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The Ideal Gas Law

The Ideal Gas Law. Bringing It All Together. 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

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The Ideal Gas Law

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  1. The Ideal Gas Law Bringing It All Together

  2. 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

  3. Introduction • When we use the combined gas law • we allow P, V, and Tto 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

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

  5. 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.

  6. 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 upto keep V & T constant • Volume goes upto keep P & T constant • Temperature goes downto keep P & V constant

  7. Application • This means that the amount (number of mols) of material varies - • directlywith pressure and volume • inverselywith 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.

  8. 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 constantwhose value depends on the units of P, V, n, and T

  9. 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 Rand entering the appropriate values. • We remember from our studies of stoichiometry that 1.00 molof 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.

  10. 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

  11. 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

  12. 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

  13. 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

  14. 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.

  15. 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 This is the value we use when pressure is expressed in atm. V = 4.00 L n = 0.125 mol R = 0.0821 L-atm/mol-K T = 430 K nRT (0.125)(0.0821)(430) PV = nRT ⇒ P = = atm (4.00) V P = 1.10321875 atm = 1.10 atm

  16. Practice Problems – Finding P • What is the pressure, in atm, of 3.50 mols of N2 gas held in 45.0 L at a temperature of 310 K? • What is the pressure, in mm Hg, of 0.0400 mols of CO2gas held in 10.0 L at a temperature of 350 K? • What is the pressure, in kPa, of 172 mols of He gas held in 675 L at a temperature of 273 K? • What is the pressure, in atm, of 0.00250 mols of Cl2gas 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

  17. 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 This is the value we use when pressure is expressed in kPa. V = ? L n = 2.50 mol R = 8.31 L-kPa/mol-K T = 315 K nRT (2.50)(8.31)(315) PV = nRT ⇒ V = = L (85.0) P P = 76.98970588 L = 77.0 L

  18. Practice Problems – Finding V • What is the volume of 1.00 mols of F2 gas at a pressure of 0.450 atm and a temperature of 298 K? • What is the volume of 40.2 molsof UF6 gas at a pressure of 645 mm Hg and a temperature of 655 K? • What is the volume of 0.0424 molsof Ne gas at a pressure of 4.53 kPa and a temperature of 242 K? • What is the volume of 3.22 mols of O2gas 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

  19. 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 This is the value we use when pressure is expressed in mm Hg. V = 22.4 L n = ? mol R = 62.4 L-mm Hg/mol-K T = 273 K PV (760)(22.4) PV = nRT ⇒ n = = mol (62.4)(273) RT P = 0.999342538 mol = 0.999 mol

  20. Practice Problems – Finding n • How many mols of CO2 gas is in 75.0 L at a pressure of 2.75 atm and a temperature of 298 K? • How many molsof SF6 gas is in 0.500 L at a pressure of 950 mm Hg and a temperature of 350 K? • How many molsof Ar gas is in 22.4 L at a pressure of 95.9 kPa and a temperature of 298 K? • How many molsof CH4 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

  21. 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 This is the value we use when pressure is expressed in atm. V = 15.0 L n = 1.60 mol R = 0.0821 L-atm/mol-K T = ? K PV (1.20)(15.0) K PV = nRT ⇒ T = = (1.60)(0.0821) nR P = 137.0280146 K = 137 K

  22. Practice Problems – Finding T • What is the temperature of 3.00 mol of CO2 gas in 75.0 L and at a pressure of 1.00 atm? • What is the temperature of 0.755 mol of He gas in 4.25 L and at a pressure of 2,320 mm Hg? • What is the temperature of 35.0 mol of CH4gas in 33.5 L and at a pressure of 2,520 kPa? • What is the temperature of 1.25 mol of CO2 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

  23. 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

  24. 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

  25. 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 This is the value we use when pressure is expressed in atm. V = 1.00 L R = 0.0821 L-atm/mol-K T = 301 K m = 5.16 g mRT (5.16)(0.0821)(301) g/mol M = = (0.974)(1.00) PV M = 130.9183121 g/mol = 131 g/mol

  26. Practice Problems – Finding M • 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? • 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? • 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? • 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

  27. 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 = mRT m RT ρRT mRT = = = PV V P P V P We replace m/V with ρ. We separate out the m/V term. We are just rearranging V and P in the denominator.

  28. 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 This is the value we use when pressure is expressed in atm. R = 0.0821 L-atm/mol-K T = 293 K P = 0.980 atm ρRT (3.42)(0.0821)(293) g/mol M = = (0.980) P M = 83.94808776 g/mol = 83.9 g/mol

  29. Practice Problems – Finding M • What is the molar mass of a gas with a density of 3.25 g/L at 150 K and 10.0 atm? • What is the molar mass of a gas with a density of 5.76 g/L at 273 K and 672 mm Hg? • What is the molar mass of a gas with a density of 5.17 g/L at 452 K and 672 kPa? • 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

  30. 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 = • ρ = ρRT If we rearrange and solve for ρ… We used this equation to solve for molar mass when we knew density. P M P RT

  31. Example 7 – Finding ρ What is the density of a sample of ammonia gas, NH3, M = 17.04 g/mol, at 0.928 atm and 336 K? M = 17.04 g/mol This is the value we use when pressure is expressed in atm. P = 0.928 atm R = 0.0821 L-atm/mol-K T = 336 K M P (17.04)(0.928) g/L ρ = = (0.0821)(336) RT ρ = 0.5732382112 g/L = 0.573 g/L

  32. Practice Problems – Finding r • What is the density of a sample of CH4, M = 16.05 g/mol, at 1.25 atm and 280 K? • What is the density of a sample of H2, M = 2.02 g/mol, at 672 mm Hg and 261 K? • What is the density of a sample of CO2, M = 44.01 g/mol, at 175 kPaand 310 K? • What is the density of a sample of UF6, M = 352.0 g/mol, at 10.2 atm and 397 K? r = 1.63 g/L r = 0.0833 g/L r = 3.00 g/L r = 110. g/L

  33. 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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