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## Chapter 5 Gases

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**The Structure of a Gas**• Gases are composed of particles that are flying around very fast in their container(s) • The particles in straight lines until they encounter either the container wall or another particle, then they bounce off • If you were able to take a snapshot of the particles in a gas, you would find that there is a lot of empty space in there**Gases Pushing**• Gas molecules are constantly in motion • As they move and strike a surface, they push on that surface • push = force • If we could measure the total amount of force exerted by gas molecules hitting the entire surface at any one instant, we would know the pressure the gas is exerting • pressure = force per unit area**The Effect of Gas Pressure**• The pressure exerted by a gas can cause some amazing and startling effects • Whenever there is a pressure difference, a gas will flow from an area of high pressure to an area of low pressure • the bigger the difference in pressure, the stronger the flow of the gas • If there is something in the gas’s path, the gas will try to push it along as the gas flows**Air Pressure**• The atmosphere exerts a pressure on everything it contacts • the atmosphere goes up about 370 miles, but 80% of the molecules are in the first 10 miles from the Earth’s surface • This is the same pressure that a column of water would exert if it were about 10.3 m high**Atmospheric Pressure Effects**• Differences in air pressure result in weather and wind patterns • The higher in the atmosphere you climb, the lower the atmospheric pressure is around you • at the surface the atmospheric pressure is 14.7 psi, but at 10,000 ft it is only 10.0 psi**Pressure Imbalance in the Ear**If there is a difference in pressure across the eardrum membrane, the membrane will be pushed out – what we commonly call a “popped eardrum”**The Pressure of a Gas**• Gas pressure is a result of the constant movement of the gas molecules and their collisions with the surfaces around them • The pressure of a gas depends on several factors • number of gas particles in a given volume • volume of the container • average speed of the gas particles**gravity**Measuring Air Pressure • We measure air pressure with abarometer • Column of mercury supported by air pressure • Force of the air on the surface of the mercury counter balances the force of gravity on the column of mercury**Practice – What happens to the height of the column of**mercury in a mercury barometer as you climb to the top of a mountain? • The height of the column increases because atmospheric pressure decreases with increasing altitude • The height of the column decreases because atmospheric pressure decreases with increasing altitude • The height of the column decreases because atmospheric pressure increases with increasing altitude • The height of the column increases because atmospheric pressure increases with increasing altitude • The height of the column increases because atmospheric pressure decreases with increasing altitude • The height of the column decreases because atmospheric pressure decreases with increasing altitude • The height of the column decreases because atmospheric pressure increases with increasing altitude • The height of the column increases because atmospheric pressure increases with increasing altitude**psi**atm mmHg Example 5.1: A high-performance bicycle tire has a pressure of 132 psi. What is the pressure in mmHg? Given: Find: 132 psi mmHg Conceptual Plan: Relationships: 1 atm = 14.7 psi, 1 atm = 760 mmHg Solution: Check: because mmHg are smaller than psi, the answer makes sense**Manometers**• The pressure of a gas trapped in a container can be measured with an instrument called a manometer • Manometers are U-shaped tubes, partially filled with a liquid, connected to the gas sample on one side and open to the air on the other • A competition is established between the pressures of the atmosphere and the gas • The difference in the liquid levels is a measure of the difference in pressure between the gas and the atmosphere**Manometer**for this sample, the gas has a larger pressure than the atmosphere, so**Boyle’s Law**Robert Boyle (1627–1691) • Pressure of a gas is inversely proportional to its volume • constant T and amount of gas • graph P vs V is curve • graph P vs 1/V is straight line • As P increases, V decreases by the same factor • P x V = constant • P1 x V1 = P2 x V2**Boyle’s Experiment**• Added Hg to a J-tube with air trapped inside • Used length of air column as a measure of volume**Boyle’s Law: A Molecular View**• Pressure is caused by the molecules striking the sides of the container • When you decrease the volume of the container with the same number of molecules in the container, more molecules will hit the wall at the same instant • This results in increasing the pressure**Boyle’s Law and Diving**• Because water is more dense than air, for each 10 m you dive below the surface, the pressure on your lungs increases 1 atm • at 20 m the total pressure is 3 atm • If your tank contained air at 1 atm of pressure, you would not be able to inhale it into your lungs • you can only generate enough force to overcome about 1.06 atm Scuba tanks have a regulator so that the air from the tank is delivered at the same pressure as the water surrounding you. This allows you to take in air even when the outside pressure is large.**Boyle’s Law and Diving**• If a diver holds her breath and rises to the surface quickly, the outside pressure drops to 1 atm • According to Boyle’s law, what should happen to the volume of air in the lungs? • Because the pressure is decreasing by a factor of 3, the volume will expand by a factor of 3, causing damage to internal organs. Always Exhale When Rising!!**V2, P1, P2**V1 A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon is now 2.78x 103 mL, what was it originally? Given: Find: V2 =2780 mL, P1 = 762 torr, P2 = 0.500 atm V1, mL Conceptual Plan: Relationships: P1∙ V1 = P2∙ V2 , 1 atm = 760 torr (exactly) Solution: Check: because P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does**Charles’s Law**Jacques Charles (1746–1823) • Volume is directly proportional to temperature • constant P and amount of gas • graph of V vs. T is straight line • As T increases, V also increases • Kelvin T = Celsius T + 273 • V = constant x T • if T measured in Kelvin**If you plot volume vs. temperature for any gas at constant**pressure, the points will all fall on a straight line If the lines are extrapolated back to a volume of “0,” they all show the same temperature, −273.15 °C, called absolute zero**Charles’s Law – A Molecular View**• The pressure of gas inside and outside the balloon are the same • At low temperatures, the gas molecules are not moving as fast, so they don’t hit the sides of the balloon as hard – therefore the volume is small • The pressure of gas inside and outside the balloon are the same • At high temperatures, the gas molecules are moving faster, so they hit the sides of the balloon harder – causing the volume to become larger**The temperature inside a balloon is raised from 25.0 °C to**250.0 °C. If the volume of cold air was 10.0 L, what is the volume of hot air? V1, T1, T2 V2 Given: Find: V1 =10.0 L, t1 = 25.0 °C L, t2 = 250.0 °C V2, L Conceptual Plan: Relationships: Solution: Check: when the temperature increases, the volume should increase, and it does**Avogadro’s Law**Amedeo Avogadro (1776–1856) • Volume directly proportional to the number of gas molecules • V = constant x n • constant P and T • more gas molecules = larger volume • Count number of gas molecules by moles • Equal volumes of gases contain equal numbers of molecules • the gas doesn’t matter**Practice — If 1.00 mole of a gas occupies 22.4 L at STP,**what volume would 0.750 moles occupy? V1, n1, n2 V2 Given: Find: V1 =22.4 L, n1 = 1.00 mol, n2 = 0.750 mol V2 Conceptual Plan: Relationships: Solution: Check: because n and V are directly proportional, when the moles decrease, the volume should decrease, and it does**Ideal Gas Law**• By combining the gas laws we can write a general equation • R is called the gas constant • The value of R depends on the units of P and V • we will use 0.08206 and convert P to atm and V to L • The other gas laws are found in the ideal gas law if two variables are kept constant • Allows us to find one of the variables if we know the other three**Standard Conditions**• Because the volume of a gas varies with pressure and temperature, chemists have agreed on a set of conditions to report our measurements so that comparison is easy – we call these standard conditions • STP • Standard pressure = 1 atm • Standard temperature = 273 K • 0 °C**P1, V1, T1, R**n P2, n, T2, R V2 A gas occupies 10.0 L at 44.1 psi and 27 °C. What volume will it occupy at standard conditions? Given: Find: V1 = 10.0L, P1 = 44.1 psi, t1 = 27 °C, P2 = 1.00 atm, t2 = 0 °C V2, L Conceptual Plan: Relationships: Solution: Check: 1 mole at STP occupies 22.4 L, because there is more than 1 mole, we expect more than 22.4 L of gas**Molar Volume**• Solving the ideal gas equation for the volume of 1 mol of gas at STP gives 22.4 L • 6.022 x 1023 molecules of gas • notice: the gas is immaterial • We call the volume of 1 mole of gas at STP the molar volume • it is important to recognize that one mole measures of different gases have different masses, even though they have the same volume**Practice — How many liters of O2 @ STP can be made from**the decomposition of 100.0 g of PbO2?2 PbO2(s)→2 PbO(s) + O2(g) g PbO2 mol PbO2 mol O2 L O2 Given: Find: 100.0 g PbO2, 2 PbO2 → 2 PbO + O2 L O2 Conceptual Plan: Relationships: 1 mol O2 = 22.4 L, 1 mol PbO2 = 239.2g, 1 mol O2≡ 2 mol PbO2 Solution: because less than 1 mole PbO2, and ½ moles of O2 as PbO2, the number makes sense Check:**Density at Standard Conditions**• Density is the ratio of mass to volume • Density of a gas is generally given in g/L • The mass of 1 mole = molar mass • The volume of 1 mole at STP = 22.4 L**Practice – Calculate the density of N2(g) at STP**MM d Given: Find: N2, dN2,g/L Conceptual Plan: Relationships: Solution: Check: the units and number are reasonable**Gas Density**• Density is directly proportional to molar mass**Molar Mass of a Gas**• One of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law**Practice — What is the molar mass of a gas if 12.0 g**occupies 197 L at 380 torr and 127 °C? n, m P, V, T, R MM n Given: Find: m=12.0 g, V= 197 L, P=380 torr, t=127°C, molar mass,g/mol m = 12.0g, V = 197 L, P = 0.50 atm, T =400 K, molar mass,g/mol Conceptual Plan: Relationships: Solution: Check: the unit is correct and the value is reasonable**Mixtures of Gases**• When gases are mixed together, their molecules behave independent of each other • all the gases in the mixture have the same volume • all completely fill the container each gas’s volume = the volume of the container • all gases in the mixture are at the same temperature • therefore they have the same average kinetic energy • Therefore, in certain applications, the mixture can be thought of as one gas • even though air is a mixture, we can measure the pressure, volume, and temperature of air as if it were a pure substance • we can calculate the total moles of molecules in an air sample, knowing P, V, and T, even though they are different molecules Tro: Chemistry: A Molecular Approach, 2/e**Partial Pressure**• The pressure of a single gas in a mixture of gases is called its partial pressure • We can calculate the partial pressure of a gas if • we know what fraction of the mixture it composes and the total pressure • or, we know the number of moles of the gas in a container of known volume and temperature • The sum of the partial pressures of all the gases in the mixture equals the total pressure • Dalton’s Law of Partial Pressures • because the gases behave independently**The partial pressure of each gas in a mixture can be**calculated using the ideal gas law**nXe, V, T, R**PXe Ptot, PXe PNe Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe Given: Find: Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol PNe, atm Conceptual Plan: Relationships: Solution: the unit is correct, the value is reasonable Check:**Mole Fraction**The fraction of the total pressure that a single gas contributes is equal to the fraction of the total number of moles that a single gas contributes The ratio of the moles of a single component to the total number of moles in the mixture is called the mole fraction, c for gases, = volume % / 100% The partial pressure of a gas is equal to the mole fraction of that gas times the total pressure**Ptot, PNe**cNe nXe, V, T, R PXe Ptot, PXe PNe Find the mole fraction of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe Given: Find: Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol PNe, atm Conceptual Plan: Relationships: Solution: the unit is correct, the value is reasonable Check:**Collecting Gases**• Gases are often collected by having them displace water from a container • The problem is that because water evaporates, there is also water vapor in the collected gas • The partial pressure of the water vapor, called the vapor pressure, depends only on the temperature • so you can use a table to find out the partial pressure of the water vapor in the gas you collect • if you collect a gas sample with a total pressure of 758.2 mmHg* at 25 °C, the partial pressure of the water vapor will be 23.78 mmHg – so the partial pressure of the dry gas will be 734.4 mmHg • Table 5.4*