Nature of Gases -...

IONIC SOLIDS

https://www.youtube.com/w

atch?v=QXT4OVM4vXI

• Ions form when electrons are

exchanged between species (atoms or

groups of atoms)

• Ionic compounds form when these

charged ions attract each other

electrostatically

• The exist as large collections of the

bound ions and do not have a single

unit like molecules

What Is an Ionic Compound?

What Holds Ionics Together?

Representing in 2D and 3D...

Factors in Crystal Formation [ Fcoulomb = k(Q1Q2)r-2 ]

❏ Charges of Ions Involved…

❏ Relative Sizes of Ions Involved...

Electrical Conductivity - Ability to pass electrical current

Solid State

Explaining Ionic Macrostate Properties with

Microstate Structure

Molten OR Aqueous

High Melting Point...

∆Hdissociation = Elattice ∝ Q1Q2 / r

This means that the energy to break down

crystalline solids is related to two things…

• Charge of the ions involved (larger

charge greater attractions)

• The distance between the ions in the

lattice (closer equals greater attractions)

Explaining Ionic Macrostate Properties with

Microstate Structure

Nature of Gases

Gases are compressible, fill their containers, and essentially all behave similarly under normal

conditions

Kinetic Molecular Theory

1. Gases are composed of large numbers of particles that are in

constant, random motion

2. The volume of the particles is negligible compared to the total

volume they occupy.

3. Attractive and repulsive forces between gas particles are

negligible.

4. Energy can be transferred between molecules during collisions,

but the average kinetic energy of the molecules is constant with

time, i.e. all collisions are perfectly elastic.

5. The average kinetic energy of the particles is proportional to the

absolute temperature. At a given temperature the particles of

all gases have the same average kinetic energy!

The Gas Laws (Boyle’s) The PV product of a gas is equal to a constant value as

long as temperature and quantity (moles) are held

constant.

PV = constant

- SO -

P1V1 = P2 V2

Q: If 700mL of a gas at 2.5atm of pressure is expanded to

2.8L, what pressure does the gas exert?

A: 𝑃2 =𝑃1𝑉1

𝑉2=

2.5 𝑎𝑡𝑚 × 0.700 𝐿

2.8 𝐿

P2 = 0.63 atm

The Gas Laws (Charles’s Law) The V/T quotient of a gas is constant as long as volume and

quantity are held constant.

V = T constant

-SO -

V1 / T1 = V2 / T2

Q: At what temperature would the volume of a balloon

containing helium at 25C reach half its original value?

A: 𝑇2 =𝑇1

𝑉1× 𝑉2 =

298 𝐾

𝑉1×

1

2𝑉1

T2 = 149 K

The Gas Laws (Avogadro’s Law)

The V/n quotient of a gas is constant as long as temperature

and pressure are held constant

V = n constant

-SO -

V1 / n1 = V2 / n2

Q: What will be the new volume of a 5.6L nitrogen balloon at

STP if an additional 0.50mol of N2 is added?

A: 𝑉2 =𝑉1

𝑛1× 𝑛2 =

5.6 𝐿

𝑛1× 𝑛1 + 0.50 𝑚𝑜𝑙

𝑛1 = 5.6 𝐿 ×1 𝑚𝑜𝑙

22.4 𝐿= 0.25 𝑚𝑜𝑙

V2 = 16.8L

The Gas Laws (Gay-Lussac’s Law)

The P/T quotient of a gas is constant as long as the volume

and quantity of the gas are held constant

P = T constant

- SO -

P1 / T1 = P2 / T2

Q: At what temperature will the pressure of a gas initially at

20C and 700torr reach one third the initial pressure?

A: 𝑇2 =𝑇1

𝑃1× 𝑃2 =

293 𝐾

700 𝑡𝑜𝑟𝑟×

1

3× 700 𝑡𝑜𝑟𝑟

T2 = 97.7K

The Ideal Gas Law (BIG ONE)

Because the following things are known, we can describe

gases with one all-encompassing equation…

V 1 / P [Boyle]

V T [Charles]

V n [Avogadro]

- SO -

V nT / P , and R is defined as the proportionality constant

Which gives us V = R (nT / P)

- OR -

PV = nRT

The Ideal Gas Law

Q: Calcium carbonate, CaCo3 (s), decomposes when

heated to give CaO (s) and Co2 (g). A sample of CaCO3 is

decomposed, and the carbon dioxide produced is

collected in a 250mL flask. After the decomposition is

complete, the gas has a pressure of 1.3atm at a

temperature of 31C. How many moles of CO2 gas were

produced?

A: We are looking for moles and are not at standard

temperature and pressure…so use PV = nRT!

𝒏 =𝑷 × 𝑽

𝑹 × 𝑻=

𝟏. 𝟑 𝒂𝒕𝒎 × 𝟎. 𝟐𝟓𝟎 𝑳

𝟎. 𝟎𝟖𝟐𝟏 𝑳 𝒂𝒕𝒎 𝒎𝒐𝒍−𝟏𝑲−𝟏 × 𝟑𝟎𝟒 𝑲

n = 0.013 mol CO2 (g)

Special Uses of Ideal Law

Gas Density and Molar Mass

let M = molar mass

PV = nRT , (rearranging) … n/V = P/RT

nM = mass , density = d = mass / V

DENSITY = d = nM / V = PM / RT , so d = PM / RT

MOLAR MASS = M = dRT / P , M = dRT / P

Special Uses of Ideal Law

Q: Determine the density of xenon gas at 75C in a

container with a pressure of 2.00atm.

A: 𝒅 =𝑷×𝑴

𝑹×𝑻=

𝟐.𝟎𝟎 𝒂𝒕𝒎 × 𝟏𝟑𝟏.𝟑 𝒈 𝒎𝒐𝒍−𝟏

𝟎.𝟎𝟖𝟐𝟏 𝑳 𝒂𝒕𝒎𝒎𝒐𝒍−𝟏𝑲−𝟏 × 𝟑𝟒𝟖 𝑲

d = 9.18 g L1

Q: What is the molar mass of a gas with a density of

2.65g/L under the same conditions?

A: 𝑴 =𝒅×𝑹×𝑻

𝑷=

𝟐.𝟔𝟓 𝒈 𝑳−𝟏 × 𝟎.𝟎𝟖𝟐𝟏 𝑳 𝒂𝒕𝒎 𝒎𝒐𝒍−𝟏 𝑲−𝟏 × 𝟑𝟒𝟖 𝑲

𝟐.𝟎𝟎 𝒂𝒕𝒎

M = 37.9 g mol1

Daltons Law of Partial Pressure Because “ideal” gases behave independently of each other, they behave

very similarly. This means that a mix of gases will behave as several ideal

gases (under normal conditions) together, which means it will all behave

similarly…the only variable is the amount of each gas present in moles (n).

PT = P1 + P2 + … + Pn

Where Pn = nn (RT/V) or Pn = (nn / nT) PT

Q: What is the total pressure exerted by a mixture of 2.00g H2

and 8.00g N2 at 273K in a 10.0L vessel?

A: First, find moles of each…

2.00𝑔 𝐻2 ×1 𝑚𝑜𝑙 𝐻2

2.02 𝑔 𝐻2= 0.990 𝑚𝑜𝑙 𝐻2

8.00𝑔 𝑁2 ×1 𝑚𝑜𝑙 𝑁2

28.02 𝑔 𝑁2= 0.286 𝑚𝑜𝑙 𝑁2

then, find combined pressure…

𝑷𝒕𝒐𝒕𝒂𝒍 =𝒏𝒕𝒐𝒕𝒂𝒍𝑹𝑻

𝑽

𝑃𝑡𝑜𝑡𝑎𝑙 =0.99𝑚𝑜𝑙 + 0.286𝑚𝑜𝑙 8.31 𝐿 𝑘𝑃𝑎 𝑚𝑜𝑙−1𝐾−1 273𝐾

10.0 𝐿

Ptotal = 289 kPa

Partial Pressure

Collecting Gas Over Water

Collection of gas over water inherently adds water vapor to the mixture

and it must be accounted for via Dalton’s Law.

Q: A sample of KClO3 is partially decomposed producing O2

gas that is collected over water. The volume of gas

collected is 0.250L at 26C and 765torr total pressure.

How many moles of O2 were collected? What mass of

KClO3 decomposed?

A: 1) Determine pressure of O2…

𝑃𝑂2= 𝑃𝑡𝑜𝑡𝑎𝑙 − 𝑃𝐻2𝑂 = 765 𝑡𝑜𝑟𝑟 − 25.2 𝑡𝑜𝑟𝑟 = 740 𝑡𝑜𝑟𝑟 = 0.974 𝑎𝑡𝑚

2) Determine moles of O2…

𝑛𝑂2=

𝑃𝑂2𝑉

𝑅𝑇=

0.974 𝑎𝑡𝑚 0.250 𝐿

0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙−1 𝐾−1 299 𝐾= 9.92 × 10−3 𝑚𝑜𝑙

3) Use stoichiometry to find masss of KClO3 …

9.92 × 10−3 𝑚𝑜𝑙 𝑂2 ×2 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂3

3 𝑚𝑜𝑙 𝑂2×

122.5 𝑔 𝐾𝐶𝑙𝑂3

1 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂3= 0.810 𝑔 𝐾𝐶𝑙𝑂3

BCE: 2KClO3 2KCl + 3O2

KEavg and urms

KEavg = 3/2 RT

Relationship of temperature and KE

urms = [3RT/M]1/2

M must be in kg in order to work

out…represents average speed, and u is not

the average velocity…but is close

Effusion and Diffusion

Both concepts rely heavily on the random motion of

gases…basically they relate the fact that while kinetic energy is

the same for molecules at the same temperature, the root-

mean-square velocity for lighter molecules is higher because

they have a lower mass.

EFFUSION: 𝑟𝑎𝑡𝑒2

𝑟𝑎𝑡𝑒1=

𝑀1

𝑀2

DIFFUSION: 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒2

𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒1=

𝑀1

𝑀2

Escape of a

gas through

a small hole

Spreading out

of a gas in space

van der Waals Equation

𝑃 = 𝑛𝑅𝑇

𝑉 − 𝑛𝑏−

𝑛2𝑎

𝑉2

Adjusts for the fact that

atoms DO have volume!

Adjusts for intermolecular

attractions!

Greatest deviations at HIGH PRESSURE and LOW

TEMPERATURE!