CHE101 Gases Instructor: HbR

# CHE101 Gases Instructor: HbR

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## CHE101 Gases Instructor: HbR

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1. CHE101 Gases Instructor: HbR

2. Elements that exist as gases at 250C and 1 atmosphere

3. Physical Characteristics of Gases • Gases assume the volume and shape of their containers. • Gases are the most compressible state of matter. • Gases will mix evenly and completely when confined to the same container. • Gases have much lower densities than liquids and solids. NO2 gas

4. Atmospheric Pressure • Atmospheric pressure is the pressure exerted by the earth’s atmosphere. • Depends on location, temperature and weather conditions. • The denser the air is, the greater the pressure it exerts. • By increasing height, density decreases. • Barometer: measure the atmospheric pressure. • Standard atmospheric pressure (1atm): pressure that supports a column of mercury exactly 760mm (or 76 cm) high at 0⁰C at sea level. Barometer

5. Force Area Pressure of a Gases Pressure = (force = mass x acceleration) Units of Pressure 1 pascal (Pa) = 1 N/m2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa

6. Atmospheric Pressure 10 miles 0.2 atm 4 miles 0.5 atm Sea level 1 atm

7. Manometers Used to Measure Gas Pressures

8. Calculate the pressure of the gas.. Pgas=755 + 24=779mmHg Pgas=763 -35=728mmHg

9. The Gas laws • Boyel’s Law: the Pressure- Volume relationship • Charls’s and Gay-Lussac’s Law: Temperature-Volume relationship • Avogadro’s Law: Volume- Amount relationship

10. Apparatus for Studying the Relationship Between Pressure and Volume of a Gas V decreases As P increases

11. Boyle’s Law Pa 1/V Constant temperatureConstant amount of gas P x V = constant P1 x V1 = P2 x V2

12. 726 mmHg x 946 mL P1 x V1 = 154 mL V2 A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P2 = = 4460 mmHg

13. Variation in Gas Volume with Temperature at Constant Pressure As T increases V increases

14. Variation of Gas Volume with Temperature at Constant Pressure Charles’ & Gay-Lussac’s Law VaT Temperature must be in Kelvin V = constant x T V1/T1 = V2 /T2 T (K) = t (0C) + 273.15

15. 1.54 L x 398.15 K V2 x T1 = 3.20 L V1 A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1 /T1 = V2 /T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? T1 = 125 (0C) + 273.15 (K) = 398.15 K T2 = = 192 K

16. Another form of Charle’s Law For constant amount of gas and volume, the pressure is proportional to the temperature.

17. Avogadro’s Law Constant temperatureConstant pressure Va number of moles (n) V = constant x n V1 / n1 = V2 / n2

18. 4NH3 + 5O2 4NO + 6H2O 1 volume NH3 1 volume NO 1 mole NH3 1 mole NO Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? At constant T and P

19. Boyle’s law: V a (at constant n and T) Va nT nT nT P P P V = constant x = R 1 P Ideal Gas Equation Charles’ law: VaT(at constant n and P) Avogadro’s law: V a n(at constant P and T) R is the gas constant PV = nRT

20. R = (1 atm)(22.414L) PV = nT (1 mol)(273.15 K) The conditions 00C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. PV = nRT R = 0.082057 L • atm / (mol • K)

21. 1 mol HCl V = n = 49.8 g x = 1.37 mol 36.45 g HCl 1.37 mol x 0.0821 x 273.15 K V = 1 atm nRT L•atm P mol•K What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273.15 K P = 1 atm PV = nRT V = 30.7 L Can you calculate the density from this equation??

22. n d = m m n V V VM M = PM m P P = = = = M RT RT RT dRT P Density (d) Calculations Again, PV = nRT,  m is the mass of the gas in g, M is the molar mass of the gas Molar Mass (M ) of a Gaseous Substance d is the density of the gas in g/L

23. d = = m M = V 4.65 g 2.10 L g 2.21 L x 0.0821 x 300.15 K 54.5 g/mol M = 1 atm g = 2.21 L dRT L•atm mol•K P A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0C. What is the molar mass of the gas? M =

24. What is the volume of CO2 produced at 37oC and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) 6 mol CO2 g C6H12O6 mol C6H12O6 mol CO2V CO2 x 1 mol C6H12O6 1 mol C6H12O6 x 180 g C6H12O6 L•atm mol•K nRT 0.187 mol x 0.0821 x 310.15 K = P 1.00 atm Gas Stoichiometry 5.60 g C6H12O6 = 0.187 mol CO2 V = = 4.76 L

25. Dalton’s Law of Partial Pressures V and T are constant P2 Ptotal= P1 + P2 P1

26. PA = nART nBRT V V PB = Consider a case in which two gases, A and B, are in a container of volume V. nA is the number of moles of A nB is the number of moles of B where n, the total no of moles; n = nA + nB, and PA, PB are partial pressures of gases A and B. Total pressure of a mixture of a gas depends on the total number of moles of gas, not on the nature of the gas.

27. In general, total pressure of a mixture of a gases, where P1, P2, P3, … are the partial pressure of the compo-nents1, 2, 3, … To see how each partial pressure is related to the total pressure, consider a mixture of gases A and B. where XA is mole fraction of A. The mole fraction is a dimensionless quantity that expresses the ratio of the number of moles of one component to the number of moles of all components present. The mole fraction of component i in a mixture,

28. ni mole fraction (Xi ) = nT So the partial pressures of A and B, PA = XA PT PB = XB PT The sum of the mole fractions for a mixture of gases must be unity. If two components are present, If a system contains more than two gases, Pi = XiPT From mole fractions and total pressure, we can calculate the partial pressures of individual components.

29. 0.116 8.24 + 0.421 + 0.116 A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)? Pi = XiPT PT = 1.37 atm = 0.0132 Xpropane = Ppropane = 0.0132 x 1.37 atm = 0.0181 atm

30. You are expected to practice relevant maths from the Book By: Raymond Chang You might get problems not discussed in the class, Presuming we can’t cover all problems in our class. So going through the book is of major important Your equations will NOT provided in questions this time As derivations were shown in our lesson! You are always more than welcome to discuss problems at my office…

31. Thank You!