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Physical States of Matter

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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

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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

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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+).

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The attractive forces between the permanent dipoles of two polar molecules

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

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

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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

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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

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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

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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

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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

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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

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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

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

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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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 =

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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

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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

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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

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

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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

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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

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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

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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

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

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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)

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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

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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

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

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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

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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

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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

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

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53 solid CO 2 gaseous CO 2

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

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

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