1. ? Gases Chapter 4

2. ? Kinetic Molecular Theory Particles in an Ideal Gases…  have no volume.  have elastic collisions.  are in constant, random, straight-line motion.  don’t attract or repel each other.  have an avg. KE directly related to Kelvin temperature.

3. ? Real Gases Particles in a REAL GAS …  have their own volume  attract each other Gas behavior is most ideal… at low pressures at high temperatures in nonpolar atoms/molecules

4. ? Characteristics of Gases - Expand to fill any container. Random motion, No attraction - Are fluids (like liquids). No attraction - Have very low densities. No volume = lots of empty space - Can be compressed. - Undergo diffusion & effusion. Random motion

5. ? A. Temperature ºF ºC K -45932212 -2730100 0273373 K = ºC + 273 Always use absolute temperature (Kelvin) when working with gases.

6. ? B. Pressure Which shoes create the most pressure?

7. ? B. Pressure (Cont.) Barometer measures atmospheric pressure Mercury Barometer Aneroid Barometer

8. ? B. Pressure (Cont.) Manometer measures contained gas pressure U-tube ManometerBourdon-tube gauge

9. ? B. Pressure (Cont.) KEY UNITS AT SEA LEVEL 101.325 kPa (kilopascal) 1 atm 760 mm Hg 760 torr 14.7 psi

10. ? F. STP Standard Temperature & Pressure 0°C 273 K 1 atm101.325 kPa -OR- STP

11. ? The Gas Laws A. Boyle’s Law B. Charle’s Law C. Combined Gas Law D. Avogadro’s principle E. Ideal Gas Law F. Dalton’s Law

12. ? A. Boyle’s Law P V Pα 1/V or PV = k The pressure and volume of a gas are inversely related at constant mass & temperature

13. ? V T B. Charles’ Law Vα T or The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure

14. ? V α T C. Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2 P 1 V 1 T 2 = P 2 V 2 T 1 1 P V α &

15. ? GIVEN: V 1 = 473 cm 3 T 1 = 36°C = 309K V 2 = ? T 2 = 94°C = 367K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 Gas Law Problems A gas occupies 473 cm 3 at 36°C. Find its volume at 94°C. CHARLES’ LAW TT VV (473 cm 3 )(367 K)=V 2 (309 K) V 2 = 562 cm 3

16. ? GIVEN: V 1 = 100. mL P 1 = 150. kPa V 2 = ? P 2 = 200. kPa WORK: P 1 V 1 T 2 = P 2 V 2 T 1 Gas Law Problems A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW PP VV (150.kPa)(100.mL)=(200.kPa)V 2 V 2 = 75.0 mL

17. ? GIVEN: V 1 = 7.84 cm 3 P 1 = 71.8 kPa T 1 = 25°C = 298 K V2 = ?V2 = ? P 2 = 101.325 kPa T 2 = 273 K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 (71.8 kPa)(7.84 cm 3 )(273 K) =(101.325 kPa) V 2 (298 K) V 2 = 5.09 cm 3 Gas Law Problems Gas Law Problems A gas occupies 7.84 cm 3 at 71.8 kPa & 25°C. Find its volume at STP. P  T  VV COMBINED GAS LAW

18. ? V n D. Avogadro’s Principle Equal volumes of gases contain equal numbers of moles at constant temp & pressure true for any gas. Vα n or

19. ? PV T PV nT E. Ideal Gas Law = k UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K R=8.315 dm 3  kPa/mol  K = R

20. ? E. Ideal Gas Law UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K R=8.315 dm 3  kPa/mol  K PV=nRT

21. ? GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821 L  atm/mol  K WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol L  atm/mol  K K P = 3.01 atm E. Ideal Gas Law Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L. IDEAL GAS LAW

22. ? GIVEN: V = ?V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa R = 8.315 dm 3  kPa/mol  K E. Ideal Gas Law Find the volume of 85 g of O 2 at 25°C and 104.5 kPa. = 2.7 mol WORK: 85 g 1 mol = 2.7 mol 32.00 g PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm 3  kPa/mol  K K V = 64 dm 3 IDEAL GAS LAW

23. ? IV. Gas Stoichiometry at Non-STP Conditions

24. ? Gas Stoichiometry Moles  Liters of a Gas STP - use 22.4 L/mol Non-STP - use ideal gas law Non- STP Problems Given liters of gas? i.start with ideal gas law Looking for liters of gas? i.start with stoichiometry conv.

25. ? F. Dalton’s Law The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases. P total = P 1 + P 2 +... P atm = P H2 + P H2O

26. ? GIVEN: P H2 = ? P total = 94.4 kPa P H2O = 2.72 kPa WORK: P total = P H2 + P H2O 94.4 kPa = P H2 + 2.72 kPa P H2 = 91.7 kPa F. Dalton’s Law Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa. Look up water-vapor pressure on p.899 for 22.5°C. Sig Figs: Round to least number of decimal places. The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H 2 and water vapor.

27. ? GIVEN: P gas = ? P total = 742.0 torr P H2O = 42.2 torr WORK: P total = P gas + P H2O 742.0 torr = P H2 + 42.2 torr P gas = 699.8 torr A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 torr. What is the partial pressure of the dry gas? DALTON’S LAW Look up water-vapor pressure on p.899 for 35.0°C. Sig Figs: Round to least number of decimal places. F. Dalton’s Law The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor.

28. ? Deviations from Ideal Gas Law Real gas: Deviate (Not Obey Gas law PV ≠ Constant Repulsive Forces Attractive Forces When: 1- T is very low and P is very high. This deviation is due to 2 factors which were ignored by kinetic theory: 1- Real gases posses attractive forces between molecules. 2- Every molecule in a real gas has a real volume.

29. ? There are 2 correction factors should be taken into consideration: I- Volume of gas molecules is not negligible: True volume (Real) = V container – non compressible volume (b) = V – b (1 mole) Actual volume = V – nb (for n moles) II- There is an attraction force between the molecules of gases (pressure): Actual Pressure = Measured pressure + Pressure due to attraction force. = P + (1 mole) Actual pressure = P + (for n moles) aV2aV2 an 2 V 2

30. ? Van der Waals Equation By introducing actual volume and actual pressure: The last equation is known as “Van der Waal’s equation” where a,b are Van der Waal’s constants depend on the nature of the gas. aV2aV2 For ideal gases: PV = nRT But in case of Real gas: PV ≠ nRT (P + )(V – b) = RT 1 mole (P + an 2 V 2 )(V – b) = nRT n mole According to Van der Waal ’ s, the pressure of real gas will be lower than that of ideal gas because attraction of neighbouring molecules tends to decrease the impact of a real molecule that it make with the wall of the container, this can be expressed as: P Real < P Ideal (Same rule but replace +ve sign with – ve sign)