chapter 9 gases: their properties and behavior. three states of matter solidsliquidsgases

Chapter 9

Gases: Their Properties and Behavior

Three States of MatterThree States of Matter

Solids Liquids Gases

Properties of Gases

• Gases mix completely with one another to form homogenous mixtures

• Gases can be compressed (keyboard cleaner)• Gases exert pressure on what ever is around

them (balloon, canister)• Gases expand into whatever volume is available

(coke bottle and balloon)• Gases are described in terms of their

temperature and pressure, the volume occupied and the amount of gas present (gas properties)

Compressibility of Gases

• Airbags fill with NAirbags fill with N22 gas in an gas in an

accident. accident. • Gas is generated by the Gas is generated by the

decomposition of sodium azide, decomposition of sodium azide, NaNNaN33..

• 2 NaN2 NaN33 2 Na + 3 N 2 Na + 3 N22

Composition of AirNitrogen

Oxygen

Argon

Carbob dioxide

Gases and Gas Pressure

Gas % by Vol Gas % by VolNitrogen 78.084 Krypton 0.0001

Oxygen 20.948 Carbon monoxide 0.00001

Argon 0.934 Xenon 0.000008

Carbon dioxide 0.033 Ozone 0.000002

Neon 0.00182 Ammonia 0.000001

Hydrogen 0.0010 Nitrogen dioxide 0.0000001

Helium 0.00052 Sulfur dioxide 0.00000002

Methane 0.0002

Factors Affecting Gas Pressure

• Temperature– Raising the temperature of a gas increases

the pressure– The faster moving particles collide with the

walls of the container more frequently– Temperature and pressure is also a linear

relationship– In contrast, decreasing the temperature of a

gas decreases the pressure

Properties of Gases

• Pressure (P)Pressure = force / area

Force = mass x acceleration

• Units of Pressure– Atmosphere (atm)– Torr – Pascals (Pa)– mmHg

• Pressure conversions– 1 atm = 1.01325 x 105 Pa– 1 atm = 760 torr– 1 atm = 760 mmHg

Atmospheric Pressure

• - pressure created from the mass of the atmosphere pressing down on the earth’s surface– Standard

atmospheric pressure at sea level – 760 mmHg

Problem

• Convert these pressure values.– 120 mmHg to atm– 100 kPa to mmHg– 270 torr to atm

Properties of Gases

• Volume (V)– mL– L– cm3

• Amount of gas (n) – moles

• Temperature (T) - Kelvins

The Gas Laws

• Gas properties– Gases are described in terms of their

temperature and pressure, the volume occupied and the amount of gas present (gas properties)

• Gas Laws can be derived using– Kinetic Molecular Theory

The Gas Laws

• Ideal Gas: A gas whose behavior follows the gas laws exactly.

• The physical properties of a gas can be defined by four variables:

P pressure

T temperature

V volume

n number of moles

The Gas Laws

• The Pressure-Volume Relationship: Boyle’s Law– The volume (V) of an ideal gas varies

inversely with the applied pressure (P) when temperature (T) and the amount (n, moles) are constant

PiVi = PfVf

PV = kVP

1

The Gas LawsBoyle’s Law

VP

1

(constant n and T)PV = k

Sample Problem UsingBoyle’s Law

A balloon contains 30.0 L of helium gas at 103kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa? (assume the temperature remains constant)

What do you think will happen to the volume at a higher temperature knowing what you know already about gases?


P1 = 103 kPa P2 = 25.0 kPaV1 = 30.0 L V2 = ? L

P1V1 = P2V2 or P1V1 / P2 = V2

V2 = (30.0 L) (103 kPa) 25.0 KPa

V2 = 1.24 x 102 L (3 sig figs)


Nitrous oxide (N2O) is used as an anesthetic. The pressure on 2.50 L of N2O changes from 105 KPa to 40.5 KPa. If the temperature does not change, what will the new volume be?

P1 = 105 kPa P2 = 40.5 kPaV1 = 2.50 L V2 = ? L

P1V1 = P2V2 or P1V1 / P2 = V2

V2 = (2.50 L) (105 kPa) 40.5 KPa

V2 = 6.48 L (3 sig figs)


A gas with a volume of 4.00 L at a pressure of 205 KPa is allowed to expand to a volume of 12.0 L. What is the pressure in the container if the temperature remains constant?

P1 = 205 kPa P2 = ? kPaV1 = 4.00 L V2 = 12.0 L

P1V1 = P2V2 or P1V1 / V2 = P2

P2 = (4.00 L) (205 kPa) 12.0 L

P2 = 68.3 KPa (3 sig figs)


The volume of a gas at 99.6 KPa and 24ºC is 4.23L. What volume will it occupy at 93.3 KPa and 24ºC?

P1 = 99.6 kPa P2 = 93.3 kPa T1 = 24ºCV1 = 4.23 L V2 = ? L T2 = 24ºC

P1V1 = P2V2 or P1V1 / P2 = V2

V2 = (4.23 L) (99.6 kPa) 93.3 kPa

V2 = 4.52 L (3 sig figs)

Problems

• A sample of nitrogen gas at 298 K and 745 torr has a volume of 37.42 L. What volume will it occupy if the pressure is increased to 894 torr at constant temperature?

– A)22.3 L – B)31.2 L – C)44.9 L – D)112 L – E)380 L

Problems

• A sample of carbon dioxide gas at 125°C and 248 torr occupies a volume of 275 L. What will the gas pressure be if the volume is increased to 321 L at 125°C?

– A)212 torr – B)289 torr – C)356 torr – D)441 torr – E)359 torr

The Gas Laws

• The Temperature-Volume Relationship – Charles Law

– The volume (V) of an ideal gas varies directly with absolute temperature (T) when pressure (P) and amount (n) are constant.

Vi / Ti = Vf / Tf

=Tfinal

Vfinal

Tinitial

Vinitial

= kT

VV T

V T

The Gas LawsCharles’ Law

(constant n and P)= kT

V

Sample Problem UsingCharles’s Law

A balloon inflated in a room at 24ºC has a volume of 4.00 L. The balloon is then heated to a temperature of 58ºC. What is the new volume if the pressure remains constant?

T1 = 24ºC or 297 K V1 = 4.00 LT2 = 58ºC or 331 K V2 = ? L

When using gas laws always express the temperatures in kelvins!

T1 = 24ºC + 273 = 297 KT2 = 58ºC + 273 = 331 K


A balloon inflated in a room at 24ºC has a volume of 4.00 L. The balloon is then heated to a temperature of 58ºC. What is the new volume if the pressure remains constant?


V1 = V2 or V1T2 = V2

T1 T2 T1

V2 = (4.00 L) (331 K) = 4.46 L297 K


If a sample of gas occupies 6.80 L at 325ºC, what will its volume be at 25ºC if the pressure does not change?


V1 = V2 or V1T2 = V2

T1 T2 T1

V2 = (6.80 L) (298 K) = 3.39 L598 K

Problems

• A sample container of carbon monoxide occupies a volume of 435 mL at a pressure of 785 torr and a temperature of 298 K. What would its temperature be if the volume were changed to 265 mL at a pressure of 785 torr?

– A)182 K – B)298 K – C)387 K – D)489 K – E)538 K

Problems

• A 0.850-mole sample of nitrous oxide, a gas used as an anesthetic by dentists, has a volume of 20.46 L at 123°C and 1.35 atm. What would be its volume at 468°C and 1.35 atm?

– A)5.38 L – B)10.9 L – C)19.0 L – D)38.3 L – E)77.9 L

The Gas Laws

• The Amount-Volume Relationship: Avogadro’s Law– The volume (V) of an ideal gas varies directly

with amount (n) when temperature (T) and pressure (P) are constant

– V1 / n1 = V2 / n2

=nfinal

Vfinal

ninitial

Vinitial= k

n

VV n

The Gas LawsAvogadro’s Law

V n (constant T and P)= k

n

V

twice as many twice as many moleculesmolecules

• Only at STP– 0ºC– 1 atm

• This way we compare gases at the same temperature and pressure.

This is where we get the fact that 22.4 L =1 mole

Summary

Boyle’s Law Charles’ Law Avogadro’s Law

constant T & n

constant P & constant P & nn

constant P & constant P & TT

V = 1/PV = 1/P V = TV = T V = nV = n

PinitialVinitial = PfinalVfinal

=Tfinal

Vfinal

Tinitial

Vinitial

=nfinal

Vfinal

ninitial

Vinitial

The Ideal Gas Law

• Boyle’s, Charles’s and Avogadro’s Laws can be combined to form the Ideal Gas Law

– PV = nRT

• R – ideal gas constant– R = 0.0821 atm L / mol K– R = 62.36 torr L / mol K– R = 8.314 J / mol K

Sample Problem Using Ideal Gas Law

When the temperature of a rigid hollow sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 1.89 x 103 kPa. How many moles of helium does the sphere contain?

P = 1.89 x 103 kPa V = 685 L T = 621 K

PV = nRT or PV / RT = n

n = (1.89 x 103 kPa) (685 L) mol · K(8.31L · kPa) (621K)

n = 251 mol He

Sample Problem Using Ideal Gas Law

A child’s lungs can hold 2.20 L. How many grams of air do her lungs hold at a pressure of 102 kPa and a body temperature of 37ºC? Use a molar mass of 29 g for air.

P = 102 kPa V = 2.20 L T = 310 K

PV = nRT or PV / RT = n

n = (102 kPa) (2.20 L) mol · K(8.31L · kPa) (310K)

n = 0.087 mol air 0.087 mol air x 29g air / mol air = 2.5 g air

Problems

• A sample of nitrogen gas is confined to a 14.0 L container at 375 torr and 37.0°C. How many moles of nitrogen are in the container?

– A)0.271 mol – B)2.27 mol – C)3.69 mo1 – D)206 mol – E)227 mol

Stoichiometric Relationships with Gases

• Various questions can be asked that relate gas laws to stoichiometry.

Problems

• A 250.0-mL sample of ammonia, NH3(g), exerts a pressure of 833 torr at 42.4°C. What mass of ammonia is in the container?

– A)0.0787 g – B)0.180 g – C)8.04 g – D)17.0 g – E)59.8 g

Stoichiometric Relationships with Gases

• The ideal gas law can be used to determine density if the molar mass of the gas is known or the molar mass if the mass of gas is known

d = m / V = PM / RT

• Density increases with molar mass

Problems

• What is the density of carbon dioxide gas at -25.2°C and 98.0 kPa?

– A)0.232 g/L – B)0.279 g/L – C)0.994 g/L – D)1.74 g/L – E)2.09 g/L

Problems

• A flask with a volume of 3.16 L contains 9.33 grams of an unknown gas at 32.0°C and 1.00 atm. What is the molar mass of the gas?

– A)7.76 g/mol – B)66.1 g/mol – C)74.0 g/mol – D)81.4 g/mol – E)144 g/mol

Problems

• Dr. I. M. A. Brightguy adds 0.1727 g of an unknown gas to a 125-mL flask. If Dr. B finds the pressure to be 736 torr at 20.0°C, is the gas likely to be methane, CH4, nitrogen, N2, oxygen, O2, neon, Ne, or argon, Ar?

– A)CH4 – B)N2 – C)Ne – D)Ar – E)O2

chapter 9 gases: their properties and behavior. three states of matter solidsliquidsgases

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