The University of Dodoma Department of Chemistry CH 111: General Chemistry

1. The University of DodomaDepartment of ChemistryCH 111: General Chemistry Lecture 5: Gas Mixtures and Partial pressures, Kinetic molecular theory Mtabazi Geofrey Sahini mtabazisahini@gmail.com, 0719116792, CNMS Office No. 225

2. Partial pressures • The pressure that each gas would exert if it were alone occupying the container is called partial pressure. • Assuming each gas behaves ideally,

3. 50 kPa 100 kPa 150 kPa 1 L oxygen 1 L nitrogen 1 L mixed gas Dalton’s law of partial pressure + 

4. Partial Pressures and Mole Fractions PT = P1 + P2 + P3 + … PA = XAPT where XA is the mole fraction XA= PA/ PT The mole fraction of a gas in a gas mixture is simply the ratio of its partial pressure to the total pressure Let ni be the number of moles of gas i exerting a partial pressure Pi then: Pi= Xi PT Where Xi is the mole fraction (ni/ nT) Note the mole fraction is a dimensionless number

5. Mole fraction (χ) For a given component in a mixture, the mole fraction 1 is For each component in the mixture Mole fraction in terms of pressure,

6. Example • The partial pressure of oxygen was observed to be 156 torr in air with a total atmospheric pressure of 743 torr. Calculate the mole fraction of O2 present.

7. Collecting Gas over Water.. • A tube from the flask which is giving off the gas is placed under water. • A test tube full of water is placed upside downabove the end of the tube to collect the gas. • As the gas bubbles into the test tube it displaces the water until the test tube is full.

8. Collecting Gas over Water.. PT = Pgas + PH2O • This equation can be used to calculate the pressure of the gas collected. Once the pressure of the collected gas is known, the number of moles of gas can be calculated using the ideal gas law: • PV= nRT Where: P = Pressure of the gas V = Volume of water displaced n = number of moles of gas R = the ideal gas constant T = the temperature of the gas

9. Example What mass of oxygen gas is collected over water at 750.1 torr (total) pressure and 25.2 °C if the volume collected is 450.8 mL? Vapor pressure of water at 25 °C is 23.76torr Use PV = nRT but first find partial pressure of O2 over water.

10. Kinetic Molecular Theory of Gases Behavior of gases: (as depicted by the gas laws or otherwise) • Volume increase with increasing temperature • Volume decrease with increasing pressure • High effusion/diffusion rates that increase with temperature • Effusion/Diffusion rates increase with decreasing molar mass • Low densities and high compressibility etc, etc.

11. Why this behavior? • The ideal gas equation describes how gases behave but doesn’t explain why they behave as they do. • The Kinetic Molecular Theory (KMT) is a theoretical model developed to account for the observed behavior of gases, based on the behavior of individual gas particles (atoms or molecules).

12. Kinetic molecular theory: Assumptions • Gas particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero) compared to the total volume of the gas. • Gas particles are in continuous random motion. As a result of their motion they undergo frequent collision with one another and with the walls of the container. • The particles exert no forces on each other. They do not interact; they neither attract nor repel each other. • Collisions between gas molecules or between the molecules and the walls of the container are elastic. That is the KE is remain constant during collisions. • The average kinetic energy of a sample of gas particles is directly proportional to the absolute temperature. (The average K.E. depends only on the temperature of the gas) • The collisions of the gas molecules with the walls of the container are the cause of the pressure exerted by the gas.

13. Using the ideal gas equation, …Boyle’s law ...Gay-Lussac’s …Charles’ law …Avogadro’s law

14. Deducing the Ideal gas Equation using KMT.. But from We get PV = NkT →PV = nNAkT PV = nRT

15. Maxwell-Boltzmann's distribution of speeds  • Molecular speed distribution depends on both molecular mass and temperature. • The higher the temperature the faster the molecules. • The heavier the gas the slower the average molecular speed.

16. Root mean square speed (rms speed) of a gas:(meaning & derivation) • For a sample of gas containing N particles, each moving with its individual speed ui, the mean speed is defined as: • Root mean square speed, urms, is defined as:

17. Maxwell-Boltzmann's distribution of speeds  • Maxwell showed theoretically that molecular speed distribution for a sample of gas takes the following form shown below.

18. Root mean square speed N.m = Total mass M = Molar mass

19. Summary According to the KMT, average Kinetic energy of a sample of gas at temperature T, is given by

20. Example Evaluate the root-mean-square speed of H2, He, N2, O2 and CO2 at 310 K (the human body temperature). Solution ūrms= (3k N T / M)1/2 = (3 R T / M) 1/2= (3 x 8.3145 x 310 / 0.002) ½ = 1966 ms-1 Note that the molecular mass of hydrogen is 0.002 kg/mol. These units are used because the constant R has been calculated using the SI units.

21. ūrms = (3k N T / M)1/2 = (3 R T / M) 1/2    = (3 x 8.3145x 310) 1/2/M1/2    = 87.9345/M1/2 ms-1

22. The root mean square speeds are:

23. Effusion and diffusion in gases • Diffusion is the spread out of a gas or vapor from a region of a relatively high pressure towards a low pressure region. Effusion—gradual movement of gas molecules through a very tiny hole into a vacuum

24. Graham’s law of effusion • The rates at which these processes occur depend on the speeds of gas molecules • Faster the molecules move, the more rapidly diffusion/effusion occur. Graham's Law The rate of diffusion/effusion of gas molecules at constant temperature and pressure is inversely proportional to the square root of the molar mass. Both diffusion and effusion rates are proportional to the mean molecular speed (or rms speed of a gas)

25. Graham’s law.. Since for gases, the density, , = MP/RT,  M (at constant P and T) It follows also that The diffusion/effusion rates for any two gases can be compared:

26. Calculations using Graham's Law Let's compare the rate off effusion of two common gases, Nitrogen and Oxygen. N2, Nitrogen, has a molecular mass of 28.0 g. O2, Oxygen, has a molecular mass of 32.0 g. Therefore, to find the ratio, the equation would be:RateN2/RateO2 = square root of 32.0 g / 28.0 g. This works out to:RateN2/RateO2 = 1.069 Adjusting to the appropriate accuracy, we find that the rate is 1.07. This tells us that N2 is 1.07 times as fast as O2.