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The Behavior of Gases Part 2. Ideal Gases Ideal Gas Law: Ideal Gas Law: The relationship PV = nRT, which describes the behavior of ideal gases. The relationship.

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Presentation on theme: "The Behavior of Gases Part 2. Ideal Gases Ideal Gas Law: Ideal Gas Law: The relationship PV = nRT, which describes the behavior of ideal gases. The relationship."— Presentation transcript:

1 The Behavior of Gases Part 2

2 Ideal Gases Ideal Gas Law: Ideal Gas Law: The relationship PV = nRT, which describes the behavior of ideal gases. The relationship PV = nRT, which describes the behavior of ideal gases. Ideal Gas Constant (R): Ideal Gas Constant (R): A term in the ideal gas law that is used to make the units work out correctly. A term in the ideal gas law that is used to make the units work out correctly.

3 Ideal Gases

4 Example: Example: When the temperature of a rigid sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 18,900 kPa. How many moles of helium does the sphere contain? When the temperature of a rigid sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 18,900 kPa. How many moles of helium does the sphere contain?

5 Ideal Gases Example: Example: When the temperature of a rigid sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 18,900 kPa. How many moles of helium does the sphere contain? When the temperature of a rigid sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 18,900 kPa. How many moles of helium does the sphere contain? V = 685 L T = 621 K P = 18,900 kPa n = ?

6 Ideal Gases Example: Example: When the temperature of a rigid sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 18,900 kPa. How many moles of helium does the sphere contain? When the temperature of a rigid sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 18,900 kPa. How many moles of helium does the sphere contain?

7 Ideal Gases

8 Real Versus Ideal Gases Do real gases behave in ideal ways? Do real gases behave in ideal ways? 1. As they are compressed, the volume of the individual real gas particles are significant 2. Many real gases have intermolecular forces 3. Most real gases can be liquefied (or solidified). An ideal gas can’t. * Ideal behavior is most often seen at high temperatures and low pressures.

9 Graham’s Law of Diffusion and Effusion Diffusion: Diffusion: The process by which a gas expands to fill the available volume; the random walks of the gas particles take them in all directions. This is why an air freshener makes the whole room smell good. The process by which a gas expands to fill the available volume; the random walks of the gas particles take them in all directions. This is why an air freshener makes the whole room smell good. Effusion: Effusion: The process by which gas particles pass through a small opening. This is why air leaves a tire with a nail hole in it. The process by which gas particles pass through a small opening. This is why air leaves a tire with a nail hole in it.

10 Graham’s Law of Diffusion and Effusion Graham’s Law: Graham’s Law: The rate at which gases diffuse or effuse is inversely proportional to the square root of their molar masses. The rate at which gases diffuse or effuse is inversely proportional to the square root of their molar masses.

11 Graham’s Law of Diffusion and Effusion Example: Example: A sample of helium effuses through a porous container 6.50 times faster than does an unknown gas. What is the molar mass of the unknown? A sample of helium effuses through a porous container 6.50 times faster than does an unknown gas. What is the molar mass of the unknown?

12 Graham’s Law of Diffusion and Effusion Example: Example: A sample of helium effuses through a porous container 6.50 times faster than does an unknown gas. What is the molar mass of the unknown? A sample of helium effuses through a porous container 6.50 times faster than does an unknown gas. What is the molar mass of the unknown? Rate 1 = 6.50·Rate 2 M 1 = 4.00 g/mol M 2 = ?

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