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Gases. Compared to the mass of a dozen eggs, the mass of air in an “empty refrigerator” is Negligible About a tenth as much About the same More. Answer:.
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Compared to the mass of a dozen eggs, the mass of air in an “empty refrigerator” is • Negligible • About a tenth as much • About the same • More
Answer: • One cubic meter of air at 0 degrees Celsius and normal atmospheric pressure has a mass of about 1.3 Kg. A medium-sized refrigerator has a volume of 0.6 cubic meters and contains 0.8 kilograms of air. A dozen large eggs weighs in at 0.75 kilograms. Why do we think of air as having no mass???
Answer: • We don’t notice the weight of air on us because we are submerged in air. If someone handed you a bag of water while you were submerged in water, you wouldn’t notice its weight either. Interesting Fact! The average mass of the atmosphere is about 5,000 trillion metric tons (or 5 x 1018 kg) according to the National Center for Atmospheric Research.
Kinetic Theory The state of matter is determined by the motion of its particles. Phases of Matter
Kinetic Theory and Gases • Gases consist of tiny particles that are in constant motion • Collisions between particles of a gas and other objects are elastic-there is no loss of kinetic energy • Average kinetic energy of particles of a gas is directly proportional to the temperature of the gas
Nature of Gases • Gases have no definite shape or volume • Gases have low density • Gases can be compressed • Gases are considered fluids • Gases are capable of diffusion-they can spread out with the absence of circulating currents
Ideal Gas • An imaginary gas that conforms perfectly to all of the assumptions of kinetic theory
Real Gas • A gas that does not obey all of the assumptions of kinetic theory • Gases will deviate from ideal gas behavior under conditions of high temperature and pressure.
Pressure • The force per unit area on a surface Pressure = Force / Area • Gas molecules exert pressure on every surface with which they collide-the pressure exerted depends on: • Number of molecules present • Temperature • Volume of container
What is normal atmospheric pressure? • At sea level, the normal atmospheric pressure is 1 atmosphere (atm) • 1 atm = 760 mm Hg • 1 atm = 760 torr • 1 atm = 101.3 kPa
Standard Temperature and Pressure (STP) (STP) is defined as: 1 atm and 0 degrees Celsius
Boyle’s Law • Relates changes in pressure and volume assuming constant temp. and molecules of gas. • States that pressure and volume are inversely proportional (If you ↓V, you ↑ P) P1V1=P2V2
Charles’ Law • Relates changes in volume and temp. assuming constant pressure and molecules of gas. • States that volume and temp. are directly proportional (increase temp, increase volume) V1 = V2 T1 T2 Varies directly with the Kelvin scale so all temps must be in Kelvin Remember… K= C + 273
Gay Lussac’s Law • Deals with changes in pressure and temperature assuming volume and amount of gas are held constant. • Pressure and temperature are directly proportional. (↑Temp, ↑Pressure) P1 = P2 T1 T2
Combined Gas Law • Relates changes in pressure, temp. and volume assuming the amount of gas is held constant. P1V1 = P2V2 T1 T2
Example Problems: • Look at what information is given to decide which gas law must be used. • Check to make sure that units are the same on both sides of the equation and that all temps. are in Kelvin
A sample of nitrogen occupies a volume of 250 mL at 25 C. What volume will it occupy at 95 C? • A sample of carbon dioxide gas occupies a volume of 3.50 L at 125 kPa. What pressure would the gas exert if the volume was decreased to 2.00 L? • A helium filled balloon has a volume of 50 L at 25 C and 820 mmHg. What volume will it occupy at 650 mm Hg and 10 C?
A sample of nitrogen occupies a volume of 250 mL at 25 C. What volume will it occupy at 95 C? Convert temps to Kelvin: K = °C + 273 V1 = 250 mL V2 = ? T1 = 25°C = 298 K T2 = 95°C = 368 K V1 = V2250 mL = V2 T1 T2 298 K 368 K ***Cross multiply, and solve for V2 (250)x(368) = (298)x(V2) V2 = 308.72 mL
A sample of carbon dioxide gas occupies a volume of 3.50 L at 125 kPa. What pressure would the gas exert if the volume was decreased to 2.00 L? V1 = 3.50 L V2 = 2.00 L P1 = 125 kPa P2 = ? P1V1=P2V2 (125)x(3.50) = (P2)x(2.00) P2 = 218.75 kPa
A helium filled balloon has a volume of 50 L at 25 C and 820 mmHg. What volume will it occupy at 650 mm Hg and 10 C? V1 = 50 L V2 = ? T1 = 25°C = 298 K T2 = 10°C = 283 K P1 = 820 mmHg P2 = 650 mm Hg P1V1 = P2V2(820)(50) = (650)(V2) T1 T2 298 K 283 K V2 = 59.9 L
Dalton’s Law of Partial Pressure • The total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. • Partial pressure- the pressure of an individual gas in a mixture of gases Pt = P1 + P2 + P3 …… Where Pt = total pressure and P1, P2, etc. = partial pressures of individual gases
Partial Pressures are used to: • Calculate water vapor pressure when a gas is collected over water. • Because water molecules at a liquid surface always evaporate, gases collected over water are not pure….they are always mixed with water vapor….so Dalton’s Law is used to calculate the partial pressure of the gas.
Example Problem-Partial Pressure • Oxygen from the decomposition of potassium chlorate was collected over water. The barometric pressure and the temp. during the experiment were 731 mm Hg and 20 C. What was the partial pressure of oxygen collected? (Use the water vapor pressure to solve) PressureTotal = PressureOxygen + Pressure Water vapor 731 mm Hg = Pressure Oxygen + 17.5 mm Hg Pressureoxygen = 714.5 mm Hg
Example Problem-Partial Pressure • A sample of gas was collected over water at 740 mm Hg and 23 C. The collecting tube is left in place, and the volume is not measured until the next day when the dry gas was found to have a pressure of 745 mm Hg and a temp. of 20 C, at which time the volume is found to be 15.3 mL. What is the original volume (from yesterday)? PressureTotal = PressureGas + Pressure Water vapor 740 mm Hg = Pressure Gas + 21.1 mm Hg PressureGas = 718.9 mm Hg (pressure from yesterday) P1 = 718.9 mm Hg, V1 = ?, T1 = 23°C = 296 K P2 = 745 mm Hg, V2 = 15.3 mL, T2 = 20°C = 293 K
…Continued • Solving for the original volume: P1 = 718.9 mm Hg, V1 = ?, T1 = 23°C = 296 K P2 = 745 mm Hg, V2 = 15.3 mL, T2 = 20°C = 293 K • Use the equation: P1V1 = P2V2 T1 T2 (718.9)(V1) = (745)(15.3) 296 293 V1 = 16.02 L
Ideal Gas Law • Expresses a relationship between pressure, volume, temp. and moles of gas PV= nRT (where P= pressure, V= volume, n= moles, R= ideal gas constant, and T= temp)
Ideal Gas Constant • The constant’ value depends on the units chosen for pressure, volume, and temp. • If 1 mole of a gas occupies 22.4 L at STP (0 degrees C and 1 atm) and these values are substituted into the ideal gas law, the value for R can be calculated: R = PV = (1atm)( 22.4 L) = 0.0821 L . atm nT (1 mol) (273 K) mol . K Since R has units of L, atm, mol, and K, all pressure, volumes, temps., and amounts must also be expressed in these units.
Examples: • What is the pressure in atm exerted by 0.50 mol sample of nitrogen in a 10 L container at 298 K? • PV = nRT • (P)(10) = (0.5)(0.0821)(298) • P = 1.2 atm
Examples: • What is the volume in liters occupied by 0.25 mol of oxygen at 20 degrees C and 740 mm Hg? • Convert all units to K, atm, and L!! • 740 mm Hg x 1 atm = 0.974 atm 760 mm Hg • 20°C + 273 = 293 K • PV = nRT • (0.974)(V) = (0.25)(0.0821)(293) • V = 6.17 L
More Examples: • At 28 C and 0.974 atm, 1.00 L of a gas has a mass of 5.16 g. What is the molar mass of the gas? (Hint.. Remember that molar mass is in units of g/mol) • PV = nRT 28°C = 301 K • (0.974)(1.00) = (n)(0.0821)(301) • n = 0.0394 mol • Molar mass: grams/mol 5.16g/0.0394 mol • Molar mass = 130.96 g/mol
More Examples: • What is the density of a sample of ammonia (NH3), if the pressure is 0.928 atm and the temp. is 63.0 C and there is 2 L of gas? • (Density = mass / volume) • PV = nRT 63°C = 336 • (0.928)(2) = (n)(0.0821)(336) • n = 0.06728 mol NH3 (Change to grams!!) • 0.06728 mol NH3 x 17.03052 g = 1.1458 g NH3 1 mol NH3 Density = mass/volume = 1.1458g/2L = 0.5729 g/L
Effusion and Diffusion • Effusion- the process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in the container • Diffusion- the spreading out of a gas due to the constant motion of gas molecules
Graham’s Law of Effusion • Rates of effusion and diffusion depend on the velocities of gas molecules. • The velocity of a gas varies inversely with its mass (lighter molecules move faster).
Suppose you wanted to compare the velocity of two gases: • Graham’s Law states that the rates of effusion of gases at the same temp. and pressure are inversely proportional to the square roots of their molar masses. Rate of effusion of A = √Mb Rate of effusion of B √Ma
Examples: Compare the rate of effusion of carbon dioxide with that of hydrochloric acid at the same temp. and pressure. • CO2 has a molar mass of 44.01 g • HCl has a molar mass of 36.46 g • Since mass and velocity are inversely related, the smaller molar mass would have the greater velocity • Therefore, HCl has a higher rate of effusion
Examples: • If a molecule of neon gas travels at an average of 400 m/s at a given temp., estimate the avg. speed of a molecule of butane (C4H10) at the same temp. • Assign Ne as “A” and C4H10 as “B” • Molar mass of C4H10 = 58.12 g = Mb • Molar mass of Ne = 20.18 g = Ma • 400 m/s = √58.12 Rate C4H10 √20.18 Rate of effusion of C4H10 = 235.70 m/s Rate of effusion of A = √Mb Rate of effusion of B √Ma
More Examples: • A sample of hydrogen effuses through a porous container about 9 times faster than an unknown gas. Estimate the molar mass of the unknown gas. • Assign H2 as “A” and Unknown as “B” • Assign rate of Unknown as “x” so rate of H2 would be “9x” 9x = √MB x √2.016 (√2.016)(9x) = (x)(√MB) The “x” cancels on both sides. (√2.016)(9) = (√MB) 12.78 = (√MB) MB = 163.3 g Rate of effusion of A = √Mb Rate of effusion of B √Ma
Stoichiometry of Gases • Remember… 1 mol of any gas at STP will occupy 22.4 L -this is called molar volume
When solving stoichiometry problems involving gases the following will be used: • Given • Unknown • Mol ratio –compares given and unknown • Molar mass-if given is in grams or unknown is asked for in grams • Molar volume- Only if at STP (used to convert moles to liters or liters to moles)
Example: C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g) • What volume in liters of oxygen would be required for the complete combustion of 0.350 L of propane at STP? 0.350 L C3H8 x1 mol C3H8 x 5 mol O2 x 22.4 L O2 = 17.5 L O2 22.4 L 1 mol C3H8 1 mol O2 b) What volume of CO2 will be produced at STP? 0.350 L C3H8 x 1 mol C3H8 x 3 mol O2 x 22.4 L O2 = 1.05 L O2 22.4 L 1 mol C3H8 1 mol O2
How many grams of calcium carbonate must be decomposed to produce 5 L of carbon dioxide at STP? CaCO3 (s) CaO (s) + CO2 (g) 5 L CO2 x 1 mol CO2 x 1 mol CaCO3 x 100.088 g CaCO3 = 22.34 g CaCO3 22.4 L 1 mol CO2 1 mol CaCO3
WO3 (s) +3H2 (g) W (s) + 3 H2O (l) How many liters of hydrogen at 35 C and 745 mm Hg are needed to react completely with 875 g of tungsten oxide? (Because we are not at STP, 1 mol of the gas will no longer occupy 22.4 L..so molar volume can not be used but we need a way to convert mol into liters to use in stoichiometry….we will have to use the Ideal Gas Law)
WO3 (s) +3H2 (g) W (s) + 3 H2O (l) How many liters of hydrogen at 35 C and 745 mm Hg are needed to react completely with 875 g of tungsten oxide? 875 g WO3 x 1 mol WO3 x 3 mol H2 = 11.32 mol H2 231.84 g WO3 1 mol WO3 745 mm Hg x 1 atm = .9802 atm 35°C = 308 K 760 mm Hg PV = nRT (0.9802)(V) = (11.32)(0.0821)(308) V = 292.03 L H2
To decide which to use first: • Write down the given and unknown • Write down the mole ratio • Write down any molar masses needed • Write down the variables for the gas law, using in pressures, volumes, temps., that are given in the problem. • If there is enough info to solve the gas law 1st, do so. You will be solving for moles of whatever the volume amount is for.
Summary: What to do for problems that are NOT at STP! • If there is not enough info to solve the gas law first, use the given to find the moles of what you are trying to find the volume of…plug moles back into the gas law and solve for volume.
2Na (s) + Cl2 (g) 2NaCl (s) How many grams of Na should be used to react with 3.54 L of chlorine at 38 C and 1.63 atm? ***Note: This is not at STP so we can’t use the 1 mol = 22.4 L conversion. Therefore, we use the Ideal Gas Law and stoichiometry. PV = nRT (1.63)(3.54) = (n)(0.0821)(311) n = 0.226 mol Cl2 0.226 mol Cl2 x 2 mol Na x 22.9898 g Na = 10.39 g Na 1 mol Cl2 1 mol Na