Presentation on theme: "Gases and Thermochemistry"— Presentation transcript:
1Gases and Thermochemistry Unit 6 - Gases IUNIT 6Gases and ThermochemistryGas Laws
2Properties of Gases One of the three physical states of matter Expand spontaneously to fill the containerWe live in a homogeneous mixture of gases called the atmosphere.
3Properties of GasesCharacterized by minimal intermolecular attractions: each gas molecule pretty much behaves as if there are no other gas molecules around.Spontaneously form homogeneous mixtures with other gasesCharacterized by pressure (P), volume (V), temperature (T), and amount (n) but not (much) by chemical identity
4Pressure Pressure is force per unit area. Depending on our use of gas, we will use one of several units of pressure.Standard Atmospheric Pressure is the typical pressure at sea level:1 atmosphere (atm)14.7 pounds per sq. in. (psi)pascal (Pa) - (The pascal is the SI unit of Pressure = 1 N/m2)bar1013 mbar29.92 inches of Hg760 mm of Hg760 torr
5PressureThe white curves are isobar lines, lines of constant atmospheric pressure.
6PressureAtmospheric pressure is measured with a barometer (a must when collecting gases over water).The weight of the atmosphere pressing on the mercury forces it up the evacuated tube until its weight equals the weight of the atmosphere (hence the unit of pressure called the mm of Hg). Blood pressures (120/80) are measured in mm of Hg.Pressures of contained gases (in cylinders or in vacuum systems) are measured with gauges suited to the specific purpose.Mercury barometer
7Converting between Units of Pressure Use the equivalents to 1 atm (in the back of your text) to convert from one unit of pressure to another.The pressure in a scanning electron microscope is 5.0 x torr. What is the pressure in atm? In pascal?These are just dimensional analysis problems:5.0 x 10-5 torr x 1 atm = 6.6 x 10-8 atm760 torr5.0 x 10-5 torr x 1 atm x Pa = 6.7 x 10-3 Pa760 torr 1 atm
8Gas Laws Boyle’s Law: PV = constant Volume is inversely proportional to the pressure of a gas if its quantity and temperature are fixed.
9Gas Laws Boyle’s Law: PV = constant Boyle’s Law: P1V1 = P2V2 Applications of Boyle’s Law: A fixed quantity of gas at 23°C exhibits a pressure of 748 torr and occupies a volume of 10.3 liters. What volume will the gas occupy at 23°C if the pressure is increased to 1.88 atm?P1V1 = P2V2: torr (10.3 L) = (1.88 atm x 760 torr) (?)1 atm? = torr (10.3 L) = L1.88 atm x 760 torr/atmPressure was increased, so the volume had to decrease.
10Gas Laws Charles’ Law: V = kT (where k is a constant) The volume of an ideal gas at 0K is zero!Volume is directly proportional to the temperature of a gas IF its quantity and pressure are fixed.
11Gas Laws Charles’ Law: V = kT Charles’ Law : V1 = V2 T1 T2 ALWAYS EXPRESS THE TEMPERATURE IN KELVIN FOR GAS LAW CALCULATIONS!!!Charles’ Law: V = kTCharles’ Law : V1 = V2T1 T2Applications of Charles’ Law: A fixed quantity of gas at 23°C exhibits a pressure of 748 torr and occupies a volume of 10.3 liters. What volume will the gas occupy if the pressure is held constant and the temperature is increased to 125°C?T1 = 23°C = 296 K T2 = 125°C = 398 KV1 = V2: L = ? LT1 T K K? = K (10.3 L) = L296 KTemperature was increased, so the volume had to increase.
12The Combined Gas Law (CGL) PV = constantT orP1V1 = P2V2T T2
13Gas LawsAvogadro’s Law: V = kn (where k is a constant and n is the number of moles of gas)Volume is directly proportional to the quantity of a gas IF its temperature and pressure are fixed.Avogadro’s hypothesis: Equal volumes of gases at the same T and P contain equal numbers of gas molecules.
14Gas Laws Avogadro’s Law: V = kn Avogadro’s Law : V1 = V2 n1 n2 STP is standard temperature and pressure.Standard temperature is 273 KStandard pressure is 1 atmAvogadro’s Law: V = knAvogadro’s Law : V1 = V2n n2Applications of Avogadro’s Law: One mole of a gas occupies a volume of 22.4 L at standard temperature and pressure (STP). What is the volume of 3.00 x 10-2 moles of that same gas at STP?V1 = V2: L = ? Ln1 n2 1 mol x 10-2 mol? = L (3.00 x mol) = L1 molThe number of gas molecules decreased, so the volume had to decrease.
15Gas Laws Ideal Gas Law: PV = nRT Boyle’s Law, Charles’ Law, and Avogadro’s Law can be combined into one gas law, theIdeal Gas Law: PV = nRTThe constants from the previous laws have been combined to form the gas constant R.An ideal gas is any gas that obeys the ideal gas law.STP is standard temperature and pressure.Standard temperature is 273 K (0°C).Standard pressure is 1 atm.1 mole of an ideal gas occupies 22.4 L at STP. This is the standard molar volume of an ideal gas.
16The Ideal Gas Law Contains the Other Gas Laws Ideal Gas Law: PV = nRTBoyle’s Law: PV = k when the quantity and temperature of a gas are constant.PV = constant, so PV = k.nRTCharles’ Law: V = kT when the quantity and pressure of a gas are constant.V = T, so V = kT.constantnRP
17The Ideal Gas Law Ideal Gas Law: PV = nRT (know!!!) The ideal gas equation can be used with reasonable success for a number of gases under a fairly large spectrum of conditions.The value for the gas constant R depends on the other units used in the ideal gas equation:R = L-atm / mol-K= J / mol-K= cal / mol-K= m3-Pa / mol-K
18Working Ideal Gas Law Problems - I What volume would 2.0 moles of He at 298 K and 1.5 atm occupy?PV=nRT(1.5 atm) V = (2.0 mol)( L-atm/mol-K)(298 K)V = (2.0 mol)( L-atm/mol-K)(298 K) = 33 L1.5 atm
19Working Ideal Gas Law Problems - I How many moles of nitrogen are in a 5.00 L container held at 100°C and 715 torr?PV = nRT(715 torr)(5.00 L) = n( L-atm/mol-K)(373 K)760 torr/atmn = (0.941 atm)(5.00 L) = mol N2( L-atm/mol-K)(373 K)
20Working Ideal Gas Law Problems - II A 2.35 L container at 25°C is pressurized to 1.2 atm with carbon dioxide. What would the pressure in the container be if its temperature is raised to 95°C?PV=nRT is not helpful in this form. Use the combined gas lawP1V1 = P2V2T T2In this problem, V1 = V2 (because the container does not change size) and the moles of gas are constant (because there is no indication that any gas was let in or out of the container). The equation then reduces toP1 = P atm = P2T T K KP2 = 1.2 atm (368 K) = 1.5 atm298 K
21Working Ideal Gas Law Problems - II A 2.35 L piston/cylinder arrangement at 25°C is pressurized to 1.2 atm with carbon dioxide. To what temperature in °C must the cylinder be raised for the CO2 pressure to reach 2.5 atm and the volume to be 4.00 L?Use the combined gas lawP1V1 = P2V2T T21.2 atm (2.35 L) = atm (4.00 L)298 K T2T2 = 2.5 atm (4.00 L) (298 K) = 1057 K = 800°C1.2 atm (2.35 L)