states of matter the solid state

States of Matter

The Solid State

Particles are tightly packed, very close together (strong cohesive forces)

Low kinetic energy (energy of motion)

Fixed shape and volume

Crystalline or amorphous structure

The Liquid State

Particles are close to each other (making them mostly incompressible)

Attractive forces keep molecules close, but not so close to restrict movement

The Gas State

•Gas particles move randomly and rapidly.

•Size of gas particles is small compared to the space between the particles.

•Gas particles exert no attractive forces on each other.

• Kinetic energy of gas particles increases with increasing temperature.

a The symbol “~” means approximately.

Gases and Pressure

•When gas particles collide with the walls of a container, they exert a pressure.

•Pressure (P) is the force (F) exerted per unit area (A).

Pressure =Force

=F

Area A

1 atmosphere (atm) =

760. mm Hg

760. torr

14.7 psi

101,325 Pa

Gas Laws

Mathematical relationships describing the behavior of gases with regard to mixing, diffusion, changes in pressure, changes in temperature

Boyle’s Law: Describes the relation between pressure and volume of a gas, under a constant temperature

PiVi = PfVf

where i = initial condition and f = final condition

Boyle’s Law: Inverse relation between Pressure and Volume

Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of a 8 L sample of Freon gas initially at 50 mm Hg after its pressure is changed to 200 mm Hg at constant T?

1. Set up a data table

Conditions 1 Conditions 2P1 = 50 mm Hg P2 = 200 mm HgV1 = 8 L V2 = ?

Example:

2. Solve Boyle’s Law for V2:

P1V1 = P2V2

V2 = V1P1

P2

V2 = 8 L x 50 mm Hg = 2 L200 mm Hg

A sample of helium gas has a volume of 6.4 L at a pressure of 0.70 atm.

What is the new volume when the pressure is increased to 1.40 atm (T constant)?

Learning Check

P1V1 = P2V2

Solve for V2:

V2 = V1P1

P2 V2 = 6.4 L x 0.70 atm = 3.2 L

1.40 atm

Volume decreases when there is an increase in the pressure (Temperature is constant).

Solution

A sample of oxygen gas has a volume of 12.0 L at 600. mm Hg. What is the new pressure when the volume changes to 36.0 L? (T and n constant.)

Learning Check

Conditions 1 Conditions 2

P1 = 600. mm Hg P2 = ?V1 = 12.0 L V2 = 36.0 L

P2 = P1 V1

V2

600. mm Hg x 12.0 L = 200. mm Hg 36.0 L

Solution

Charles’ Law: Describes relation between temperature and volume of a gas, under constant pressure

Vi/Ti = Vf/Tf

Charles’s Law: Direct relationship between Volume and Temperature

Charles’ Law

Example:

A balloon has a volume of 785 mL at 21°C. If the temperature drops to 0°C, what is the new volume of the balloon (P constant)?

1. Set up data table:Conditions 1 Conditions 2

V1 = 785 mL V2 = ?

T1 = 21°C = 294 K T2 = 0°C = 273 K

Be sure that you always use the Kelvin (K)temperature in gas calculations!

2. Solve Charles’ law for V2

V1 = V2

T1 T2

V2 = V1 T2

T1

V2 = 785 mL x 273 K = 729 mL

294 K

A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. At what temperature (in °C) will the volume of the oxygen be 640 mL (P and n constant)?

Learning Check

T2 = T1V2

V1

T2 = 291 K x 640 mL = 443 K

420 mL

= 443 K – 273 K = 170°C

Solution

Combined Gas Law: Describes relation between pressure, temperature and volume of a gas

PiVi/Ti = PfVf/Tf

A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm (n constant)?

1. Set up Data TableConditions 1 Conditions 2

P1 = 0.800 atm P2 = 3.20 atmV1 = 0.180 L (180 mL) V2 = 90.0 mLT1 = 29°C + 273 = 302 K T2 = ?

Example:

2. Solve for T2 P1 V1 = P2 V2

T1 T2

T2 = T1 P2V2

P1V1

T2 = 302 K x 3.20 atm x 90.0 mL = 604 K

0.800 atm 180.0 mL

T2 = 604 K – 273 = 331 °C

A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the volume(mL) of the gas at –95°C and a pressure of 802 mm Hg (n constant)?

Learning Check

Data TableT1 = 308 K T2 = -95°C + 273 = 178KV1 = 675 mL V2 = ?P1 = 646 mm Hg P2 = 802 mm Hg

Solve for V2

V2 = V1 P1 T2

P2T1

V2 = 675 mL x 646 mm Hg x 178K = 314 mL 802 mm Hg x 308K

Solution

Gay–Lussac’s Law: Describes the relation between pressure and temperature of a gas, at a constant volume

•Pressure and temperature are directly related

Pressure

Temperature= constant

P

T= k

Note: Temperature must be expressed in kelvins.

P1

T1

=P2

T2

Avogadro’s Law: Equal volumes of gases measured at the same temperature and pressure contain equal number of molecules

Vi/ni = Vf/nf

where n = number of moles

Avogadro’s Law Example:

If 0.75 mole of helium gas occupies a volume of 1.5 L, what volume will 1.2 moles of helium occupy at the same temperature and pressure?

Conditions 1 Conditions 2V1 = 1.5 L V2 = ?n1 = 0.75 mole He n2 = 1.2 moles He

V1/n1 =V2/n2

V2 = V1n2

n1

V2 = 1.5 L x 1.2 moles He = 2.4 L 0.75 mole He

Ideal Gas Law: Describes relation between pressure, volume, temperature and the number of molecules in an ideal gas sample

PV = nRT

where R = universal gas constant (0.0821 L atm/K mol)

A cylinder contains 5.0 L of O2 at 20.0°C and 0.85 atm. How many grams of oxygen are in the cylinder?

Ideal Gas Law Example:

P = 0.85 atm, V = 5.0 L, T = 293 K, n (or g =?)

PV = nRT n = PV

RT

= (0.85 atm)(5.0 L)(mole K) = 0.18 mole O2

(0.0821atm L)(293 K)

= 0. 18 mole O2 x 32.0 g O2 = 5.8 g O2

1 mole O2

Partial Pressure: Pressure an individual gas in a mixture would exert were it alone in the same container

Dalton’s Law: Total pressure exerted by a mixture of gases equals the sum of the partial pressures

P(total) = P(gas 1) + P(gas 2) + P(gas 3) etc.

Dalton’s Law of Partial Pressures

Summary of Gas Laws

Boyle: PiVi = PfVf Charles: Vi/Ti = Vf/Tf

Avogradro: Vi/ni = Vf/nf Gay-Lussac: Pi/Ti = Pf/Tf

Dalton: P(total) = P(gas 1) + P(gas 2) + P(gas 3)

*Combined: PiVi/Ti = PfVf/Tf

*Ideal: PV = nRT

* Memorize for exam

Intermolecular Forces

Intermolecular forces: attractive forces that exist between molecules.

In order of increasing strength, these are:

•London dispersion forces

•dipole–dipole interactions

•hydrogen bonding

London Dispersion Forces

London dispersion forces: weak interactions due to the momentary changes in electron density in a molecule.

•Change in electron density creates a temporary dipole.

•All covalent compounds exhibit London dispersion forces.

•The weak interaction between these temporary dipoles constitutes London dispersion forces.

•The larger the molecule, the larger the attractive force, and the stronger the intermolecular forces.

Dipole-dipole interaction: attraction between positive end of one polar molecule and negative end of a different polar molecule

Hydrogen bonding: Specific type of dipole-dipole force, between the partial positive charge on H and partial negative charge on an electronegative element such as O, N, F

Intermolecular Forces:Boiling Point and Melting Point

Boiling point: temperature at which a liquid is converted to a gas

Melting point: temperature at which a solid is converted to a liquid

The stronger the intermolecular forces on a substance, the higher its boiling point and melting point are.

Examples of Intermolecular Forces andBoiling, Melting Points:

•Both molecules have London dispersion forces and nonpolar bonds.

•In this case, the larger molecule will have stronger attractive forces.

Vapor Pressure

Evaporation: the conversion of liquids into the gas phase.

•Evaporation is endothermic—it absorbs heat from the surroundings.

Condensation: the conversion of gases into the liquid phase.

•Condensation is exothermic—it gives off heat to the surroundings.

Viscosity and Surface Tension

Viscosity: a measure of a fluid’s resistance to flow freely

•Compounds with strong intermolecular forces tend to be more viscous than compounds with weaker forces.

•Substances composed of large molecules tend to be more viscous, too, because large molecules do not slide past each other as freely.

Surface tension: a measure of the resistance of a liquid to spread out.

•Interior molecules in a liquid are surrounded by intermolecular forces on all sides.

•Surface molecules only experience intermolecular forces from the sides and from below.

The stronger the intermolecular forces, the higher the surface tension.

The Solid State: Types of Solids

Crystalline solid: has a regular arrangement of particles—atoms, molecules, or ions—with a repeating structure.

There are four different types of crystalline solids— ionic, molecular, network, and metallic.

Crystalline Solids

Ionic solid: composed of oppositely charged ions

Molecular solid: composed of individual molecules arranged regularly

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Network solid: composed of a vast number of atoms covalently bonded together (SiO2).

Metallic solid: a lattice of metal cations surrounded by a cloud of e− that move freely (Cu).

Amorphous Solids

•They can be formed when liquids cool too quickly for regular crystal formation.

•Very large covalent molecules tend to form amorphous solids, because they can become folded and intertwined.

Examples: rubber, glass, and plastic.

Amorphous solid: has no regular arrangement of its closely packed particles.