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Chapter 5 the Gaseous state

- Vanessa Prasad-Permaul
- Valencia Community College
- CHM 1045

GAS LAWS PRESSURE AND MEASUREMENT

Elements that exist as gases at 250C and 1

atmosphere

GAS LAWS PRESSURE AND MEASUREMENT

Physical Characteristics of Gases

- Gases assume the volume and shape of their

containers. - Gases are the most compressible state of matter.
- Gases will mix evenly and completely when

confined to the same container. - Gases have much lower densities than liquids and

solids.

GAS LAWS PRESSURE AND MEASUREMENT

a) Gas is a large collection of particles moving

at random throughout a volume b) Collisions of

randomly moving particles with the walls of the

container exert a force per unit area that we

perceive as gas pressure

GAS LAWS PRESSURE AND MEASUREMENT

HOW IS PRESSURE DEFINED? The force the gas

exerts on a given area of the container in which

it is contained. The SI unit for pressure is

the Pascal, Pa.

- If youve ever inflated a tire, youve probably

made a pressure measurement in pounds (force) per

square inch (area).

GAS LAWS PRESSURE AND MEASUREMENT

Units of Pressure

1 pascal (Pa) 1 kg/ms2 1 atm 760 mmHg 760

torr 1 atm 101,325 Pa 1 bar 105 Pa 1 atm

14.69 lb/in2

- Hg is used instead of H2O
- more dense
- better visibility
- accuracy

GAS LAWS PRESSURE AND MEASUREMENT

- EXERCISE 5.1
- A GAS CONTAINER HAD A MEASURED PRESSURE OF
- 57kPa. CALCULATE THE PRESSURE IN UNITS OF ATM
- AND mmHg.
- First, convert to atm (57 kPa 57 x 103 Pa).
- 57 x 103 Pa x 1 atm

0.562 0.56 atm - 1.01325 x 105 Pa
- Next, convert to mmHg.
- 57 x 103 Pa x 760 mmHg 427.5

4.3 x 102 mmHg - 1.01325 x 105 Pa

GAS LAWS EMPIRICAL GAS LAWS

EMPIRICAL GAS LAWS

You can predict the behavior of gases based on

the following properties

If two of these physical properties are held

constant, it is possible to show a simple

relationship between the other two

GAS LAWS EMPIRICAL GAS LAWS

Boyles experiment A manometer to study the

relationship between pressure (P) Volume (V)

of a gas

As P (h) increases

V decreases

GAS LAWS EMPIRICAL GAS LAWS

BOYLES LAW the volume of a sample of gas at a

given temperature varies inversely with the

applied pressure So. For a given amount of gas

(n) _at_ constant temperature (T) If pressure

(P) increases, the volume (V) of the gas

decreases P1 V1 P2 V2

Boyles Law

- PressureVolume Law (Boyles Law)

GAS LAWS EMPIRICAL GAS LAWS

Boyles Law

Constant temperature Any given amount of gas

GAS LAWS EMPIRICAL GAS LAWS

EXERCISE 5.2 A volume of carbon dioxide gas

equivalent to 20.0 L was collected _at_ 23oC and

1.00atm pressure. What would be the volume of gas

at constant temperature and 0.830atm? P1 V1

P2 V2 Application of Boyles law gives V2

V1 x P1 20.0 L x 1.00atm 24.096

24.1 L P2

0.830atm

GAS LAWS EMPIRICAL GAS LAWS

Charles Law relating volume and temperature

As T increases

V increases

Charles Law

- TemperatureVolume Law (Charles Law)

GAS LAWS EMPIRICAL GAS LAWS

Charles Law

Variation of gas volume with temperature

V a T

For a given amount of gas at constant pressure

Temperature must be in Kelvin

T (K) t (0C) 273.15

V / T constant

GAS LAWS EMPIRICAL GAS LAWS

CHARLES LAW the volume occupied by any sample

of gas at a constant pressure is directly

proportional to the absolute temperature So. Fo

r a given amount of gas (n) _at_ constant

pressure (P) If temperature (T) increases, the

volume (V) of the gas increases

V1 V2 T1 T2

GAS LAWS EMPIRICAL GAS LAWS

EXERCISE 5.3 A chemical reaction is expected to

produce 4.3dm3 of oxygen at 19oC and 101kPa. What

will the volume be at constant pressure and

25oC? First, convert the temperatures to the

Kelvin scale. Ti (19 273) 292 K Tf

(25 273) 298 K Following is the data

table. Vi 4.38 dm3 Pi 101 kPa Ti 292

K Vf ? Pf 101 kPa Tf 298 K Apply

Charless law to obtain Vf Vi x Tf 4.38

dm3 x 298K 4.470 4.47 dm3

Ti 292K

GAS LAWS EMPIRICAL GAS LAWS

COMBINED GAS LAW Relating Volume, Temperature

and Pressure Taking Boyles Law and Charles

Law The volume occupied by a given amount of

gas is proportional to the absolute temperature

divided by the pressure V constant x T or

PV constant (for a given amount of gas)

P T P1V1 P2V2 T1 T2

Combined Gas Law

We can combine Boyles and charles law to come

up with the combined gas law Use Kelvins

for temp, any pressure, any volume

GAS LAWS EMPIRICAL GAS LAWS

EXERCISE 5.4 A balloon contains 5.41dm3 of helium

at 24oC and 10.5kPa. Suppose the gas in the

balloon is heated to 35oC and the pressure is

now 102.8kPa, what is the volume of the

gas? First, convert the temperatures to

kelvins. Ti (24 273) 297 K Tf (35

273) 308 K Following is the data

table. Vi 5.41 dm3 Pi 101.5 kPa Ti

297 K Vf ? Pf 102.8 kPa Tf 308

K Apply both Boyles law and Charless law

combined to get Vf Vi x Pi x Tf 5.41 dm3

x 101.5kPa x 308K 5.539 5.54 dm3

Pf Ti

102.8kPa 297K

AVOGADROS LAW relating volume and amount

GAS LAWS EMPIRICAL GAS LAWS

equal volumes of any two gases at the same

temperature pressure contain the same number

of molecules

Constant temperature Constant pressure

Avogadros Law

- The VolumeAmount Law (Avogadros Law)

Avogadros Law

- The VolumeAmount Law (Avogadros Law)
- At constant pressure and temperature, the volume

of a gas is directly proportional to the number

of moles of the gas present. - Use any volume and moles
- V1 V2
- n1 n2

GAS LAWS EMPIRICAL GAS LAWS

Avogadros Number one mole of any gas contains

the same number of molecules 6.023 x 1023 Must

occupy the same volume at a given temperature and

pressure

The conditions 0 0C and 1 atm are called standard

temperature and pressure (STP).

IDEAL GAS LAW

GAS LAWS THE IDEAL GAS LAW

Charles law V a T (at constant n and P)

Avogadros law V a n (at constant P and T)

R is the molar gas constant

PV nRT

GAS LAWS THE IDEAL GAS LAW

IDEAL GAS EQUATION

PV nRT

R 0.082057 L atm / (mol K) R 8.3145

J / (mol K) R 1.9872 cal / (mol K)

The units of pressure times volume are the units

of energy joules (J) or calories (cal)

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 LatmK1mol1
- P has to be in atm
- V has to be in L
- T has to be in K

GAS LAWS THE IDEAL GAS LAW

EXERCISE 5.5 Show that the moles of gas are

proportional to the pressure for constant volume

and temperature

Use the ideal gas law, PV nRT, and solve for

n n PV RT n V x P

RT

Note that everything in parentheses is constant.

Therefore, you can write n constant x P Or,

expressing this as a proportion, n ? P

GAS LAWS THE IDEAL GAS LAW

EXERCISE 5.6 What is the pressure in a 50.0L gas

cylinder that contains 3.03kg of oxygen at

23oC? T 23oC 273K 296K V 50.0L R

0.08206 L . atm/K . mol n 3.03kg x 1000g x

1mol 94.688 mol O2

1 kg 31.998g P ? PV nRT P nRT

0.0347mol x 0.08206 L . atm x 296K V

K . mol

46.0 atm

50.0L

The Ideal Gas Law

- 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.

GAS LAWS THE IDEAL GAS LAW

EXERCISE 5.7 Calculate the density of helium

(g/L) at 21oC and 752mmHg. The density of air

under these conditions is 1.188g/L. What is the

difference in mass between 1 liter of air and 1

liter of helium?

Variable Value

P 752 mmHg x 1 atm 0.98947 atm 760mmHg

V 1 L (exact number)

T (21 273) 294 K

n ?

32

GAS LAWS THE IDEAL GAS LAW

Using the ideal gas law, solve for n, the moles

of helium. n PV 0.98947 atm x

1L RT

0.08206 L . atm/ K . mol x 294K

0.4101 mol

Now convert mol He to grams. 0.04101 mol He x

4.00g He 0.16404g He 0.164g/L

1.00 mol He

1 L

Therefore, the density of He at 21C and 752 mmHg

is 0.164 g/L. The difference in mass between one

liter of air and one liter of He mass air -

mass He 1.188 g - 0.16404 g 1.02396

1.024 g difference

GAS LAWS THE IDEAL GAS LAW

EXERCISE 5.8 A sample of a gaseous substance at

25oC and 0.862 atm has a density of 2.26 g/L.

What is the molecular mass of the

substance? Variable Value

P

0.862 atm

V 1 L (exact

number)

T (25 273) 298 K

n

? From the ideal gas law, PV

nRT, you obtain n PV

0.862 atm x 1 L 0.03525

mol RT 0.08206 L . atm/K . Mol x

298K

34

34

GAS LAWS THE IDEAL GAS LAW

Dividing the mass of the gas by moles gives you

the mass per mole (the molar mass). Molar mass

grams of gas 2.26g

64.114 g/mol moles

of gas 0.03525mol Therefore, the

molecular mass is 64.1 amu.

GAS LAWS GAS MIXTURES

Daltons Law of Partial Pressures

V and T are constant

P1

P2

Ptotal P1 P2

Daltons Law of Partial Pressures

- For a two-component system, the moles of

components A and B can be represented by the mole

fractions (XA and XB).

- Mole fraction is related to the total pressure

by

GAS LAWS GAS MIXTURES

EXERCISE 5.10 A 10.0L flask contains 1.031g O2

and 0.572g CO2 at 18oC. What are the partial

pressures of each gas? What is the total

pressure? What is the mole fraction of oxygen in

this mixture? Each gas obeys the ideal gas law.

1.031 g O2 x 1 mol 0.0322188 mol O2

32.00g P nRT 0.0322mol x

0.0821L.atm/K.mol x 291K 0.076936 atm

V

10.0L 0.572 g CO2 x 1 mol 0.012997 mol

CO2 44.01g P nRT

0.0130mol x 0.0821L.atm/K.mol x 291K

0.031036 atm V

10.0L

GAS LAWS GAS MIXTURES

The total pressure is equal to the sum of the

partial pressures PT PO2 PCO2

0.076936atm 0.031036atm 0.10797

0.1080

atm The mole fraction of oxygen in the mixture

is Mole fraction O2 PO2 0.076936atm

0.7122 0.712

PT 0.1080atm 0.712 x 100 71.2

mole of O2 in this gas mixture

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 constant random motion
- Gas collisions are perfectly elastic
- Gas molecules experience no attraction or

repulsion

Kinetic Molecular Theory

- Average Kinetic Energy (KE) is given by

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

- The RootMeanSquare Speed is a measure of the

average molecular speed.

Taking square root of both sides gives the

equation

Example 17

- Calculate the rootmeansquare speeds of helium

atoms and nitrogen molecules in m/s at 25C.

Kinetic Molecular Theory

- Maxwell speed distribution curves.

Grahams Law

- Diffusion is the mixing of different gases by

random molecular motion and collision.

Grahams Law

- Effusion is when gas molecules escape without

collision, through a tiny hole into a vacuum.

Grahams Law

- Grahams Law Rate of effusion is proportional to

its rms speed, urms. - For two gases at same temperature and pressure

Behavior of Real Gases

- At higher pressures, particles are much closer

together and attractive forces become more

important than at lower pressures.

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.

Behavior of Real Gases

- Corrections for non-ideality require van der

Waals equation.

Excluded Volume

IntermolecularAttractions

Example 1 Boyles 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)

Example 2 Charles Law

- A sample of CO2(g) at 35?C has a volume of 8.56
- x10-4 L. What would the resulting volume be if
- we increased the temperature to 85?C? (at
- constant moles and pressure)

Example 3 Avogadros 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)

Example 4Combined 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?

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?

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 69.5C.

Example 7 Ideal Gas Law

- What is the volume (in liters) occupied by 7.40 g

of CO2 at STP?

Example 8 Density MM

- What is the molar mass of a gas with a density of

1.342 g/L at STP?

Example 9 Density MM

- What is the density of uranium hexafluoride, UF6,

(MM 352 g/mol) under conditions of STP?

Example 10 Density MM

- The density of a gaseous compound is 3.38 g/L at

40C and 1.97 atm. What is its molar mass?

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?

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?

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?

Example 14 Partial Pressure

- On a humid day in summer, the mole fraction of

gaseous H2O (water vapor) in the air at 25C 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?

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)

Example 15

- All of the mole fractions of elements in a given

compound must add up to? - 100
- 1
- 50
- 2

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)

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?

Example 19 Diffusion

- What are the relative rates of diffusion of the

three naturally occurring isotopes of neon 20Ne,

21Ne, and 22Ne?

Behavior of Real Gases

- 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.

Example 20 Ideal Vs. Van Der Waals

- Given that 3.50 moles of NH3 occupy 5.20 L at

47C, calculate the pressure of the gas (in atm)

using - the ideal gas equation
- (b) the van der Waals equation. (a 4.17, b

0.0371)

Example 21 Ideal Vs. Van Der Waals

- 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 L2atm mol2, and b 0.0387

L/mol.