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Chapter 8 Gases

Chapter 8 Gases. 8.8 Ideal Gas Law PV = nRT. Ideal Gas Law. The relationship between the four properties (P, V, n, and T) of gases can be written equal to a constant R. PV = R nT Rearranging this expression gives the expression called the ideal gas law. PV = nRT.

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Chapter 8 Gases

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  1. Chapter 8 Gases 8.8 Ideal Gas Law PV = nRT

  2. Ideal Gas Law • The relationship between the four properties (P, V, n, and T) of gases can be written equal to a constant R. PV = R nT • Rearranging this expression gives the expression called the ideal gas law. PV = nRT

  3. Universal Gas Constant, R • The universal gas constant, R, can be calculated using the molar volume of a gas at STP. • At STP (273 K and 1.00 atm), 1 mole of a gas occupies 22.4 L. P V R = PV = (1.00 atm)(22.4 L) nT (1 mole) (273K) n T = 0.0821 L atm mole K • Note there are four units associated with R.

  4. Learning Check Another value for the universal gas constant is obtained using mm Hg for the STP pressure. What is the value of R when a pressure of 760 mm Hg is placed in the R value expression?

  5. Solution What is the value of R when the STP value for P is 760 mmHg? R = PV = (760 mm Hg) (22.4 L) nT (1 mole) (273K) = 62.4 L mm Hg mole K

  6. Learning Check Dinitrogen oxide (N2O), laughing gas, is used by dentists as an anesthetic. If a 20.0 L tank of laughing gas contains 2.8 moles N2O at 23°C, what is the pressure (mm Hg) in the tank?

  7. Solution 1. Adjust the units of the given properties to match the units of R. V = 20.0 L, T = 296 K, n = 2.8 moles, P = ? 2. Rearrange the ideal gas law for P. P = nRT V P = (2.8 moles)(62.4 L mm Hg)(296 K) (20.0 L) (mole K) = 2.6 x 103 mm Hg

  8. Learning Check A cylinder contains 5.0 L of O2 at 20.0°C and 0.85 atm. How many grams of oxygen are in the cylinder?

  9. Solution 1. Determine the given properties. P = 0.85 atm, V = 5.0 L, T = 293 K, n (or g =?) 2. Rearrange the ideal gas law for n (moles). n = PV RT = (0.85 atm)(5.0 L)(mole K) = 0.18 mole O2 (0.0821atm L)(293 K) 3. Convert moles to grams using molar mass. = 0. 18 mole O2 x 32.0 g O2 = 5.8 g O2 1 mole O2

  10. Molar Mass of a Gas What is the molar mass of a gas if 0.250 g of the gas occupy 215 mL at 0.813 atm and 30.0°C? 1. Solve for the moles (n) of gas. n = PV = (0.813 atm) (0.215 L) RT (0.0821 L atm/mole K)(303K) = 0.00703 mole 2. Set up the molar mass relationship. Molar mass = g = 0.250 g = 35.6 g/mole mole 0.00703 mole

  11. Gases in Equations • The amounts of gases reacted or produced in a chemical reaction can be calculated using the ideal gas law and mole factors. Problem: What volume (L) of Cl2 gas at 1.2 atm and 27°C is needed to completely react with 1.5 g of aluminum? 2Al(s) + 3Cl2 (g) 2AlCl3(s)

  12. Gases in Equations (continued) 2Al(s) + 3Cl2 (g) 2AlCl3(s) 1.5 g ? L 1.2 atm, 300K 1. Calculate the moles of Cl2 needed. 1.5 g Al x 1 mole Al x 3 moles Cl2 = 0.083 mole Cl2 27.0 g Al 2 moles Al 2. Place the moles Cl2 in the ideal gas equation. V = nRT = (0.083 mole Cl2)(0.0821 Latm/moleK)(300K) P 1.2 atm = 1.7 L Cl2

  13. Learning Check What volume (L) of O2 at24°C and 0.950 atm are needed to react with 28.0 g NH3? 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

  14. Solution 1. Calculate the moles of O2 needed. 28.0 g NH3 x 1 mole NH3 x 5 mole O2 17.0 g NH3 4 mole NH3 = 2.06 mole O2 2. Place the moles O2 in the ideal gas equation. V = nRT = (2.06 moles)(0.0821 L atm/moleK)(297K)P 0.950 atm = 52.9 L O2

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