the gaseous state of matter

The gasesThe Gaseous State of Matter

The air in a hot air balloon expands When it is heated. Some of the airescapes from the top of the balloon, lowering the air density inside the balloon, making the balloon buoyant.

The atmosphere

Copyright 2011 John Wiley & Sons, Inc 12-2

The atmosphere protects the planet and provides chemicals necessary for life.

The Atmosphere


Each of the gases, N2, O2, CO2 and H2O among others in the atmosphere, serve a purpose

O2 supports metabolism

N2 dilutes the O2 so explosive combustion does not take place and is also part of proteins

CO2 and H2O trap heat

General Properties

Gases • Have an indefinite volume

Expand to fill a container• Have an indefinite shape

Take the shape of a container• Have low densities

• Have high kinetic energies


2

air

H O

d 1.2 g / L at 25 C

d 1.0 g / mL

Kinetic Molecular Theory (KMT)

Assumptions of the KMT and ideal gases include:

1. Gases consist of tiny particles

2. The distance between particles is large compared with the size of the particles.

3. Gas particles have no attraction for each other

4. Gas particles move in straight lines in all directions, colliding frequently with each other and with the walls of the container.


Kinetic Molecular Theory

Assumptions of the KMT (continued):

5. Collisions are perfectly elastic (no energy is lost in the collision).

6. The average kinetic energy for particles is the same for all gases at the same temperature.

7. The average kinetic energy is directly proportional to the Kelvin temperature.


21KE = where is mass and is velocity

2mv m v

Diffusion


Effusion

Gas molecules pass through a very small opening from a container at higher pressure of one at lower pressure.

Graham’s law of effusion:


rate of effusion of gas A density B molar mass B= =

rate of effusion of gas B density A molar mass A

© 2013 John Wiley & Sons, Inc. All rights reserved.

Graham’s law of effusion

Rates of effusion of gases are inversely proportional to the square roots of their molar masses

Measurement of Pressure

ForcePressure =

Area


Pressure depends on the• Number of gas molecules• Temperature of the gas• Volume the gas occupies

Atmospheric Pressure

Atmospheric pressure is due to the mass of the atmospheric gases pressing down on the earth’s surface.


Barometer


Pressure Conversions

Convert 675 mm Hg to atm. Note: 760 mm Hg = 1 atm


1 atm675 mm Hg = 0.888 atm

760 mm Hg

Convert 675 mm Hg to torr. Note: 760 mm Hg = 760 torr.

760 torr675 mm Hg = 675 torr

760 mm Hg

Dependence of Pressure on Number of Molecules


P is proportional to n (number of molecules) at Tc (constant T) and Vc (constant V).The increased pressure is due to more frequent collisions with walls of the container as well increased force of each collision.

Dependence of Pressure on Temperature


P is proportional to T at nc

(constant number of moles) and Vc.

The increased pressure is due to• more frequent collisions• higher energy collisions

Boyle’s Law


1 1 2 2

1At and : α or c cT n V PV PV

P

What happens to V if you double P?• V decreases by half!

What happens to P if you double V?• P decreases by half!

Boyle’s Law

A sample of argon gas occupies 500.0 mL at 920. torr. Calculate the pressure of the gas if the volume is increased to 937 mL at constant temperature.


1 12

2

PV

PV

1 1 2 2 PV PV

2

920. torr 500. mL = = 491 torr

937 mLP

Knowns V1 = 500 mL P1 = 920. torr V2 = 937 mL

Calculate

Set-Up

Boyle’s Law

Another approach to the same problem:

Since volume increased from 500. mL to 937 ml, the pressure of 920. torr must decrease.

Multiply the pressure by a volume ratio that decreases the pressure:


2

500. mL

937 mL = 920. torr = 491 torrP

Charles’ Law


1 2

1 2

At and :

α or

c cP n

V VV T

T T

• The volume of an ideal gas at absolute zero (-273°C) is zero.

• Real gases condense at their boiling point so it is not possible to have a gas with zero volume.

• The gas laws are based on Kelvin temperature.

• All gas law problems must be worked in Kelvin!

Charles’ Law

A 2.0 L He balloon at 25°C is taken outside on a cold winter day at -15°C. What is the volume of the balloon if the pressure remains constant?


1 2

1 2

V V

T T

1 22

1

rearranged gives V T

VT

2

(2.0 L)(258 K) = = 1.7 L

298 KV

Knowns V1 = 2.0 L T1 = 25°C= 298 K T2 = -15°C = 258 K

Calculate

Set-Up 1 2

1 2

V V

T T

Charles’ Law


Another approach to the same problem:

Since T decreased from 25°C to -15°C, the volume of the 2.0L balloon must decrease.

Multiply the volume by a Kelvin temperature ratio that decreases the volume:

2

258K

298 = 2.0L = 1.7L

KP

Gay-Lussac’s Law


1 2

1 2

At and : α or c c

P PV n P T

T T

Combined Gas Laws


Used for calculating the results of changes in gas conditions.

1 1 2 2

1 2

PV PV

T T

• Boyle’s Law where Tc

• Charles’ Law where Pc

1 2

1 2

V V

T T

1 1 2 2 PV PV

• Gay Lussacs’ Law where Vc

1 2

1 2

P P

T T

P1 and P2 , V1 and V2 can be any units as long as they are the same. T1 and T2 must be in Kelvin.

Combined Gas Law


If a sample of air occupies 500. mL at STP, what is the volume at 85°C and 560 torr?

STP: Standard Temperature 273K or 0°CStandard Pressure 1 atm or 760 torr

1 1 22

1 2

PV T

VT P

Knowns V1 = 500. mL T1 =273K P1= 760 torrT2 = 85°C = 358K P2= 560 torr

Set-Up

2

(760 torr)(500. mL)(358K) = 890. ml

(273K)(560 torr)V Calculate

1 1 2 2

1 2

PV PV

T T

Combined Gas Law

A sample of oxygen gas occupies 500.0 mL at 722 torr and –25°C. Calculate the temperature in °C if the gas has a volume of 2.53 L at 491 mmHg.


1 2 22

1 1

T PVT

PV

1 1 2 2

1 2

PV PV

T T

2

491 torr 2530 ml 248K= =853K 580 C

722 torr 500.0 mlT

Knowns V1 = 500. mL T1 = -25°C = 248K P1= 722 torr V2 = 2.53 L = 2530 mL P2= 560 torr

Set-Up

Calculate

Dalton’s Law of Partial Pressures

The total pressure of a mixture of gases is the sum of the partial pressures exerted by each of the gases in the mixture.

PTotal = PA + PB + PC + ….

Atmospheric pressure is the result of the combined pressure of the nitrogen and oxygen and other trace gases in air.


2 2 2 2.Air N O Ar CO H OP P P P P P

Collecting Gas Over Water

• Gases collected over water contain both the gas and water vapor.

• The vapor pressure of water is

constant at a given temperature• Pressure in the bottle is

equalized so that the Pinside = Patm


2 atm gas H OP P P

Avogadro’s Law

Equal volumes of different gases at the same T and P contain the same number of molecules.


1 volume1 molecule

1 mol

1 volume1 molecule

1 mol

2 volumes2 molecules

2 mol

The ratio isthe same:

Mole-Mass-Volume Relationships

Molar Volume: One mole of any gas occupies 22.4 L at STP.

Determine the molar mass of a gas, if 3.94 g of the gas occupied a volume of 3.52 L at STP.


Knowns m = 3.94 g V = 3.52 L T = 273 K P = 1 atm

Set-Up

Calculate

22.4 L1 mol = 22.4 L so the conversion factor is

1mol

3.94 g

1.52 L

22.4 L

1 mol

= 58.1g/mol

Density of Gases

Calculate the density of nitrogen gas at STP.

Note that densities are always cited for a particular temperature, since gas densities decrease as temperature increases.


mass gd = =

volume L

STP

1 mold = molar mass

22.4 L

STP

28.02 g 1 mold = = 1.25g/L

1 mol 22.4 L

Ideal Gas Law

Calculate the volume of 1 mole of any gas at STP.


PV nRT R

L atm where = 0.0821

mol K

Knowns n = 1 mole T = 273K P = 1 atm

Set-Up

Calculate

nRT

VP

L atm(1 mol)(0.0821 )(273 K)

mol K = (1 atm)

V

Molar volume!

= 22.4 L

Ideal Gas Law

How many moles of Ar are contained in 1.3L at 24°C and 745 mm Hg?


Knowns V = 1.3 L T = 24°C = 297 K P = 745 mm Hg = 0.980 atm

Set-Up

Calculate

PV

nRT

(0.980 atm)(1.3 L) = =0.052 mol

L atm(0.0821 )(297 K)

mol K

n

PV nRT R

L atm where = 0.0821

mol K

Ideal Gas Law

Calculate the molar mass (M) of an unknown gas, if 4.12 g occupy a volume of 943mL at 23°C and 751 torr.


Knowns m =4.12 g V = 943 mL = 0.943 L T = 23°C = 296 K P = 751 torr = 0.988 atm

Set-Up

Calculate

g g = so =

M Mn PV RT

L atm(4.12 g)(0.0821 )(296 K)

mol KM = =107 g/mol(0.988 atm)(0.943 L)

R T

P V

g M =

Gas Stoichiometry


• Convert between moles and volume using the Molar Volume if the conditions are at STP : 1 mol = 22.4 L.

• Use the Ideal Gas Law if the conditions are not at STP.

Gas Stoichiometry

Calculate the number of moles of phosphorus needed to react with 4.0L of hydrogen gas at 273 K and 1 atm.

P4(s) + 6H2(g) 4PH3(g)


2 4.0 L H2 1 mol H

22.4L

4

2

1 mol P

6 mol H

4= 0.030 mol P

Knowns V = 4.0 L T = 273 K P = 1 atm

Solution Map L H2 mol H2 mol P4

Calculate

Gas Stoichiometry

What volume of oxygen at 760 torr and 25°C are needed to react completely with 3.2 g C2H6?

2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(l)


Solution Map m C2H6 mol C2H6 mol O2 volume O2

Knowns m = 3.2 g C2H6 T = 25°C = 298K P = 1 atm

2 63.2g C H 2 6

2 6

1 mol C H

30.08g C H

2

2 6

7 mol O

2 mol C H

2= 0.37mol OCalculate

L atm(0.37 mol)(0.0821 )(298 K)

mol K = = 9.1 L(1 atm)

V

Volume-Volume Calculations


Calculate the volume of nitrogen needed to react with 9.0L of hydrogen gas at 450K and 5.00 atm.

N2(g) + 3H2(g) 2NH3(g)

2 9.0 L H2

2

1 L N

3 L H

2= 3.0 L N

Knowns V = 9.0 L T = 450K P = 5.00 atm

Solution Map Assume T and P for both gases are the same.Use volume ratio instead of mole ratio!L H2 L N2

Calculate

Real Gases

Most real gases behave like ideal gases under ordinary temperature and pressure conditions.

Conditions where real gases don’t behave ideally:• At high P because the distance between particles is too

small and the molecules are too crowded together. • At low T because gas molecules begin to attract each

other.

High P and low T are used to condense gases.


The law of effusion and diffusion

The postulate of the kinetic theory states that the temperature of a system is proportional to the average kinetic energy of its particles and nothing else.

In other words, two gases at the same temperature, such as and must have the same average kinetic energy, hence

multiplying by 2 gives:


Law of Effusion and diffusion

Then rearranging gives:

Taking out the square root both sides we get


the gaseous state of matter

Education

Transcript of the gaseous state of matter