1. Chapter 11 Molecular Composition of Gases

2. Avogadro’s Law Equal Volumes of Gases at the Same Temperature & Pressure contain the Same Number of “Particles.”

3. Balloons Holding 1.0 L of Gas at 25º C and 1 atm. Each balloon contains 0.041 mole of gas or 2.5 x 10 22 molecules.

4. Avogadro’s Law For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures). V = an a = proportionality constant V = volume of the gas n = number of moles of gas

5. What are the four quantities needed to describe a gas? Pressure Volume Temperature Moles of gas In Chapter 10, which three of these quantities did we vary? Pressure Volume Temperature Now we will consider the moles of gas.

6. Moles of Gas The number of moles of gas will always affect at least one of the other three quantities.

7. Ideal Gas Law 4 Mathematical relationship among 4 pressure, 4 volume, 4 temperature and 4 the number of moles.

8. Ideal Gas Law 4 Mathematical equation coming from the combination of coming from the combination of - Boyle’s Law - Charles’ Law & - Avogadro’s Law

9. Ideal Gas Law 4 An equation of state for a gas. 4 “state” is the condition of the gas at a given time. PV = nRT An Ideal Gas is a hypothetical substance. Ideal Gas Law is an empirical equation.

10. Ideal Gas Law PV = nRT R = proportionality constant R = proportionality constant = ideal gas constant = 0.0821 L atm   mol  = 0.0821 L atm   mol  See Table 11-1 on page 342 for additional values for R.

11. Ideal Gas Law PV = nRT Because of the units of R, P = pressure in atm P = pressure in atm V = volume in liters V = volume in liters n = moles n = moles T = temperature in Kelvins T = temperature in Kelvins Holds closely at P < 1 atm Holds closely at P < 1 atm

12. Standard Temperature and Pressure “STP” P = 1 atmosphere T = C

13. Ideal Gas Law PV = nRT If gas is at STP, R = 0.0821 L atm   mol  R = 0.0821 L atm   mol  P = 1.00 atm P = 1.00 atm T = 273 Kelvin T = 273 Kelvin Then Then V /n = volume in liters/ mole V /n = volume in liters/ mole = RT/ P = RT/ P = 0.0821 L atm x 273 K = 0.0821 L atm x 273 K mole K mole K 1.00 atm 1.00 atm = 22.4 L/ mole = 22.4 L/ mole

14. Standard Molar Volume of a Gas The volume occupied by one mole of a gas at STP. 22.4 L/ mole

15. A Mole of Any Gas Occupies a Volume of Approximately 22.42 L at STP

16. Problem What is the volume of 77.0 g of nitrogen dioxide at STP?

17. Solution 77.0 g NO 2 x 1 mole NO 2 x 22.4 L 46.01 g NO 2 1 mole 46.01 g NO 2 1 mole = 37. 5 L NO 2

18. Problem What is the mass of 1.33 x 10 4 mL of oxygen gas at STP?

19. Solution 1.33 x 10 4 mL x 1 L x 1 mol 1000 mL 22.4 L 1000 mL 22.4 L x 32.00 g O 2 x 32.00 g O 2 1 mole O 2 1 mole O 2 = 19.0 g O 2

20. Problem At STP, 3 L of chlorine is produced during a chemical reaction. What is the mass of this gas?

21. Solution 3 L x 1 mole x 70.90 g Cl 2 3 L x 1 mole x 70.90 g Cl 2 22.4 L 1 mole Cl 2 22.4 L 1 mole Cl 2 = 9 g Cl 2

22. Problem What pressure, in atmospheres, is exerted by 0.325 mole of hydrogen gas in a 4.08 L container at 35  C? Not at STP!!! Use Ideal Gas Equation!!!

23. Solution: P V = n R T P = nRT/V T (K) = 35 + 273 = 308 K P = 0.325 mole x 0.0821 L atm x 308 K mole K mole K 4.08 L 4.08 L P = 2.01 atm P = 2.01 atm

24. Problem A sample that contains 4.38 mole of a gas at 250 K has a pressure of 0.857 atm. What is the volume?

25. Solution: PV = nRT V = nRT/P V = 4.38 mole x 0.0821 L atm x 250 K mole K mole K 0.857 atm 0.857 atm V = 105 L V = 105 L

26. Molar Mass of a Gas n = # of moles = grams of gas= (m) molar mass molar mass molar mass molar mass

27. P V = n R T P V = (m/molar mass) R T Therefore, molar mass = m R T P V

28. Problem At 28  C and 0.974 atm, 1.00 L of gas has a mass of 5.16 g. What is the molar mass of this gas?

29. Solution molar mass = m R T P V Molar Mass = 5.16 g x 0.0821 L atm x 301 K mole K 0.974 atm x 1.00 L Molar Mass = 130.92 g/mole = 131 g/mole

30. Also, since density (d) = m V Then, since molar mass = m R T P V molar mass = d R T P

31. Density = (molar mass) x P R T Temperature in K!! Gas density usually measured in g/L

32. Problem What is the density of a sample of ammonia gas, NH 3, if the pressure is 0.928 atm and the temperature is 63.0  C? What is the density of a sample of ammonia gas, NH 3, if the pressure is 0.928 atm and the temperature is 63.0  C?

33. Solution Density = (molar mass) x P R T Density = 17.04 g/mole x 0.928 atm 0.0821 L atm x 336 K mole K Density = 0.573 g/ L NH 3

34. Problem The density of dry air at sea level (1 atm) is 1.225 g/L at 15  C. (1 atm) is 1.225 g/L at 15  C. What is the average molar mass of the air?

35. Solution Density = (molar mass) x P R T Therefore, Molar Mass = Density x R x T P

36. Solution Molar Mass = Density x R x T P Molar Mass = 1.225 g x x x x 0.0821 L atm x 288 K L mole K ___ 1 atm Molar Mass = 28.96 g/mole = 29.0 g/mole

37. Homework Read pages 338-339. Complete Worksheet.