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2. 2 Real Gases An ideal gas adheres to the Kinetic Theory exactly in all situations. Real gases deviate from ideal behavior at high pressures and low temperatures. –When the pressure is high, it becomes more difficult to compress a gas because the particles actually have a volume of their own. –When the temperature is low, gas particles slow down and attractions between them become significant as they clump together and form liquids.

3. 3 We must define some terms: n = moles of gas particles V = volume (of the container) T = temperature (must be in Kelvin) P = pressure (You will see these variables in a variety of gas laws)

4. 4 Kelvin is the only temperature scale that measures absolute speed of particles. K = o C + 273 All temperatures in gas problems must be in Kelvin. Temperature

5. 5 1 atm = 101.325 kPa = 760 torr = 760 mmHg 1 atm is the normal atmospheric pressure at sea level. Pressure changes with altitude. Air pressure is measured with a barometer. Pressure

6. 6 Standard Temperature and Pressure (STP) Standard Temperature = 0 o C = 273 K Standard Pressure = 1 atm or equivalent

7. 7 Pressure, Volume, & Temperature Boyle’s Law –Pressure and volume are inversely proportional if the temperature remains constant P 1 V 1 = P 2 V 2 Robert Boyle

8. 8 Charles’ Law –Volume and temperature are directly proportional if pressure remains constant Temperature must be in Kelvin Jacques Charles V 1 = V 2 T 1 T 2

9. 9 Gay-Lussac’s Law –Pressure and temperature are directly proportional if volume remains constant Temperature must be in Kelvin Joseph Louis Gay-Lussac P 1 = P 2 T 1 T 2

10. 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 = volume of the gas V = volume of the gas n = number of moles of gas n = number of moles of gas n 1 /V 1 = n 2 /V 2

11. Avogadro’s Principle At STP, 1 mole of gas is equal to 22.4 L At STP, 1 mole of gas is equal to 22.4 L 11

12. 12 Combined Gas Law Temperature must be in Kelvin Cross out any constants P 1 V 1 = P 2 V 2 T 1 T 2

13. 13 Sample Problem At conditions of 785 torr of pressure and 15.0 o C temperature, a gas occupies a volume of 45.5 mL. What will be the volume of the same gas at 745 torr and 30.0 o C? T 1 = 15.0 o C + 273 = 288 K T 2 = 30.0 o C + 273 = 303 K

14. 14 P 1 V 1 = P 2 V 2 T 1 T 2 T 1 T 2 P 1 V 1 T 2 = P 2 V 2 T 1 P 1 = V 1 = T 2 = P 2 = V 2 = T 1 = 785 torr 45.5 mL 288 K 745 torr ? 303 K P 1 V 1 T 2 = V 2 P 2 T 1 P 2 T 1 (785 torr)(45.5 mL)(288 K) (745 torr)(303 K) V 2 = 50.4 mL

15. On a cold morning (10.0 o C) a group of hot-air balloonists start filling their balloon with air. After the balloon is three-fourths filled, they turn on the propane burner to heat the air. At what Celsius temperature will the air completely fill the envelope to its maximum capacity of 1700. m 3 ? P 1 V 1 = P 2 V 2 T 1 T 2 T 1 T 2 Pressure is constant V 1 = ¾ x 1700 m 3 V 2 = 1700. m 3 T 1 = 283.15 K T 2 = ? T 2 = V 2 T 1 V 1 V 1 V 1 T 2 = V 2 T 1 T 2 = (1700. m 3 )(283 K) 1275 m 3 1275 m 3 T 2 = 377 K T 2 = 377 K – 273 = 104 o C

16. 16 Ideal Gas Law R is the universal gas constant. An “R” value is picked based upon the unit being used to measure pressure. P V = n R T (R = 0.08206 atm L/mol K) (R = 8.314 kPa L/mol K) (R = 62.4 mmHg L/mol K)

17. 17 How many moles of a gas at 100. o C does it take to fill a 1.00-L flask to a pressure of 1.50 atm? n = 0.0490 mol PV = nRT P = V = n = R = T = 1.50 atm 1.00L? 0.0821 Latm/molK 373 K n = PV RT RT n = (1.50 atm)(1.00 L) (0.0821 Latm/molK)(373 K)

18. 18 What is the volume occupied by 9.45 g of C 2 H 2 at STP? 9.45 g C 2 H 2 x 1 mol C 2 H 2 = 0.3629 mol C 2 H 2 26.036 g C 2 H 2 26.036 g C 2 H 2 PV = nRT P = V = n = R = T = 1.00 atm ? 0.03629 mol 0.0821 Latm/molK 273 K V = nRT P P V = (0.363 mol)(0.0821 Latm/molK)(273 K) 1.00 atm 1.00 atm V = 8.14 L

19. 19 Gas Stoichiometry Only gas volumes at STP (Avogadro’s Principle 1 mol = 22.4 L) can be entered into a stoichiometry equation If gas is at a different temperature & pressure, use PV=nRT to convert liters to moles and then continue with stoichiometry

20. 20 3 H 2 + N 2  2 NH 3 3 H 2 + N 2  2 NH 3 A chemist performs this reaction (Haber process) in a chamber at 327 o C under a pressure of 900. mm Hg. How many grams of ammonia would be produced from 166.3 L of hydrogen at the above conditions? PV = nRT n = PV RT RT n = (900 mmHg)(166.3 L) (62.4 L mmHg/mol K)(600 K) (62.4 L mmHg/mol K)(600 K) n = 4.00 mol H 2 4.00 mol H 2 x 2 mol NH 3 x 17g NH 3 = 45.3 g NH 3 3 mol H 2 1 mol NH 3