter 5: Gases

Renee Y. Becker

Valencia Community College

CHM 1045

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

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

• Units of pressure: atmosphere (atm)

Pa (N/m2, 101,325 Pa = 1 atm) Torr (760 Torr = 1 atm)

bar (1.01325 bar = 1 atm)

mm Hg (760 mm Hg = 1 atm)

lb/in2 (14.696 lb/in2 = 1 atm)

in Hg (29.921 in Hg = 1 atm)

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• Pressure–Volume Law (Boyle’s Law):

Boyle’s Law

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Pressure1

Volume

V1P1 k15

Boyle’s Law

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

• Temperature–Volume Law (Charles’ Law):

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Charles’ Law

V T

V1

T1=k1

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Charles’ Law

Example 2: Charles’ 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? (at constant moles and pressure)

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• 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

Example 3: Avogadro’s Law

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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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

Example 4:Combined Gas Law

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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Example 5: Gas Laws

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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The Ideal Gas Law

• 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

TRnVP

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Example 6: Ideal Gas Law

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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Example 7: Ideal Gas Law

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

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• 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

Example 8: Density & MM

What is the molar mass of a gas with a density

of 1.342 g/L at STP?

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Example 9: Density & MM

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

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Example 10: Density & MM

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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Dalton’s Law of Partial Pressures

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• In a mixture of gases the total pressure, Ptot, is the

sum of the partial pressures of the gases:

• Dalton’s law allows us to work with mixtures of

gases.

• T has to be in K

• V has to be in L

nV

RTP total

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Dalton’s Law of Partial Pressures

Example 11: Dalton

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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Example 12: Dalton

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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• 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

Example 13: Mole Fraction

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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Example 14: Partial Pressure

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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Gas Stoichiometry & Example

• 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)30

Example 15:

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

1. 100

2. 1

3. 50

4. 2

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Example 16:

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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Kinetic Molecular Theory

• 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 rapid random motion

• Gas collisions are perfectly elastic

• Gas molecules experience no attraction or

repulsion 33

Kinetic Molecular Theory

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• Average Kinetic Energy (KE) is given

by:

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

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

MM

RTu

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Taking square root of both sides gives the equation

MM

RTurms

3

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Example 17:

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

Example 18:

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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Example 19: Diffusion

• 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.

Behavior of Real Gases

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•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

Example 20: Ideal Vs. Van Der Waals

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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Example 21: Ideal Vs. Van Der Waals