Physical States of Matter

 Existing as a gas, liquid, or solid depends on: ◦ Balance between the kinetic energy of it particles ◦ The strength of the interactions between the particles

 Attractive forces that exist between molecules  In order of increasing strength, these are: ◦ London dispersion forces ◦ Dipole-dipole interactions ◦ Hydrogen bonding ◦ Ion-Dipole and Ion-Ion interactions  The strength of the intermolecular forces determines whether a compound has a high or low melting point/boiling point, and thus whether it is a solid, liquid, or gas at a given temperature

 London dispersion forces are very weak interactions due to the momentary changes in electron density in a molecule.  More e − density in one region creates a partial negative charge (δ−). Less e − density in one region creates a partial positive charge (δ+).

 The attractive forces between the permanent dipoles of two polar molecules

 Occurs when a hydrogen atom bonded to O, N, or F is electrostatically attracted to an O, N, or F atom in aonther molecule

 Attractions between ion and charged end of polar molecules ◦ Negative ends of water dipoles surround cation ◦ Positive ends of water dipoles surround anion

 The boiling point is the temperature at which a liquid is converted to the gas phase  The melting point is the temperature at which a solid is converted to the liquid phase  The stronger the intermolecular forces, the higher the boiling point and melting point

 A gas consists of particles that move randomly and rapidly  The size of gas particles is small compared to the space between the particles  Gas particles exert no attractive forces on each other  The kinetic energy of gas particles increases with increasing temperature  When gas particles collide with each other, they rebound and travel in new directions

 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 AreaA 1 atmosphere (atm) = 760. mm Hg 760. torr 14.7 psi 101,325 Pa

 Boyle’s law: For a fixed amount of gas at constant temperature, the pressure and volume of the gas are inversely related ◦ If one quantity increases, the other decreases ◦ The product of the two quantities is a constant, k Pressure x Volume = constant PxV=k

 If the volume of a cylinder of gas is halved the pressure of the gas inside the cylinder doubles P 1 V 1 = P 2 V 2 initial conditions new conditions

 If a 4.0-L container of helium gas has a pressure of 10.0 atm, what pressure does the gas exert if the volume is increased to 6.0 L?  Identify the known quantities and the desired quantity P 1 = 10.0 atm V 1 = 4.0 L V 2 = 6.0 L known quantities P 2 = ? desired quantity

 Write the equation and rearrange it to isolate the desired quantity on one side  Solve the problem P 1 V 1 = P 2 V 2 Solve for P 2 by dividing both sides by V 2. P1V1P1V1 V2V2 = P 2 P1V1P1V1 V2V2 P 2 == (10.0 atm)(4.0 L) (6.0 L) Liters cancel =6.7 atm Answer

 Inhalation:  The rib cage expands and diaphragm lowers  Lung V increases  Causes P to decrease  Air is drawn in to equalize the pressure

 Rib cage contracts and diaphragm rises  Lung V decreases  Causes P to increase  Air is expelled to equalize pressure

 For a fixed amount of gas at constant pressure, the volume of the gas is proportional to its Kelvin temperature ◦ If one quantity increases, the other increases as well ◦ Dividing V by T is a constant k Volume Temperature = constant V T = k

 If the temperature of the cylinder is doubled, the volume of the gas inside the cylinder doubles V1V1 T1T1 = V2V2 T2T2 initial conditions new conditions

 For a fixed amount of gas at constant volume, the pressure of a gas is proportional to its Kelvin temperature ◦ If one quantity increases, the other increases as well ◦ Dividing P by T is a constant, k Pressure Temperature = constant P T = k

 Increasing the T increases the kinetic energy of the gas particles, causing the pressure exerted by the particles to increase P1P1 T1T1 = P2P2 T2T2 initial conditions new conditions

 Combining all three as laws into one equation  This equation is used for determining the effect of changing two factors (e.g., P and T) on the third factor (V) P1V1P1V1 T1T1 = P2V2P2V2 T2T2 initial conditions new conditions

 When P and T are held constant, the V of a gas is proportional to the number of moles present ◦ If one quantity increases, the other increases as well ◦ Dividing the volume by the number of moles is a constant, k Volume Number of moles = constant V n = k

 If the number of moles of a gas in a cylinder is increased, the volume of the cylinder will increase as well V1V1 n1n1 = V2V2 n2n2 initial conditions new conditions

 Often amounts of gas are compared at a set of standard conditions of temperature and pressure, abbreviated as STP  STP conditions are:  At STP, 1 mole of any gas has a volume of 22.4 L  22.4 L is called the standard molar volume 1 atm (760 mm Hg) for pressure 273 K (0 o C) for temperature

HOW TO Convert Moles of Gas to Volume at STP Example How many moles are contained in 2.0 L of N 2 at standard temperature and pressure? Step [1] Identify the known quantities and the desired quantity. 2.0 L of N 2 original quantity ? moles of N 2 desired quantity

HOW TO Convert Moles of Gas to Volume at STP Step [2] Write out the conversion factors L 1 mol 22.4 L or Choose this one to cancel L Step [3] Set up and solve the problem. 2.0 Lx 1 mol 22.4 L = mol N 2 Liters cancel Answer

 All four properties of gases (i.e., P, V, n, and T) can be combined into a single equation called the ideal gas law  R is the universal gas constant PV = nRT For atm: R= L atm mol K For mm Hg: R = 62.4 L mm Hg mol K

Boyle’s LawT 1 = T 2 P 1 V 1 = P 2 V 2 Charles’ LawP 1 = P 2 Gay-Lussac’s Law V 1 = V 2 Avagadro’s Law Combined Gas Law V n = k V1V1 n1n1 = V2V2 n2n2

33 HOW TO Carry Out Calculations with the Ideal Gas Law Example How many moles of gas are contained in a typical human breath that takes in 0.50 L of air at 1.0 atm pressure and 37 o C? Step [1] Identify the known quantities and the desired quantity P = 1.0 atm V = 0.50 L T = 37 o C known quantities n = ? mol desired quantity

HOW TO Carry Out Calculations with the Ideal Gas Law Step [2] Convert all values to proper units and choose the value of R that contains these units. The pressure is given in atm, so use the following R value: R = L atm mol K Temperature is given in o C, but must be in K: K = o C K = 37 o C K = 310 K

Step [3] HOW TO Carry Out Calculations with the Ideal Gas Law Write the equation and rearrange it to isolate the desired quantity on one side. PV = nRTSolve for n by dividing both sides by RT. PV RT = n Step [4] Solve the problem. PV RT n == (1.0 atm) (0.50 L) L atm mol K (310 K) mol Answer =

 Dalton’s law: The total pressure (P total ) of a gas mixture is the sum of the partial pressures of its component gases  For a mixture of three gases A,B, and C: P total =P A + P B + P C partial pressures of A, B, and C

37 Sample Problem 7.9 A sample of exhaled air contains four gases with the following partial pressures: N 2 (563 mm Hg), O 2 (118 mm Hg), CO 2 (30 mm Hg), and H 2 O (50 mm Hg). What is the total pressure of the sample? P total = P N 2 + P O 2 + P CO 2 + P H 2 O P total = P total = 761 mm Hg Answer

 Vapor pressure is the pressure exerted by gas molecules in equilibrium with the liquid phase  Vapor pressure increases with increasing temperature  The boiling point o a liquid is the temperature at which its vapor pressure = 780 mm Hg

 The stronger the intermolecular forces, the lower the vapor pressure at a given temperature

 Viscosity is 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

 The 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 stronger the surface molecules are pulled down toward the interior of a liquid and the higher the surface tension  Water has a very high surface tension because of its strong intermolecular hydrogen bonding  When small objects seem to float on the surface of water they are held up by the surface tension only Photo by Jeffffd

 Solids can be either crystalline or amorphous  Crystalline solids have a regular arrangement of particles- atoms, molecules, or ions – with a repeating structure  Amorphous solids have no regular arrangement of its closely packed particles  There are four different types of crystalline solids – ionic, molecular, network, and metallic

 An ionic solid is composed of oppositely charged ions (NaCl)  A molecular solid is composed of individual molecules arranged regularly (H 2 0)

 A network solid is composed of a vast number of atoms covalently bonded together (SiO 2 )  A metallic solid is a lattice of metal cations surrounded by a cloud of e - that move freely (Cu)

 Amorphous solids have no regular arrangement of their particles  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

SOLID LIQUID GAS fusion freezing evaporation condensation deposition sublimation endothermic exothermic System absorbs energy from surrounds in the form of heat o Requires the addition of heat System releases energy into surrounds in the form of heat or light o Requires heat to be decreased

 The amount of energy needed to melt 1 gram of a substance is called its heat of fusion 48 solid water liquid water

 How much energy in calories is absorbed when 50.0 g of ice cubes melt? The heat of fusion of H 2 O is 79.7 cal/g [1] Identify original quantity and desired quantity: 50.0 g original quantity ? calories desired quantity

79.7 cal 1 g  How much energy in calories is absorbed when 50.0 g of ice cubes melt? The heat of fusion of H 2 O is 79.7 cal/g [2] Write out the conversion factors: The heat of fusion is the conversion factor. Choose this factor to cancel the unwanted unit, g. 1 g 79.1 cal

 How much energy in calories is absorbed when 50.0 g of ice cubes melt? The heat of fusion of H 2 O is 79.7 cal/g [3] Solve the problem: 79.7 cal 1 g 50.0 g x Unwanted unit cancels = 3,990 cal

 The amount of energy needed to vaporize 1 gram of substance is called its heat of vaporization 52 liquid water gaseous water

53 solid CO 2 gaseous CO 2

 A heating curve shows how a substance’s temperature changes as heat is added 54

 A cooling curve shows how a substance’s temperature changes as heat is removed 55