1 / 10

A.P. Chemistry

A.P. Chemistry. Chapter 5 Gases Part 2.

acacia
Télécharger la présentation

A.P. Chemistry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. A.P. Chemistry Chapter 5 Gases Part 2

  2. Van der Waal’s Equation: (p. 222-224) Due to deviation from ideal behavior, corrections (adjustments) are made to the ideal gas law. As pressure increases, attractive forces become significant enough to affect the motion of the molecules. As temperature decreases, kinetic energy is decreased. Correction term [P + a(n/V)2] correction term (V –nb) For pressure for volume Therefore, Van der Waal’s constants for some common gases Gas a b He 0.034 0.0237 Ne 0.211 0.0171 Ar 1.34 0.0322 Kr 2.32 0.0398 Xe 4.19 0.0266 H2 0.2440.0266 N2 1.39 0.0391 O2 1.36 0.0318 Cl2 6.49 0.0562 CO2 3.59 0.0427 CH4 2.25 0.0428 CCl4 20.4 0.138 NH3 4.17 0.0371 H2O 5.46 0.0305

  3. Problem: Calculate the pressure exerted by 0.3000 mol of He in a 0.2000 L container at –25oC, (1.) using the ideal gas law; and (2.), using the van der Waal’s equation.

  4. Dalton’s Law of Partial Pressures: (p. 205-210)the total pressure of a mixture of gases is just the sum of the pressures that each gas would exert if it were present alone. PT = p1 + p2 + p3 + … + pn Problem: A volume of 2.0 L of He at 46oC and 1.2 atm pressure was added to a vessel that contained 4.5 L of N2 at STP. What is the total pressure and partial pressure of each gas at STP after the He is added? (1st- calculate # of moles of He; 2nd- use ideal gas law to change pressure of He to STP; 3rd- use Dalton’s Law)

  5. Mole Fraction: (p. 208) 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 is given by Xi = ni/nT Where ni and nT are the number of moles of component i and the total number of moles present, respectively. Mole fraction is always smaller than 1 (one). Partial pressure can be calculated, using mole fraction, by Pi = Xi/PT Problem: The vapor pressure of water in air at 28oC is 28.3 torr. Calculate the mole fraction of water in a sample of air at 28oC and 1.03 atm.

  6. Standard Conditions (p. 201) (STP) Standard atmospheric pressure (1 atm) is equal to the pressure that supports a column of mercury exactly 760 mm high at 0oC at sea level. Molar Gas Volume: (p. 201-205) one mole of any gas will occupy 22.4 L at STP. Density of a Gas (p. 205)(usually expressed in grams per liter, because the density of gases is very low under atmospheric conditions. Problem: What is the density of 1 mole of argon at STP?

  7. Molar Mass of a Gaseous Substance: (p. 204-205) Problem: A gas at 34.0oC and 1.75 atm has a density of 3.40 g/L. Calculate the molar mass of the gas. Problem: Consider the following combustion reaction where all reactants and products are at the same temperature: 2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(l) If 10.0 l of C2H6 are burned at STP, how many liters of oxygen would be required for a complete reaction at the same conditions?

  8. Kinetic Theory (p. 211-212) 1. A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules are considered to be “points”; that is, they possess mass but have negligible volume. 2. Gas molecules are in constant motion in random directions, and they frequently collide with each other. Collisions among molecules are perfectly elastic. In other words, energy can be transferred from one molecule to another as a result of a collision. Nevertheless, the total energy of all the molecules in a system remains the same. 3. Gas molecules exert neither repulsive nor attractive forces on one another. 4. The average kinetic energy of the molecules is proportional to the temperature of the gas in Kelvins. Any two gases at the same temperature will have the same average kinetic energy. The average kinetic energy of a molecule is given by KE = ½ mu2 (where m is the mass of the molecule and u is its speed).

  9. Root-mean-square (rms) speed (urms) (p. 217-218) The root-mean-square speed of a gas increases with the square root of its temperature (in Kelvins). Because molar mass appears in the denominator, the heavier the gas, the more slowly the molecules move. Substitute 8.314 J/K mol for R (universal gas constant), convert molar mass to kg/mol, then urms will be calculated in meters per second (m/s). urms = 3RT/M Problem: Calculate the root-mean-square speed of O2 molecules at 300.K.

  10. Graham’s Law: (p. 219)under the same conditions of temperature and pressure, rates of diffusion for gases are inversely proportional to the square roots of their molar masses. Problem: How many times faster than He would NO2 gas effuse? Effusion: (p. 219) process by which a gas under pressure escapes from one compartment of a container to another by passing through a small opening. Diffusion: (p. 220) gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.

More Related