The Gas Laws =13WUqWd_Yk8 =13WUqWd_Yk8

Gas Laws The gas laws describe the behavior of “ideal” gases which approximate the behavior of real gases. The gas laws describe the behavior of “ideal” gases which approximate the behavior of real gases. The gas laws describe the relationship between pressure, volume, temperature, and number of gas particles. The gas laws describe the relationship between pressure, volume, temperature, and number of gas particles.

Gas Laws Standard pressure for a gas Standard pressure for a gas kPa = 760 mm of Hg, or 1kPa = 7.50 mm of Hg kPa = 760 mm of Hg, or 1kPa = 7.50 mm of Hg

Boyle’s Law Boyle’s law describes the relationship between a gas volume and its pressure. Boyle’s law describes the relationship between a gas volume and its pressure. What happens to the volume of a gas as the pressure on it increases? Decreases? What happens to the volume of a gas as the pressure on it increases? Decreases? If the pressure is doubled, the volume decreases to ½ of its original size. If the pressure is doubled, the volume decreases to ½ of its original size. If the pressure is decreased to ½ its original amount, then the volume doubles. If the pressure is decreased to ½ its original amount, then the volume doubles.

Boyle’s Law

Boyle’s Law Mathematically, this says that pressure and volume are inversely related, or that the product of the pressure and volume is a constant. Mathematically, this says that pressure and volume are inversely related, or that the product of the pressure and volume is a constant. Therefore Boyle’s Law can be stated: Therefore Boyle’s Law can be stated: P 1 V 1 = P 2 V 2

Boyle’s Law – Example Problems 1. If 400 cm 3 of oxygen are collected at a pressure of 9.80 kPa, then what volume will the gas occupy at 9.40 kPa? 2. What is the volume of hydrogen gas at a pressure of 106 kPa if 200 cm 3 of hydrogen were collected at a pressure of 100 kPa? 3. Calculate the pressure of a gas which occupies 100 cm 3, if at a pressure of 95 kPa, it occupies a volume of 200 cm 3.

Dalton’s Law of Partial Pressures

Dalton’s Law of Partial Pressures Chapter 12 (red book) page 409 In a system with more than one gas, the total pressure is the sum of the individual pressures. In a system with more than one gas, the total pressure is the sum of the individual pressures. Total P vap =  P = P gas1 + P gas2 +… Total P vap =  P = P gas1 + P gas2 +…

Dalton’s Law of Partial Pressures Normally in a lab setting, gases are collected by water displacement (gas bubbles through water as they are collected) Normally in a lab setting, gases are collected by water displacement (gas bubbles through water as they are collected) Water vapor pressure becomes part of the P total this is the “wet gas” pressure Water vapor pressure becomes part of the P total this is the “wet gas” pressure To calculate the “dry gas” value you must account for the water vapor pressure To calculate the “dry gas” value you must account for the water vapor pressure Use table 9-1 pg. 44 as needed. Use table 9-1 pg. 44 as needed.

Dalton & Boyle combo. examples In a series of lab experiments, different gases were collected over water. Correct the following volume to the volume the dry gas would occupy at standard pressure: 63 cm 3 gas at 20 o C and 95.6 kPa In a series of lab experiments, different gases were collected over water. Correct the following volume to the volume the dry gas would occupy at standard pressure: 63 cm 3 gas at 20 o C and 95.6 kPa

Charles’ Law Describes the relationship between the temperature of a gas and its volume (at a constant pressure). Describes the relationship between the temperature of a gas and its volume (at a constant pressure). In this relationship, note that the temperature must be in Kelvin. Why? Think about what happens if you divide a number by zero

Charles’ Law

This relationship shows that the temperature and volume of a gas at a constant pressure are directly related, or in other words their quotient is a constant. This relationship shows that the temperature and volume of a gas at a constant pressure are directly related, or in other words their quotient is a constant. Mathematically stated… Mathematically stated… V 1 = T 1 or V 1 = V 2 or V 2 = V 1 T 2 V 2 T 2 T 1 T 2 T 1

Charles’ Law – Example Problems 1. What volume will a sample of nitrogen occupy at 27 o C if the gas occupies a volume of 400 cm 3 at a temperature of 0 o C? Assume the pressure remain constant 2. What is the volume of a gas at -20 o C if the gas occupied 50.0 cm 3 at a temperature of 0 o C? 3. If a gas occupies a volume of 700 cm 3 at 10 o C, at what temperature will it occupy a volume of 1000 cm 3 if the pressure remain constant?

Combined Gas Law The Combined Gas Law – does just that; it combines Boyle’s Law (P 1 V 1 = P 2 V 2 ) and Charles’ Law (V 1 /T 1 = V 2 /T 2 ). The Combined Gas Law – does just that; it combines Boyle’s Law (P 1 V 1 = P 2 V 2 ) and Charles’ Law (V 1 /T 1 = V 2 /T 2 ). Used when a pressure and temperature change occur

Combined Gas Law new volume = old volume x pressure ratio x K temperature ratio new volume = old volume x pressure ratio x K temperature ratio Each ratio is considered independently… Each ratio is considered independently… –Ask each question separately –Use “Old-New” table OldNew What happens to the gas volume? Pressure (P) Volume (V) Temp. (T) in Kelvins

Combined Gas Law - Example Calculate the volume of a gas at STP if 500 cm 3 of the gas are collected at 27 o C and 96.0 kPa. Calculate the volume of a gas at STP if 500 cm 3 of the gas are collected at 27 o C and 96.0 kPa. STP = Standard temperature & pressure: STP = Standard temperature & pressure: 273 K (0 o C)273 K (0 o C) kPa101.3 kPa

Combined Gas Law - Example If 400 cm 3 of oxygen is collected over water at 20 o C, and the atmospheric pressure is 97,000 Pa, what is the volume of the dry oxygen at STP? If 400 cm 3 of oxygen is collected over water at 20 o C, and the atmospheric pressure is 97,000 Pa, what is the volume of the dry oxygen at STP? STP = Standard temperature & pressure: STP = Standard temperature & pressure: 273 K (0 o C)273 K (0 o C) kPa101.3 kPa

Gas Density uMLqQ&feature=related uMLqQ&feature=related uMLqQ&feature=related uMLqQ&feature=related

Gas Density Density = mass ÷ volume Density = mass ÷ volume Since the volume of gas varies with pressure (inversely) and temperature (directly), these affect the density of a gas. Since the volume of gas varies with pressure (inversely) and temperature (directly), these affect the density of a gas. What happens to the density as the pressure on a gas increases? Decreases? What happens to the density as the pressure on a gas increases? Decreases? What happens to the density as the temperature on a gas increases? Decreases? What happens to the density as the temperature on a gas increases? Decreases?

Gas Density THEREFORE: The density varies directly with the pressure, and it varies inversely with the temperature. THEREFORE: The density varies directly with the pressure, and it varies inversely with the temperature.  VERY important that you know these relationships! Units for gas density  grams/dm 3 or grams/liter (1 dm 3 = 1 liter) Units for gas density  grams/dm 3 or grams/liter (1 dm 3 = 1 liter) Grams/cm 3 would give very small numbers for most gases. Grams/cm 3 would give very small numbers for most gases.

Gas Density - Examples What is the density of a gas which has a mass of 4.50 g and occupies 2.50 dm 3 ? What is the density of a gas which has a mass of 4.50 g and occupies 2.50 dm 3 ?Solution: 4.50 g = 1.80g/dm 3 = 1.80g/dm dm dm 3

Gas Density - Examples If the density of helium is g/dm 3 at STP (273 K and kPa), what is its density at 99.0 kPa and 27 o C? If the density of helium is g/dm 3 at STP (273 K and kPa), what is its density at 99.0 kPa and 27 o C? How to solve… How to solve… New density = old density x pressure ratio (new/old) x temp. ratio (old/new) New density = old density x pressure ratio (new/old) x temp. ratio (old/new) (direct relationship)(inverse relationship) (direct relationship)(inverse relationship) OldNew What happens to the gas density? Pressure (P) Temp (T) Density (D)

Gas Density - Examples Solution Solution New density = (0.179 g/dm 3 )x(99 kPa/101.3 kPa)x(273 K/300 K) New density = (0.179 g/dm 3 )x(99 kPa/101.3 kPa)x(273 K/300 K) = g/dm 3

Graham’s Law Only a few physical properties of gases depend on the identity of the gas. Only a few physical properties of gases depend on the identity of the gas. Diffusion - The rate at which two gases mix. Diffusion - The rate at which two gases mix. Effusion - The rate at which a gas escapes through a pinhole into a vacuum. Effusion - The rate at which a gas escapes through a pinhole into a vacuum.

Graham’s Law of Diffusion The rate at which gases diffuse is inversely proportional to the square root of their densities, or … The rate at which gases diffuse is inversely proportional to the square root of their densities, or …

Graham’s Law of Diffusion Since volumes of different gases contain the same number of particles (see Avogadro's Hypothesis), the number of moles per liter at a given T and P is constant. Therefore, the density of a gas is directly proportional to its molar mass (MM), and… Since volumes of different gases contain the same number of particles (see Avogadro's Hypothesis), the number of moles per liter at a given T and P is constant. Therefore, the density of a gas is directly proportional to its molar mass (MM), and… Avogadro's Hypothesis Avogadro's Hypothesis

Graham’s Law of Effusion The rate of effusion of a gas is inversely proportional to the square root of either the density or the molar mass of the gas. The rate of effusion of a gas is inversely proportional to the square root of either the density or the molar mass of the gas.

Graham’s Law of Diffusion - restated

Graham’s Law of Diffusion - Example Example 1. What is the ratio of the velocity of helium atoms to the velocity of radon atoms when both gases are at the same temperature? Example 1. What is the ratio of the velocity of helium atoms to the velocity of radon atoms when both gases are at the same temperature? Solution: the ratio of V 1 to V 2 = 7.45:1 Solution: the ratio of V 1 to V 2 = 7.45:1

Upcoming Dates to Keep in Mind Graham’s Law Lab –Tuesday, April 10 th 2012 Graham’s Law Lab –Tuesday, April 10 th 2012 Chapter 9 Test – Thursday, April 12 th 2012 Chapter 9 Test – Thursday, April 12 th 2012 –Assignment #2 is due that day!!