1. Avogadro’s Hypothesis Recall: 1.00 mol of any gas @ SATP occupies a volume of 24.8 L; or 2.00 mol of any gas @ SATP occupies 49.6 L; or 0.500 mol of any gas @ SATP occupies 12.4 L.

2. How will a graph of V vs n look @SATP? Linear, with slope = 24.8 L/mol. Therefore: V α n, at fixed T, P; V 1 = V 2 (P, T unchanged), or n 1 n 2 V = constant (P, T unchanged) n

3. Putting Avogadro’s Hypothesis into the mix: gas laweq’nunchanged Boyle’sP 1 V 1 = P 2 V 2 n, T Charles’V 1 = V 2 n, P T 1 T 2 Gay-Lussac’sP 1 = P 2 n, V T 1 T 2 Avogadro’sV 1 = V 2 P, T Hypothesisn 1 n 2

4. P 1 V 1 = P 2 V 2 = P 3 V 3 = constant value n 1 T 1 n 2 T 2 n 3 T 3 PV = R  ideal gas constant Tn What is the value of R? Use the molar volume of any gas @ SATP to calculate R.

5. P = 101.3 kPa T = (25 + 273) = 298 K V = 24.8 L n = 1.00 mol R = PV/(nT) = 8.31 kPa*L*mol -1 *K -1 What if P = 760 mmHg? Value of R will be... R = 62.4 mmHg*L*mol -1 *K -1 What if P = 1.00 atm? Value of R will be... R = 0.0821 atm*L*mol -1 *K -1

6. Ideal Gas Law Recall PV = R cross-multiplying gives Tn PV = nRT  IDEAL GAS LAW Fine print: Valid for any ideal gas, BUT NOT at high P or low T—because gases will liquefy.

7. sample problem #1 What volume is occupied by 25.2 g of Xenon contained at a pressure of 862 kPa at 30 o C? PV = nRT n = 25.2 g/131.3 g*mol -1 = 0.192 mol P = 862 kPa T = 30 + 273 = 303 K R = 8.314 L*kPa*mol -1 K -1 V = nRT/P = 0.561 L

8. sample problem #2 What is the pressure (in atm) of propane, C 3 H 8, at 20 o C, contained in a 1.0 L tank, if the tank holds 465 g of C 3 H 8 ? (101.3 kPa = 1.00 atm) n = 465 g/(44 g*mol -1 ) = 10.6 mol T = (20 + 273) = 293 K R = 0.0821 L*atm*mol -1 *K -1 V = 1.0 L P = nRT/V = 255 atm

9. Holy $%^%!! We calculated the P insideto be 255 atm! Can this be right?

10. Shake the cylinder. What do you hear? Liquid propane sloshing around. What does this suggest about the actual P inside? Less than calculated—C 3 H 8 not all in gas phase.

11. Aside: A propane explosion looks like this... http://www.youtube.com/watch?v=__1Ym_F94 CE

12. sample problem #3 (gas stoichiometry) What volume (L) of hydrogen gas, can be expected from the reaction of 12.3 g of Mg with 450 mL of 0.976 mol/L HCl at 25 o C at a pressure of 98.4 kPa? Where do we start? Mg(s) + 2 HCl(aq)  MgCl 2 (aq) + H 2 (g) Now what? Find limiting reagent.

13. Mg(s)+2 HCl(aq) 12.3 g0.450 L ↓/24.3 g*mol -1 * 0.976 mol/L = 0.506 mol= 0.439 mol ↑limiting reagent

14. 2 HCl(aq)  H 2 (g) 0.439 mol  *1/2  0.220 mol expected Use PV = nRT, or V = nRT/P to get the volume of H 2 = [ (0.220 mol)*(8.314 L*kPa*mol -1 K -1 )*(298 K)] 98.4 kPa = 5.54 L of H 2 expected

15. HW PP on p 556—do a few (see previous worked examples if you need help) PP on p 560 #31- 38