# A.P. Chemistry Chapter 5 Gases Part 2. Van der Waal’s Equation: (p. 222-224) Due to deviation from ideal behavior, corrections (adjustments) are made.

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A.P. Chemistry Chapter 5 Gases Part 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 pressurefor volume Therefore, Van der Waal’s constants for some common gases Gasab He0.0340.0237 Ne0.2110.0171 Ar1.340.0322 Kr2.320.0398 Xe4.190.0266 H 2 0.2440.0266 N 2 1.390.0391 O 2 1.360.0318 Cl 2 6.490.0562 CO 2 3.590.0427 CH 4 2.250.0428 CCl 4 20.40.138 NH 3 4.170.0371 H 2 O5.460.0305

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

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. P T = p 1 + p 2 + p 3 + … + p n Problem: A volume of 2.0 L of He at 46 o C and 1.2 atm pressure was added to a vessel that contained 4.5 L of N 2 at STP. What is the total pressure and partial pressure of each gas at STP after the He is added? (1 st - calculate # of moles of He; 2 nd - use ideal gas law to change pressure of He to STP; 3 rd - use Dalton’s Law)

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 byX i = n i /n T Where n i and n T 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 P i = X i /P T Problem: The vapor pressure of water in air at 28 o C is 28.3 torr. Calculate the mole fraction of water in a sample of air at 28 o C and 1.03 atm.

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 0 o C 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?

Molar Mass of a Gaseous Substance: (p. 204-205) Problem: A gas at 34.0 o C 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: 2C 2 H 6 (g) + 7O 2 (g)  4CO 2 (g) + 6H 2 O(l) If 10.0 l of C 2 H 6 are burned at STP, how many liters of oxygen would be required for a complete reaction at the same conditions?

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 byKE = ½ mu 2 (where m is the mass of the molecule and u is its speed).

Root-mean-square (rms) speed (u rms ) (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 u rms will be calculated in meters per second (m/s). u rms = 3RT/M Problem: Calculate the root-mean-square speed of O 2 molecules at 300.K.

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 NO 2 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.

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