Physical chemistry

Gas Laws

Elements that exist as gases at 250C and 1 atmosphere

•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.

Physical Characteristics of Gases

Units of Pressure

1 pascal (Pa) = 1 N/m2

1 atm = 760 mmHg = 760 torr

1 atm = 101,325 Pa

Barometer

Pressure = Force

Area

Sea level 1 atm

4 miles 0.5 atm

10 miles 0.2 atm

BOYLE‘S LAW:

At a constant temperature, the volume occupied by a

fixed quantity of gas is inversely proportional to

the applied pressure.

This can be expressed mathematically as:

•V α 1/P •V = constant . 1/p.•PV = constant.

As P (h) increases V decreases

P a 1/V

P x V = constant

P1 x V1 = P2 x V2

Boyle’s Law

Constant temperatureConstant amount of gas

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant

temperature to 154 mL?

P1 x V1 = P2 x V2

P1 = 726 mmHg

V1 = 946 mL

P2? =

V2 = 154 mL

P2 = P1 x V1

V2

726 mmHg x 946 mL

154 mL= =4460 mmHg

Charles´ law:At a constant pressure, the

volume of a fixed mass of gas is proportional to its temperature.

V α T (P constant) or,

V/ T = constant

It is most important to use Kelvin scale here.

As T increases V increases

Variation of gas volume with temperatureat constant pressure.

V a T

V = constant x T

V1/T1 = V2/T2 T (K) = t (0C) + 273.15

Charles’ & Gay-Lussac’s Law

Temperature must bein Kelvin

We can use -273 oC point on the graph to define the zero of new scale of temperature

(Kelvin scale (K)).

• We can convert between oC and K • T (K) = t (oC) + 273 K • We can combine Boyle`s and Charle`s laws

by using the equation: • PV/T = constant.• The name is given constant is the gas

constant, (R). Its value is 8.314 J K-1 mole-1

(standard value)

A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a

volume of 1.54 L if the pressure remains constant?

V1 = 3.20 L

T1 = 398.15 K

V2 = 1.54 L

T2? =

T2 = V2 x T1

V1

1.54 L x 398.15 K

3.20 L= =192 K

V1/T1 = V2/T2

Avogadro's law:

Under conditions of constant temperature and pressure equal volumes of gas

contain equal number of molecules. (Since equal numbers of molecules

means equal numbers of moles, the number of moles of any gas is related

directly to its volume .

V α n

Avogadro’s Law

V a number of moles (n)

V = constant x n

V1/n1 = V2/n2

Constant temperatureConstant pressure

Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature

and pressure?

4NH3 + 5O2 4NO + 6H2O

1 mole NH3 1 mole NO

At constant T and P

1 volume NH3 1 volume NO

Ideal Gas Equation

Charles’ law: V a T (at constant n and P)

Avogadro’s law: V a n (at constant P and T)

Boyle’s law: V a (at constant n and T)1

P

V a nT

P

V = constant x = RnT

P

nT

P

R is the gas constant

PV = nRT

The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).

PV = nRT

R = PV

nT=

)1 atm)(22.414L(

)1 mol)(273.15 K(

R = 0.082057 L • atm / (mol • K)

Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

What is the volume (in liters) occupied by 49.8 g of HCl at STP?

PV = nRT

V = nRT

P

T = 0 0C = 273.15 K

P = 1 atm

n = 49.8 g x 1 mol HCl36.45 g HCl

=1.37 mol

V= 1 atm

1.37 mol x 0.0821 x 273.15 K L•atmmol•K

V = 30.6 L

Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of

argon in the lightbulb (in atm)?

PV = nRT n, V and R are constant

nR

V=

P

T =constant

P1

T1

P2

T2

=

P1 = 1.20 atm

T1 = 291 K

P2? =

T2 = 358 K

P2 = P1 x T2

T1

=1.20 atm x 358 K

291 K =1.48 atm

Density (d) Calculations

d = m

V=

PM

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

PM= d is the density of the gas in g/L

Combined gas law:

• In ideal gas equation we assume we have a gas under two different sets of conditions Pi, Vi, Ti and Pf, Vf, Tf

• Pi Vi/ Ti = Pf Vf/ Tf

Ex(3): what would be the volume of a gas at (S.T.P) if it was found to occupy a volume

of 255 ml at 25oC and 650 torr ?

• (i) (f) • V 255 ml ? • P 650 torr 760 torr • T 298 k 273 k • Vf =255 mlx(650torr/760torr)x(273 k/298k) • The volume at ST.P ~ 200 ml.

Dalton’s Law of Partial Pressures

V and T are

constant

P1 P2 Ptotal = P1 + P2

Dalton`s law of partial pressures:

• Total pressure in a container is equal to the sum of the partial pressures of the component gases.

P total = Pa + Pb + Pc + ……..

• Pa, Pb, Pc is the partial pressure gases a, b, c respectively

when two or more gases that do not react chemically are

placed in the same container, the pressure exerted by each

gas in the mixture is the same as it would be if it were the only gas in the container.

Ex (2) : A ten liter flask at 25oC contains a gaseous solution of carbon monoxide and

carbon dioxide at a total pressure of 2.0 atm. If 2 mole of carbon monoxide is present, find

its partial pressure and also that of the carbon dioxide.

•By Dalton`s law Pt = Pco + Pco2 = 2.0 atm

• Pco = n RT/V =

•2 mol x 0.0821 lit atm mol-1 K-1 x 298K/10 lit• = 0.49 atm • Pco2= Pt- Pco = 2.0 - 0.49 = 1.51atm

Ex (1): If 200 ml of nitrogen gas at 25 oC and a pressure of 250 torr are mixed with 350 ml of oxygen at 25 oC and a pressure of 300 torr, so that the resulting volume

is 300ml, what would be the final

pressure of the mixture at 25 oC.?

There is no temperature change, we simply a Boyle`s law calculation for each gas.

For N2 for O2

(i ) (f) (i) (f)

P 250 torr ? 300 torr ? V 200 ml 300 ml 350 ml 300ml

for N2 : P = 250 torr x 200 ml

300 ml = 167 torrforO2: P =300 torr x 350 ml

300 ml = 350 torrthe total pressure of the mixture is the

sum of partial pressuresPt = P O2+ P N2 = 167 + 350 = 517 torr.

Physical Properties of Solutions

12.1

A solution is a homogenous mixture of 2 or more substances

The solute is(are) the substance(s) present in the smaller amount(s)

The solvent is the substance present in the larger amount

A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature.

An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature.

A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature.

Sodium acetate crystals rapidly form when a seed crystal isadded to a supersaturated solution of sodium acetate.

12.1

Colligative Properties are those properties of a liquid that may be altered by the presence of a solute.

Examples• vapor pressure•melting point • boiling point • osmotic pressure.

VAPOUR PRESSURE

The term "vapour" is applied to the gas of any compound that would normally be found as a liquid at room temperature and pressure

For example, water, gasoline, rubbing alcohol, and finger nail polish remover (ethyl acetate) are all normally liquids, but they all evaporate to give a gas.

The molecules that are found on the upper side have only about half as many neighbors as do molecules inside the liquid so to begin with the forces holding them in the liquid

are slightly lower than molecules in the bulk liquid. The rate at which a given volume of liquid will evaporate is

determined by the surface area.

Equilibrium Vapour Pressure

If we put the beaker in a closed chamber a slightly different phenomenon occurs. Now molecules from the liquid evaporate as before, but some of these evaporated molecules may also return to the liquid .

When a nonvolatile solute is added to a liquid to form a solution, the vapour pressure above the solution decreases

On the surface of the pure solvent (shown on the left) there are more solvent molecules at the surface than in the right-hand solution flask. Therefore, it is more likely that solvent molecules escape into the gas phase on the left than on the right. Therefore, the solution should have a lower vapour pressure than the pure solvent.

Raoult's law states that the vapor pressure of a solution, P, equals the mole fraction of the solvent, csolvent, multiplied by the vapor

pressure of the pure solvent, Po .

P = csolvent * Po

The French chemist Francois Raoult discovered the law that mathematically describes the vapor pressure lowering phenomenon.

Raoult`s law: (Vapour pressure of solutions). the vapour pressure of any solution (P total) is the sum of the partial pressure of the components (PA, PB).

Pt = PA + PBPA = XA PAO, PB = XB PBO Pt = XA PAO + XB PBO

(PAO, PBO is the vapour pressure of the component A&B)

Ex: If heptane and octane formed ideal solution at 40 oC of a solution containing 1.0 mole of heptane and 4.0 mole of octane. At 40 oC the vapour pressure of heptane is 0.121 atm, and the vapour pressure of octane is 0.041 atm, what is the vapour pressure of the solution?

• X heptane = 1/5 and X octane = 4/5

Hence,• Pt= 1/5(0.121atm) + 4/5(0.041atm)

• = 0.057 atm.

The change in the vapor pressure that occurs when a solute is added to a solvent is therefore a colligative property. If it depends on the mole fraction of the solute, then it must depend on the ratio of the number of particles of solute to solvent in the solution but not the identity of the solute .

Colligative Properties:Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

Vapor-Pressure Lowering

Raoult’s law

If the solution contains only one solute:

X1 = 1 – X2

P 10 -P1 = DP = X2 P 1

0

P 10 =vapor pressure of pure solvent

X1 = mole fraction of the solvent

X2 = mole fraction of the solute

12.6

P1 = X1 P 10

Boiling point and freezing point of solutions

•The boiling point of a liquid is defined as the temperature at which the vapour pressure of the liquid is equal to the atmospheric pressure.

*At the freezing point the vapour pressure of solid and liquid are equal.

Boiling-Point Elevation

DTb = Tb – T b0

Tb > T b0 DTb > 0

T b is the boiling point of the pure solvent

0

T b is the boiling point of the solution

DTb = Kb m

m is the molality of the solution

Kb is the molal boiling-point elevation constant (0C/m)

12.6

Freezing-Point Depression

DTf = T f – Tf0

T f > Tf0 DTf > 0

T f is the freezing point of the pure solvent

0

T f is the freezing point of the solution

DTf = Kf m

m is the molality of the solution

Kf is the molal freezing-point depression constant (0C/m)

12.6

12.6

What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g.

DTf = Kf m

m =

moles of solute

mass of solvent (kg) =2.41 m =

3.202 kg solvent

478 g x 1 mol

62.01 g

Kf water = 1.86 0C/m

DTf = Kf m =1.86 0C/m x 2.41 m = 4.48 0C

DTf = T f – Tf0

Tf = T f – DTf0 =0.00 0C – 4.48 0C = -4.48 0C

12.6

Ex: What is the boiling point and freezing point of a solution prepared by dissolving 2.4g of biphenyl(C6H12) in 75g of benzene? The molecular weight of biphenyl is 154 (the molal boiling point elevation and freezing point depression of benzene is:(b = 80.1oC, Kb = 2.53 oC & f = 5.5oC, Kf = -5.12oC)

Solution:• m = 1000 x 2.4/(75 x 154) = 0.208 m• Δtb = m kb

• = 0.208 x 2.53 = 0.526 oC

• The boiling point of the solution is 80.1 + 0.526 = 80.6 oC

• Δtf = m kf

• =0.208 x -5.12 = -1.06oC

• The freezing point of the solution is 5.5oC – 1.1oC = 4.4oC

Colligative Properties of Solutions

Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

12.6

Vapor-Pressure Lowering P1 = X1 P 10

Boiling-Point Elevation DTb = Kb m

Freezing-Point Depression DTf = Kf m

Osmotic Pressure (p) p = MRT

Osmotic Pressure (p)

12.6

Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one.

A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules.

Osmotic pressure (p) is the pressure required to stop osmosis.

dilutemoreconcentrated

Semi permeable membrane such as cellophane that permits some

molecules but not all to pass through it.

Osmosis plays an important rule in plants and animal physiological

processes, the passage of substances through the semi permeable walls of living cell. The action of the kidneys

and the rise of sap in trees.

HighP

LowP

Osmotic Pressure (p)

p = MRT

M is the molarity of the solution

R is the gas constant

T is the temperature (in K)12.6

Chemistry In Action: Desalination