Gases and atmospheric

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Gases and atmospheric

Chemistry

States of Matter

STATE	PROPERTIES	EXAMPLES
SOLID – particles held tightly together.	-definite shape & volume -virtually incompressible -does not flow easily	-crystal of sugar
LIQUID – particles held more loosely together	-assumes shape of container but has a definite volume - virtually incompressible -flows readily	-coloured water
GAS – appears to be an absence of forces between molecules	-assumes shape & volume of container -highly compressible -flows readily	-gas station air pump

Kinetic Molecular Theory

all substances contain particles that are in constant, random

motion

3 types of motion;

translational (straight line)
rotational (spinning)
vibrational (back & forth motion of atoms in a molecule)

solids – vibrational-bonds hold molecules tightly

liquids – all 3 types of motion – loosely held

gases – translational (mostly) – very loosely held

9.2 Gas Laws

pressure – Force per unit area

atmospheric pressure – Force per unit area exerted by air on an object

SI UNIT – for atmospheric pressure is the Pascal (Pa) which equals a force 1N/m² (newton/square meter)

* we normally use kilopascals (kPa)

1 kPa=1000Pa=1kN/m²

Standard Pressure=101.325 kPa=1 atm

Standard Ambient Pressure= 100kPa

STP= 0ºC and 101.325 kPa

SATP=25ºC and 100kPa

UNIT NAME	UNIT SYMBOL	DEFINITION/CONVERSION
Pascal	Pa	1 Pa = 1N/m²
Atmosphere	atm	1 atm = 101.325kPa
mm of mercury	mm Hg	760mm of Hg=1 atm=101.325kPa
torr	torr	1 torr = 1mm of Hg

*** Know how to convert these units

e.g. Convert 100 kPa to atm and mm of Hg

a) 100kPa x 1 atm = 0.987 atm

101.325 kPa

b) 100kPa x 760 mm of Hg = 750.mm of Hg

101.325kPa

BOYLE’S LAW

discovered by Robert Boyle

“As the pressure on a gas increases, the volume of the gas

decreases proportionally, provided that the temperature and

amount of gas remain constant: The volume and pressure of a

gas is inversely proportional”

Boyle’s Law mathematically is simply;

P₁V₁ = P₂V₂

Sample Problem 1

Q: A 2.0L party balloon at 98kPa is taken to the top of a mountain

where the pressure is 75kPa. Assume the temperature is

constant. What is the new volume of the balloon?

A: V₁=2.0L

P₁=98kPa

P₂=75kPa

V₂= ?

P₁V₁= P₂V₂

V₂ = P₁V₁

P₂

= 98kPa x 2.0L

75kPa

V₂ = 2.6L

Therefore the new volume is 2.6L.

KELVIN TEMPERATURE SCALE

when we work with gases we use a temperature scale called the Kelvin Temperature Scale. K

When the line is extrapolated downward the line meets the temperature line (X-axis) at a point where the temperature is
273ºC.

this temperature is called Absolute Zero and is the lowest possible temperature

absolute zero is the basis for the Kelvin temperature scale

Absolute zero (-273ºC) is Zero Kelvin (0 K)

To convert ºC to K. add 273

To convert K to ºC, subtract 273

Do questions 12-14 on page 432.

* For convenience we use 273K for STP

and 298K for SATP

i.e. SATP = 100kPa and 298K

STP = 101.325kPa and 273K

CHARLES’ LAW

“As the temperature of a gas increases, the volume increases proportionally, provided that the pressure & amount of gas remain constant”

V₁ = k V₂ =k

T₁T₂

where k= constant value

therefore V₁ = V₂

T₁ T₂

Sample Problem 3

Q: A gas inside a cylinder with a movable piston is to be heated to

315ºC. The volume of the gas in the cylinder is 0.30L at 25ºC.

What is the final volume when the temperature is 315ºC?

A: V₁=0.30L

T₁=25ºC=298K

V₂=?

T₂= 315ºC=588K

V₁ = V₂

T₁ T₂

V₂=V₁T₂

T₁

V₂=0.30L x 588K

298K

V₂=0.59L

Therefore the final volume of the gas at 315ºC is 0.59L.

Pressure and Temperature Law

sometimes called Gay-Lussac’s Law

P₁ = k P₂ = k

T₁ T₂

therefore, P₁ = P₂

T₁ T₂

this represents a direct relationship i.e. heat an aerosol can and

the pressure will increase until the can ruptures.

Sample Problem 4

Q: A sealed storage tank contains argon gas at 18ºC and a pressure

of 875kPa at night. What is the new pressure of the tank and

its contents when it warms up to 32ºC during the day?

A: T₁=18ºC=291K

P₁=875kPa

P₂=?

T₂=32ºC=305K

P₁ = P₂

T₁ T₂

P₂= P₁T₂

e 435&436

T₁

= 875kPa x 305K

291K

P₂= 917kPa

Therefore the new pressure is 917kPa.

COMBINED GAS LAW

P₁V₁ = P₂V₂

T₁ T₂

Sample Problem 5

Q: A balloon containing hydrogen gas at 20ºC and a pressure of

100kPa has a volume of 7.50L. Calculate the volume of the

balloon after it rises 10km into the upper atmosphere where

the temperature is -36ºC and the outside air pressure is 28kPa.

Assume that no hydrogen gas escapes and that the balloon

expands, etc.

A: T₁=20ºC=293K

P₁=100kPa

V₁=7.50L

T₂=-36ºC=237K

P₂=28kPa

V₂= ?

P₁V₁ = P₂V₂

T₁ T₂

V₂=P₁V₁ x T₂

T₁ P₂

=100kPa x 7.50L x 237K

293K x 28kPa

V₂=22L

Therefore the new volume of the balloon is 22L.

THE IDEAL GAS LAW

An ideal gas is a hypothetical gas that obeys the gas laws perfectly under all conditions i.e. won’t condense into a liquid when cooled, graphs are all perfect straight lines, etc.

An ideal gas fits the following criteria;

The volume of a gas is inversely proportional to pressure

v 1/P (Boyle’s Law)

The volume of a gas is directly proportional to the absolute Kelvin temperature V T (Charles’ Law)

The pressure of a gas is directly proportional to the absolute Kelvin temperature P T (Pressure-Temperature Law)

The greater the amount of gas, in moles (temperature & pressure constant), the greater the volume V n

Therefore V 1/P x T x n

V=R[constant] x 1/P x T x n

V= nRT

Or PV=nRT

PV=nRT is the Ideal Gas Law.

Where R= the gas constant

To find the value of R, we would substitute known values and solve for R.

Example: At STP 1.00mol of an ideal gas would occupy a volume of

22.414L.

V=22.414L

n=1.00mol

P=101.325kPa

T=0ºC=273.15K

PV=nRT

R=PV

=101.325kPa x 22.414L

1.00mol x 273.15K

R= 8.31kPa*L

mol * K

If the mass of gas was given, and not moles, we would have to

calculate n first.

Example: If 0.78g of hydrogen at 22ºC and 125kPa is produced, what

volume of hydrogen would be expected?

Step 1: Find moles of hydrogen

n=m/M = 0.78g/2.02g/mol = 0.39mol

Step 2: Use the Ideal Gas Law to find the volume of hydrogen

expected.

PV=nRT

V_H2=nRT

V_H2= 0.39mol x 8.31kPA*L/mol*K x 295K

125kPa

V_H2= 7.6L

Therefore the volume of hydrogen gas expected was 7.6L.

Sample Problem 1

Q: What mass of neon gas should be introduced into an evacuated

0.88L tube to produce a pressure of 90kPa at 30ºC?

V=0.88L

P=90kPa

T=30ºC=303K

m_Ne=?

PV=nRT

n_Ne=PV

n_Ne=90kPa x 0.88L

8.31kPa*L/mol*K x 303K

n_Ne=0.031mol

n=m/M , m_Ne=n x M

m_Ne=0.031mol x 20.18g/mol

=0.63g

Therefore the mass of neon gas needed was 0.63g.

Source : http://www.leafrancis.net/Chem9Notes.doc

Web site link: http://www.leafrancis.net/

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