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Introduction into: Dalton’s Law of Partial Pressures

Introduction into: Dalton’s Law of Partial Pressures.

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Introduction into: Dalton’s Law of Partial Pressures

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  1. Introduction into: Dalton’s Law of Partial Pressures So far, we’ve assumed that all the gases we’ve been working with are pure. However, this isn’t true most of the time. Most gases we use are actually mixtures of several different elements or compounds. To deal with this, we will examine the idea of partial pressure.

  2. Dalton’s Law of Partial Pressures • To work with mixtures, we usually treat each gas as if it were alone, rather than in a mixture. This is a reasonable assumption because gases don’t interact with each other much anyway. As a result, the total pressure of the mixture of gases would be the same as the sums of all the partial pressures. • Dalton’s Law of Partial Pressures: • Ptot= P1 + P2 + P3 + … Pn • Where n is the total number of gases in the mixture.

  3. Partial Pressure • Partial pressure: The pressure that each of the components in a mixture of gases would exert if the other gases in the mixture were removed. • Let’s back up at little: • For any pure gas (let's use helium), PV = nRT holds true. Therefore, P is directly proportional to n if V and T remain constant. As n goes up, so would P. Or the reverse. • Suppose you were to double the moles of helium gas present. What would happen? • Answer: the gas pressure doubles.

  4. Partial Pressure Continued • However, suppose the new quantity of gas added was a DIFFERENT gas. Suppose that, instead of helium, you added neon. • What would happen to the pressure? • Answer: the pressure doubles, same as before. • Dalton's Law immediately follows from this example since each gas is causing 50% of the pressure. Summing their two pressures gives the total pressure. • Written as an equation, it looks like this: • PHe + PNe = Ptotal

  5. Example Problems • Example 1: A container contains three gases. If the partial pressure of carbon dioxide is 0.50 atm, the partial pressure of nitrogen is 0.30 atm, and the partial pressure of methane is 0.40 atm, what’s the total pressure inside the container? • Ptot = 0.50 atm + 0.30 atm + 0.40 atm = 1.20 atm • Example 2: A container has a total internal pressure of 1.50 atm. This container contains two gases, nitrogen (which has a partial pressure of 0.80 atm) and oxygen (with unknown partial pressure). Given this information, what’s the partial pressure of oxygen in the container? • Ptot = Pnitrogen + Poxygen • 1.50 atm = 0.80 atm + X • X = 0.70 atm

  6. Graham’s Law • Have you ever noticed when air freshener is sprayed into the air, it can take seconds before you notice the smell filling the room? You begin noticing the smell because the molecules diffuse through the surrounding air. • To begin to think about this law, please consider samples of two different gases at the same Kelvin temperature. • Since temperature is proportional to the kinetic energy of the gas molecules, the kinetic energy (KE) of the two gas samples is also the same. • In equation form, we can write: KE1 = KE2

  7. Graham’s Law Continued • Since KE = ½mv2, • (m = mass and v = velocity) we can write the following equation: • m1v12 = m2v22**Note that the value of one-half cancels. • The equation above can be rearranged algebraically into the following: • v2= √(m1) v1= √(m2) • Graham's Law is often stated as follows: • Rate of diffusion of gas1 = √(Mm2) Rate of diffusion of gas2= √(Mm1) • where MM means the molar mass of the substance in question

  8. Graham’s Law • Diffusion is the process in which particles in a system move from an area of high concentration to an area of low concentration. • Diffusion continues until a uniform concentration of particles is reached throughout the system. • The rate at which gas molecules diffuse, at constant temperature, decreases as the molar mass of the gas increases(convince yourself of this statement based on the equation given on the previous slide).

  9. Graham’s Law Practice Problems • How much faster does hydrogen gas escape through a porous container than sulfur dioxide? • 2.278 x 10-4mol of an unidentified gaseous substance effuses through a tiny hole in 95.70 s. Under identical conditions, 1.738 x 10-4mol of argon gas takes 81.60 s to effuse. What is the molar mass of the unidentified substance?

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