GasLawsPPT

Properties of Gases

 Variable volume and shape

 Expand to occupy volume available  Volume, Pressure, Temperature,

and the number of moles present are interrelated

 Can be easily compressed  Exert pressure on whatever

surrounds them

 Easily diffuse into one another

Mercury Barometer



Used to define and

measure atmospheric

pressure



On the average at sea

level the column of

mercury rises to a height of

about 760 mm.



This quantity is equal to 1

atmosphere



It is also known as

Barometer



The mercury barometer was

the basis for defining

pressure, but it is difficult to

use or to transport



Furthermore Mercury is very

toxic and seldom used

anymore



Most barometers are now

aneroid barometers or

electronic pressure sensors,

Pressure Units &

Conversions

 The above represent some of the more common units for

measuring pressure. The standard SI unit is the Pascal or

kilopascal. (kPa)

 The US Weather Bureaus commonly report atmospheric

pressures in inches of mercury.

 Pounds per square inch or PSI is widely used in the United

States.

Standard Temperature and

Pressure



S

tandard

T

emperature and

P

ressure or

STP

= 0

C or 273.15 Kelvin and a pressure

of 1 atmosphere.



This is used as a reference point when

comparing quantities of gases



Gases are seldom measured at exactly these

conditions.



We need to be able to compute the volume

= 340 kPa

If the pressure of helium gas in a balloon

has a volume of 4.00 dm

at 210 kPa, what

will the pressure be at 2.50 dm

?

P

V

= P

V

(210 kPa) (4.00 dm

) = P

2

(2.50 dm

)

P

= (210 kPa) (4.00 dm

)

(2.50 dm

)

Sample Problem 1:

Boyle’s Law

 According to Boyle’s Law

the pressure and volume of a gas are inversely

proportional at constant pressure.

 PV = constant.  P₁V₁ = P₂V₂

Boyle’s Law

 A graph of pressure and volume gives an inverse

function

 A graph of pressure and the reciprocal of volume

Standard Temperature and

Pressure (STP)

The volume of a gas varies with temperature

and pressure. Therefore it is helpful to have

a convenient reference point at which to

compare gases.

For this purpose standard temperature and

pressure are defined as:

Temperature = 0

C 273 K

Pressure = 1 atmosphere

= 760 torr

= 101.3 kPa

This point is often called STP

Charles’ Law

According to Charles’ Law the volume of a

gas is proportional to the Kelvin temperature

as long as the pressure is constant

V = kT

V₁ = T₁ V₂ T₂

Note: The temperature for gas laws must

Charles’ Law

 A graph of temperature and volume yields a straight line.  Where this line crosses the x axis (x intercept) is defined

as absolute zero

Sample Problem 2

A gas sample at 40

C occupies a volume of

2.32 dm

. If the temperature is increased to 75

C, what will be the final volume?

2.58 dm

13

V₁ = V₂

T₁ T₂

Convert temperatures to Kelvin. 40oC = 313K

75oC = 348K

2.32 dm3 = V 2

313 K 348K

(313K)( V₂) = (2.32 dm3) (348K)

Gay-Lussac’s Law

Gay-Lussac’s Law defines the relationship between pressure and temperature of a gas.

The pressure and temperature of a gas are directly

proportional

P

= P

T

T

Sample Problem 3:

The pressure of a gas in a tank is 3.20 atm at 22 oC. If the temperature rises to 60oC, what will be the pressure in the tank?

3.6 atm

15

P₁ = P₂

T₁ T₂

Convert temperatures to Kelvin. 22oC = 295K

60oC = 333K

3.20 atm = P₂

295 K 333K

(295K)( P₂) = (3.20 atm)(333K)

The Combined Gas Law

1. If the amount of the gas is constant, then

Boyle’s Charles’ and Gay-Lussac’s Laws

can be combined into one relationship

2. P

V

= P

V

T

T

Sample Problem 4:

A gas at 110 kPa and 30

C fills a container

at 2.0 dm

. If the temperature rises to 80

C

and the pressure increases to 440 kPa, what

is the new volume?

V

0.58 dm

17

P1V₁ = P₂V2

T₁ T₂

Convert temperatures to Kelvin. 30oC = 303K

80oC = 353K

V2 = V₁ P1 T₂

P₂ T₁

= (2.0 dm3) (110 kPa ) (353K)

Advogadro’s Law

 Equal volumes of a gas under the same temperature

and pressure contain the same number of particles.

 If the temperature and pressure are constant the

volume of a gas is proportional to the number of moles of gas present

V = constant * n

where n is the number of moles of gas V/n = constant

V₁/n₁ = constant = V₂ /n₂

V₁/n₁ = V₂ /n₂

Universal Gas Equation

 Based on the previous laws there are four factors

that define the quantity of gas: Volume, Pressure, Kelvin Temperature, and the number of moles of gas present (n).

 Putting these all together:

PV

nT = Constant = R

Universal Gas Equation

The Universal gas equation is usually written as

PV = nRT

Where P = pressure

V = volume

T = Kelvin Temperature n = number of moles

The numerical value of R depends on the pressure unit (and perhaps the energy unit) Some common values of R include:

R = 62.36 dm3 torr mol-1 K-1 = 0.0821 dm3 atm mol-1 K-1 = 8.314 dm3kPa mol-1 K-1

Sample Problem 5

Example:

at 20oC and a pressure of 0.880 atmospheres?

:

V = (0.781 mol)(0.08205 dm-3 atm mol-1 K-1)(293K)

0.880 atm

V = 21.3 dm3

(25.0 g)

n = --- = 0.781 mol (32.0 g mol-1)

V =? P = 0.880 atm; T = (20 + 273)K = 293K

R = 0.08205 dm-3 atm mol-1 K-1

PV = nRT so V = nRT/P

Data

Formula Calculation

Density (

d

) Calculations

d = m

V

=

P

M

RT

m

is the mass of the gas in g

M

is the molar mass of the gas

Molar Mass (

M

) of a Gaseous Substance

dRT

P

M

=

d

is the density of the gas in g/L

Universal Gas

Equation –Alternate

Forms

A 2.10 dm3 vessel contains 4.65 g of a gas at 1.00

atmospheres and 27.0oC. What is the molar mass of

the gas?

dRT

P

M

=

d = m

_V

2.10 dm3

=

=

M

=

g dm3

1 atm

x

x

mol•K

M

=

Dalton’s Law of Partial

Pressures

The total pressure of a mixture of gases is

equal to the sum of the pressures of the

individual gases (partial pressures).

P

= P

+ P

+ P

+ P

+

where P_T = total pressure

P₁ = partial pressure of gas 1 P₂ = partial pressure of gas 2 P₃ = partial pressure of gas 3 P₄ = partial pressure of gas 4

Dalton’s Law of Partial

Pressures

 Applies to a

mixture of gases

 Very useful

correction when collecting gases over water

since they inevitably

Sample Problem 7

Henrietta Minkelspurg generates Hydrogen gas and collected it over water.

If the volume of the gas is 250 cm3 and the

barometric pressure is 765.0 torr at 25oC, what

is the pressure of the “dry” hydrogen gas at STP?

(P_H2O = 23.8 torr at 25o_C)

Sample Problem 8

 Henrietta Minkelspurg generated Hydrogen gas and collects it

over water. If the volume of the gas is 250 cm3 and the

barometric pressure is 765 torr at 25oC, what is the volume of

the “dry” oxygen gas at STP?

 From the previous calculation the adjusted pressure is 742.2

torr

V₂ = (250 cm3)(742.2 torr)(273K)

(298K)(760.torr)

V₂ = 223.7 cm3

P₁= P_H2 = 742.2 torr; P₂= Std Pressure = 760 torr T₁= 298K; T₂= 273K; V₁= 250 cm3; V

2= ?

Kinetic Molecular Theory



Matter consists of particles (atoms or molecules)

that are in continuous, random, rapid motion



The Volume occupied by the particles has a

negligibly small effect on their behavior



Collisions between particles are elastic



Attractive forces between particles have a

negligible effect on their behavior



Gases have no fixed volume or shape, but take

the volume and shape of the container



The average kinetic energy of the particles is

proportional to their Kelvin temperature

Maxwell-Boltzman

Distribution

 Molecules are in constant motion  Not all particles have the same energy

 The average kinetic energy is related to the temperature

 An increase in temperature

spreads out the

distribution and the mean speed is

The distribution of speeds for nitrogen gas molecules

at three different temperatures

The distribution of speeds of three different gases at the same temperature

u

=

3

RT

M



Velocity of a Gas

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.

NH₃

17.0 g/mol

HCl

36.5 g/mol

NH₄Cl

Diffusion

DIFFUSION AND EFFUSION



Diffusion

Diffusion

is the

is the

gradual mixing of

gradual mixing of

molecules of

molecules of

different gases.

different gases.



Effusion

Effusion

is the

is the

movement of

movement of

molecules through

molecules through

a small hole into an

a small hole into an

empty container.

empty container.

Graham’s Law

Graham’s law governs

Graham’s law governs

effusion and diffusion of

effusion and diffusion of

gas molecules.

gas molecules.

KE=1/2 mv

Thomas Graham, 1805-1869.

Thomas Graham, 1805-1869.

Professor in Glasgow and London.

Professor in Glasgow and London.

The rate of effusion is

The rate of effusion is

inversely proportional

inversely proportional

to its molar mass.

to its molar mass.

The rate of effusion is

The rate of effusion is

inversely proportional

inversely proportional

Scheffler

Sample Problem 9

1 mole of oxygen gas and 2 moles of ammonia are placed in a

container and allowed to react at 850o_{C according to the equation:}

4 NH₃(g) + 5 O₂(g)  4 NO(g) + 6 H₂O(g)

Using Graham's Law, what is the ratio of the effusion rates of NH₃(g) to O₂(g)?

Sample Problem 10

takes 4.55 sec to effuse out of a container?

Scheffler

Sample Problem 11

What is the molar mass of gas X if it effuses

0.876 times as rapidly as N

(g)?

Ideal Gases v Real Gases



Ideal gases are gases that obey the Kinetic

Molecular Theory perfectly.



The gas laws apply to ideal gases, but in

reality there is no perfectly ideal gas.



Under normal conditions of temperature and

pressure many real gases approximate ideal

gases.



Under more extreme conditions more polar

In an Ideal Gas

--- The particles (atoms or molecules) in continuous,

random, rapid motion.

 The particles collide with no loss of momentum

 The volume occupied by the particles is essentially zero

when compared to the volume of the container

 The particles are neither attracted to each other nor

repelled

 The average kinetic energy of the particles is proportional

to their Kelvin temperature

At normal temperatures and pressures gases closely approximate idea behavior

Real Gases

These deviations occur because

 Real gases do not actually have zero volume  Polar gas particles do attract if compressed

 For ideal gases the product of pressure and

van der Waals Equation

(P + n

a

/V

)(V - n

b

) = nRT

The van der Waals equation shown below includes corrections added to the universal gas law to account for these deviations from ideal behavior

where a => attractive forces between molecules b => residual volume or molecules

The van der Waals constants for some elements are shown below

Substance _{a (dm}6atm mol-2) b (dm3 mol-1)

He 0.0341 0.02370 CH₄ 2.25 0.0428 H₂O 5.46 0.0305

Sample Problem 12

What is the volume of CO₂ produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:

C₆H₁₂O₆ (s) + 6O₂ (g) 6CO₂ (g) + 6H₂O (l)

g C₆H₁₂O₆ mol C₆H₁₂O₆ mol CO₂ V CO₂

5.60 g C₆H₁₂O₆ 1 mol C6H12O6 180 g C₆H₁₂O₆

x 6 mol CO2

1 mol C₆H₁₂O₆

x = 0.187 mol CO₂

V = nRT P

0.187 mol x 0.0821 x 310.15 Kdm3•atm mol•K

1.00 atm