130 likes | 290 Views
The Behavior of Gases. Part 2. Ideal Gases. Ideal Gas Law: The relationship PV = nRT, which describes the behavior of ideal gases. Ideal Gas Constant (R): A term in the ideal gas law that is used to make the units work out correctly. Ideal Gases. Ideal Gases. Example:
E N D
The Behavior of Gases Part 2
Ideal Gases • Ideal Gas Law: • The relationship PV = nRT, which describes the behavior of ideal gases. • Ideal Gas Constant (R): • A term in the ideal gas law that is used to make the units work out correctly.
Ideal Gases • 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?
Ideal Gases • 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? V = 685 L T = 621 K P = 18,900 kPa n = ?
Ideal Gases • 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?
Real Versus Ideal Gases • Do real gases behave in ideal ways? • As they are compressed, the volume of the individual real gas particles are significant • Many real gases have intermolecular forces • 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.
Graham’s Law of Diffusion and Effusion • 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. • 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.
Graham’s Law of Diffusion and Effusion • Graham’s Law: • The rate at which gases diffuse or effuse is inversely proportional to the square root of their molar masses.
Graham’s Law of Diffusion and Effusion • 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?
Graham’s Law of Diffusion and Effusion • 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? Rate1 = 6.50·Rate2 M1 = 4.00 g/mol M2 = ?