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The Ideal Gas Law Bringing It All Together

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Objectives When you complete this presentation, you will be able to state the ideal gas law derive the ideal gas law constant and discuss its units use the ideal gas law to calculate pressure, volume, temperature, or amount of gas in a system calculate molar mass or density of a gas

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Introduction When we use the combined gas law we allow P, V, and T to vary. we keep the amount of gas constant. If we vary the amount of gas as well as the pressure, volume, and temperature we will use the Ideal Gas Law

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Introduction When we use the phrase Ideal Gas Law, we are talking about an ideal gas we are not talking about a real gas An ideal gas obeys all of the assumptions of the kinetic theory of gases - small particles no attraction moving rapidly perfectly elastic collisions

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Introduction Under most common circumstances, real gases act like ideal gases. Only under conditions of - low temperature or high pressure will real gases deviate from ideality.

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Application Applying the kinetic theory, as we add an amount of gas to a container of gas, we are introducing additional particles to collide with the walls of the container. Pressure goes up to keep V & T constant Volume goes up to keep P & T constant Temperature goes down to keep P & V constant

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Application This means that the amount ( number of mols ) of material varies - directly with pressure and volume directly with pressure and volume inversely with temperature inversely with temperature We will use the equation - PV = nRT P is pressure, V is volume, n is mols of gas, T is temperature, and R is the gas constant.

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Application PV = nRT P: measured in atm, kPa, or mm Hg V: measured in L n: measured in mol T: measured in K R: a constant whose value depends on the units of P, V, n, and T

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The Value of R R is a constant whose value depends on the units of P, V, n, and T (mostly on the units of pressure ) We can find R by solving the ideal gas law for R and entering the appropriate values. We remember from our studies of stoichiometry that 1.00 mol of a gas has a volume of 22.4 L at STP (standard temperature and pressure). This gives us enough information to find R for each unit of pressure.

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The Value of R Where P is in atm: STP is 1.00 atm at 273 K pV = nRT R = PV/nT R = [(1.00 atm)(22.4 L)]/[(1.00 mol)(273 K)] R = 0.0821 L-atm/mol-K

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The Value of R Where P is in mm Hg: STP is 760 mm Hg at 273 K pV = nRT R = PV/nT R = [(760 mm Hg)(22.4 L)]/[(1.00 mol)(273 K)] R = 62.4 L-mm Hg/mol-K

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The Value of R Where P is in kPa: STP is 101.3 kPa at 273 K pV = nRT R = PV/nT R = [(101.3 kPa)(22.4 L)]/[(1.00 mol)(273 K)] R = 8.314 L-mm Hg/mol-K

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The Value of R PV = nRT If P is measured in atm : R = 0.0821 atm -L/mol-K If P is measured in kPa : R = 8.314 kPa -L/mol-K If P is measured in mm Hg : R = 62.4 mm Hg -L/mol-K

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The Value of R PV = nRT Remember: The value of R is dependent on the units of pressure. Always use the correct value of R. All appropriate values for R will be given to you for any test or quiz.

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Example 1 – Finding P What is the pressure, in atm, of 0.125 mols of helium in a 4.00 L container at a temperature of 430 K? P = ? atm V = 4.00 L n = 0.125 mol R = 0.0821 L-atm/mol-K T = 430 K PV = nRT P = nRT V = (0.125)(0.0821)(430) (4.00) atm P = 1.10321875 atm = 1.10 atm This is the value we use when pressure is expressed in atm.

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Practice Problems – Finding P 1.What is the pressure, in atm, of 3.50 mols of N 2 gas held in 45.0 L at a temperature of 310 K? 2.What is the pressure, in mm Hg, of 0.0400 mols of CO 2 gas held in 10.0 L at a temperature of 350 K? 3.What is the pressure, in kPa, of 172 mols of He gas held in 675 L at a temperature of 273 K? 4.What is the pressure, in atm, of 0.00250 mols of Cl 2 gas held in 0.100 L at a temperature of 455 K? P = 0.198 atm P = 87.4 mm Hg P = 578 kPa P = 0.934 atm

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Example 2 – Finding V What is the volume of 2.50 mols of oxygen at a pressure of 85.0 kPa and a temperature of 315 K? P = 85.0 kPa V = ? L n = 2.50 mol R = 8.31 L-kPa/mol-K T = 315 K PV = nRT V = nRT P = (2.50)(8.31)(315) (85.0) L P = 76.98970588 L = 77.0 L This is the value we use when pressure is expressed in kPa.

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Practice Problems – Finding V 1.What is the volume of 1.00 mols of F 2 gas at a pressure of 0.450 atm and a temperature of 298 K? 2.What is the volume of 40.2 mols of UF 6 gas at a pressure of 645 mm Hg and a temperature of 655 K? 3.What is the volume of 0.0424 mols of Ne gas at a pressure of 4.53 kPa and a temperature of 242 K? 4.What is the volume of 3.22 mols of O 2 gas at a pressure of 4.67 atm a temperature of 273 K? V = 54.4 L V = 2,550 L V = 18.8 L V = 15.5 L

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Example 3 – Finding n How many mols of nitrogen is contained in a volume of 22.4 L at a pressure of 760 mm Hg and a temperature of 273 K? P = 760 mm Hg V = 22.4 L n = ? mol R = 62.4 L-mm Hg/mol-K T = 273 K PV = nRT n = PV RT = (760)(22.4) (62.4)(273) mol P = 0.999342538 mol = 0.999 mol This is the value we use when pressure is expressed in mm Hg.

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Practice Problems – Finding n 1.How many mols of CO 2 gas is in 75.0 L at a pressure of 2.75 atm and a temperature of 298 K? 2.How many mols of SF 6 gas is in 0.500 L at a pressure of 950 mm Hg and a temperature of 350 K? 3.How many mols of Ar gas is in 22.4 L at a pressure of 95.9 kPa and a temperature of 298 K? 4.How many mols of CH 4 gas is in 0.0782 L at a pressure of 32.5 atm a temperature of 653 K? n = 8.43 mol n = 0.0217 mol n = 0.867 mol n = 0.0474 mol

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Example 4 – Finding T What is the temperature of 1.60 mols of neon contained in a volume of 15.0 L at a pressure of 1.20 atm? P = 1.20 atm V = 15.0 L n = 1.60 mol R = 0.0821 L-atm/mol-K T = ? K PV = nRT T = PV nR = (1.20)(15.0) (1.60)(0.0821) K P = 137.0280146 K = 137 K This is the value we use when pressure is expressed in atm.

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Practice Problems – Finding T 1.What is the temperature of 3.00 mol of CO 2 gas in 75.0 L and at a pressure of 1.00 atm? 2.What is the temperature of 0.755 mol of He gas in 4.25 L and at a pressure of 2,320 mm Hg? 3.What is the temperature of 35.0 mol of CH 4 gas in 33.5 L and at a pressure of 2,520 kPa? 4.What is the temperature of 1.25 mol of CO 2 gas in 25.0 L and at a pressure of 1.70 atm? T = 305 K T = 209 K T = 290 K T = 414 K

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Application The ability to measure the amount of the gas from pressure, volume, and temperature measurements is a powerful tool for exploring other properties of gases. If we can also measure the mass of the gas, we can determine the molar mass of the gas the density of the gas

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The Molar Mass of a Gas We can determine the number of mols, n, of a gas by using pressure, volume, and temperature measurements and the ideal gas law. n = PV/RT The molar mass, M, is the mass, m, divided by the number of mols. M =m/n Putting the two equations together M = m/(PV/RT) = mRT/PV

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Example 5 – Finding M At 301 K and 0.974 atm, 1.00 L of a gas has a mass of 5.16 g. What is the molar mass of this gas? P = 0.974 atm V = 1.00 L R = 0.0821 L-atm/mol-K T = 301 K m = 5.16 g M = mRT PV = (5.16)(0.0821)(301) (0.974)(1.00) g/mol M = 130.9183121 g/mol = 131 g/mol This is the value we use when pressure is expressed in atm.

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Practice Problems – Finding M 1.At 302 K and 1.05 atm, 1.81 L of a gas has a mass of 5.42 g. What is the molar mass of this gas? 2.At 260 K and 695 mm Hg, 5.41 L of a gas has a mass of 10.2 g. What is the molar mass of this gas? 3.At 285 K and 97.2 kPa, 95.6 L of a gas has a mass of 329 g. What is the molar mass of this gas? 4.At 310 K and 4.15 atm, 0.350 L of a gas has a mass of 3.32 g. What is the molar mass of this gas? M = 70.0 g/mol M = 44.0 g/mol M = 83.9 g/mol M = 58.2 g/mol

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The Density of a Gas The density of a gas is the mass of the gas divided by its volume ρ = m/V If we measure the density of a gas and know its pressure and temperature, we can find the molar mass, M. M =M =M =M = mRT PV m RT V P = m RT V P = ρ RT P = We are just rearranging V and P in the denominator. We separate out the m/V term. We replace m/V with ρ.

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Example 6 – Finding M What is the molar mass of a gas with a density of 3.42 g/L at 293 K and 0.980 atm? ρ = 3.42 g/L R = 0.0821 L-atm/mol-K T = 293 K P = 0.980 atm M = ρ RT P = (3.42)(0.0821)(293) (0.980) g/mol M = 83.94808776 g/mol = 83.9 g/mol This is the value we use when pressure is expressed in atm.

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Practice Problems – Finding M 1.What is the molar mass of a gas with a density of 3.25 g/L at 150 K and 10.0 atm? 2.What is the molar mass of a gas with a density of 5.76 g/L at 273 K and 672 mm Hg? 3.What is the molar mass of a gas with a density of 5.17 g/L at 452 K and 672 kPa? 4.What is the molar mass of a gas with a density of 0.508 g/L at 345 K and 0.450 atm? M = 4.00 g/mol M = 146 g/mol M = 352 g/mol M = 32.0 g/mol

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The Density of a Gas If we know the molar mass of a gas, then we can calculate the density of that gas under specific conditions of pressure and temperature. M =M =M =M = ρ =ρ =ρ =ρ = M P RT ρ RT P If we rearrange and solve for ρ … We used this equation to solve for molar mass when we knew density.

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Example 7 – Finding ρ What is the density of a sample of ammonia gas, NH 3, M = 17.04 g/mol, at 0.928 atm and 336 K? M = 17.04 g/mol P = 0.928 atm R = 0.0821 L-atm/mol-K T = 336 K ρ = M P RT = (17.04)(0.928) (0.0821)(336) g/L ρ = 0.5732382112 g/L = 0.573 g/L This is the value we use when pressure is expressed in atm.

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Practice Problems – Finding Practice Problems – Finding 1.What is the density of a sample of CH 4, M = 16.05 g/mol, at 1.25 atm and 280 K? 2.What is the density of a sample of H 2, M = 2.02 g/mol, at 672 mm Hg and 261 K? 3.What is the density of a sample of CO 2, M = 44.01 g/mol, at 175 kPa and 310 K? 4.What is the density of a sample of UF 6, M = 352.0 g/mol, at 10.2 atm and 397 K? = 1.63 g/L = 0.0833 g/L = 3.00 g/L = 110. g/L

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Summary The ideal gas law allows us to calculate the pressures, volumes, amounts, and temperatures of gases. PV = nRT The value of the ideal gas constant, R, depends on the units of pressure. For units of atm, R = 0.0821 L-atm/mol-K For units of kPa, R = 8.314 L-kPa/mol-K For units of mm Hg, R = 62.4 L-mm Hg/mol-K We can also use the ideal gas law to calculate the molar masses and densities of gases.

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The Ideal Gas Law PV = nRT.

The Ideal Gas Law PV = nRT.

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