The Gas Laws v =13WUqWd_Yk8 v =13WUqWd_Yk8.

The Gas LawsThe Gas Laws

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Gas LawsGas Laws

The gas laws describe the behavior The gas laws describe the behavior of “ideal” gases which approximate of “ideal” gases which approximate the behavior of real gases.the behavior of real gases.

The gas laws describe the The gas laws describe the relationship between relationship between pressurepressure, , volumevolume, , temperaturetemperature, and , and number of gas particlesnumber of gas particles..

Gas LawsGas Laws Standard pressure for a gasStandard pressure for a gas

•101.325 kPa101.325 kPa = 760 mm of Hg, or = 760 mm of Hg, or 1kPa = 7.50 mm of Hg1kPa = 7.50 mm of Hg

Boyle’s LawBoyle’s Law

Boyle’s law describes the Boyle’s law describes the relationship between a gas relationship between a gas volumevolume and its and its pressurepressure..

What happens to the What happens to the volumevolume of a gas as of a gas as the the pressurepressure on it on it increasesincreases? ? DecreasesDecreases??

If the pressure is doubled, the volume If the pressure is doubled, the volume decreases to ½ of its original size. decreases to ½ of its original size.

If the pressure is decreased to ½ its If the pressure is decreased to ½ its original amount, then the volume original amount, then the volume doubles.doubles.

Boyle’s LawBoyle’s Law http://www.asc-csa.gc.ca/images/http://www.asc-csa.gc.ca/images/

neemo_graph_boyles_law.jpgneemo_graph_boyles_law.jpg

Boyle’s LawBoyle’s Law Mathematically, this says Mathematically, this says

that pressure and volume that pressure and volume are are inverselyinversely related, or related, or that the product of the that the product of the pressure and volume is a pressure and volume is a constant.constant.

Therefore Boyle’s Law can Therefore Boyle’s Law can be stated:be stated:

PP11VV11 = P = P22VV22

Boyle’s Law – Example Boyle’s Law – Example ProblemsProblems

1.1. If 400 cmIf 400 cm33 of oxygen are collected at a of oxygen are collected at a pressure of 9.80 kPa, then what pressure of 9.80 kPa, then what volume will the gas occupy at 9.40 volume will the gas occupy at 9.40 kPa?kPa?

2.2. What is the volume of hydrogen gas at What is the volume of hydrogen gas at a pressure of 106 kPa if 200 cma pressure of 106 kPa if 200 cm33 of of hydrogen were collected at a pressure hydrogen were collected at a pressure of 100 kPa?of 100 kPa?

3.3. Calculate the pressure of a gas which Calculate the pressure of a gas which occupies 100 cmoccupies 100 cm33, if at a pressure of 95 , if at a pressure of 95 kPa, it occupies a volume of 200 cmkPa, it occupies a volume of 200 cm33..

Dalton’s Law of Partial Dalton’s Law of Partial PressuresPressures


Chapter 12 (red book) page Chapter 12 (red book) page 409409 In a system with more than one In a system with more than one

gas, the gas, the total pressuretotal pressure is the is the sumsum of the individual pressures. of the individual pressures.

Total PTotal Pvapvap = = P = PP = Pgas1gas1 + P + Pgas2gas2 +… +…


Normally in a lab setting, gases are Normally in a lab setting, gases are collected by water displacement (gas collected by water displacement (gas bubbles through water as they are bubbles through water as they are collected)collected)

Water vapor pressure becomes part of Water vapor pressure becomes part of the Pthe Ptotaltotal this is the “wet gas” pressure this is the “wet gas” pressure

To calculate the “dry gas” value you To calculate the “dry gas” value you must account for the water vapor must account for the water vapor pressurepressure

Use table 9-1 pg. 44 as needed.Use table 9-1 pg. 44 as needed.

Dalton & Boyle combo. Dalton & Boyle combo. examplesexamples

In a series of lab In a series of lab experiments, experiments, different gases were different gases were collected over water. collected over water. Correct the following Correct the following volume to the volume to the volume the dry gas volume the dry gas would occupy at would occupy at standard pressure: standard pressure: 63 cm 63 cm33 gas at 20 gas at 20 ooC and 95.6 kPaC and 95.6 kPa

Charles’ LawCharles’ Law

Describes the Describes the relationship relationship between the between the temperaturetemperature of of a gas and its a gas and its volumevolume (at a (at a constant constant pressure).pressure).

In this relationship, note that the temperature must be in Kelvin. Why?

• Think about what happens if you divide a number by zero


This relationship shows that the This relationship shows that the temperature and volume of a gas at a temperature and volume of a gas at a constant pressure are directly related, or constant pressure are directly related, or in other words their quotient is a constant.in other words their quotient is a constant.

Mathematically stated…Mathematically stated…

VV11 = T = T11 oror V V1 1 = V= V22 oror V V2 2 = V= V1 1 TT22

VV22 T T2 2 TT11 T T2 2 TT11

Charles’ Law – Example Charles’ Law – Example ProblemsProblems

1.1. What volume will a sample of nitrogen What volume will a sample of nitrogen occupy at 27 occupy at 27 ooC if the gas occupies a C if the gas occupies a volume of 400 cmvolume of 400 cm33 at a temperature of at a temperature of 00ooC? Assume the pressure remain C? Assume the pressure remain constantconstant

2.2. What is the volume of a gas at -20 What is the volume of a gas at -20 ooC if the C if the gas occupied 50.0 cmgas occupied 50.0 cm33 at a temperature of at a temperature of 0 0 ooC?C?

3.3. If a gas occupies a volume of 700 cmIf a gas occupies a volume of 700 cm33 at at 10 10 ooC, at what temperature will it occupy a C, at what temperature will it occupy a volume of 1000 cmvolume of 1000 cm33 if the pressure remain if the pressure remain constant?constant?

Combined Gas LawCombined Gas Law

The Combined Gas Law – does just The Combined Gas Law – does just that; it combines Boyle’s Law (Pthat; it combines Boyle’s Law (P11VV11 = = PP22VV22) and Charles’ Law (V) and Charles’ Law (V11/T/T11 = = VV22/T/T22).).

Used when a pressure and temperature change occur

Combined Gas LawCombined Gas Law new volume = old volume x pressure ratio x K temperature new volume = old volume x pressure ratio x K temperature

ratioratio

Each ratio is considered independently…Each ratio is considered independently…– Ask each question separately Ask each question separately – Use “Old-New” tableUse “Old-New” table

OldOld NewNew What happens What happens to the gas to the gas volume?volume?

Pressure Pressure (P)(P)

Volume (V)Volume (V)

Temp. (T) in Temp. (T) in KelvinsKelvins

Combined Gas Law - Combined Gas Law - ExampleExample

Calculate the volume of a gas at Calculate the volume of a gas at STPSTP if 500 cmif 500 cm33 of the gas are collected at of the gas are collected at 27 27 ooC and 96.0 kPa.C and 96.0 kPa.

STP = Standard temperature & STP = Standard temperature & pressure:pressure:• 273 K (0 273 K (0 ooC)C)• 101.3 kPa101.3 kPa

Combined Gas Law - Combined Gas Law - ExampleExample

If 400 cmIf 400 cm33 of oxygen is collected over of oxygen is collected over water at 20water at 20ooC, and the atmospheric C, and the atmospheric pressure is 97,000 Pa, what is the pressure is 97,000 Pa, what is the volume of the dry oxygen at STP?volume of the dry oxygen at STP?

STP = Standard temperature & STP = Standard temperature & pressure:pressure:• 273 K (0 273 K (0 ooC)C)• 101.3 kPa101.3 kPa

Gas DensityGas Density

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Density = mass Density = mass ÷ volume÷ volume Since the volume of gas varies with Since the volume of gas varies with

pressure (inversely) and temperature pressure (inversely) and temperature (directly), these affect the density of a gas.(directly), these affect the density of a gas.

What happens to the density as the What happens to the density as the pressure on a gas increases? Decreases?pressure on a gas increases? Decreases?

What happens to the density as the What happens to the density as the temperature on a gas increases? temperature on a gas increases? Decreases?Decreases?


THEREFORE: The THEREFORE: The densitydensity varies varies directlydirectly with the pressure, and it with the pressure, and it varies varies inverselyinversely with the temperature. with the temperature.

VERYVERY important that you know these important that you know these relationships!relationships!

Units for gas density Units for gas density grams/dm grams/dm33 or or grams/liter (1 dmgrams/liter (1 dm33 = 1 liter) = 1 liter)

Grams/cmGrams/cm33 would give very small would give very small numbers for most gases.numbers for most gases.

Gas Density - ExamplesGas Density - Examples

What is the density of a gas which has a What is the density of a gas which has a mass of 4.50 g and occupies 2.50 dmmass of 4.50 g and occupies 2.50 dm33??

Solution:Solution:

4.50 g 4.50 g = 1.80g/dm= 1.80g/dm33

2.50 dm2.50 dm33


If the density of helium is 0.179 g/dmIf the density of helium is 0.179 g/dm33 at STP at STP (273 K and 101.3 kPa), what is its density at (273 K and 101.3 kPa), what is its density at 99.0 kPa and 27 99.0 kPa and 27 ooC?C?

How to solve…How to solve… New density = old density x pressure ratio (new/old) x temp. ratio New density = old density x pressure ratio (new/old) x temp. ratio

(old/new)(old/new)

(direct relationship)(direct relationship) (inverse (inverse relationship)relationship) OldOld NewNew What happens to What happens to

the gas density?the gas density?

Pressure (P)Pressure (P)

Temp (T)Temp (T)

Density (D)Density (D)


SolutionSolution New density = (0.179 g/dmNew density = (0.179 g/dm33)x(99 )x(99

kPa/101.3 kPa)x(273 K/300 K) kPa/101.3 kPa)x(273 K/300 K)

= = 0.159 g/dm0.159 g/dm33

Graham’s LawGraham’s Law

Only a few physical properties of gases Only a few physical properties of gases depend on the identity of the gas. depend on the identity of the gas.

DiffusionDiffusion - The rate at which two - The rate at which two gases mix. gases mix.

Effusion Effusion - The rate at which a gas - The rate at which a gas escapes through a pinhole into a escapes through a pinhole into a vacuum. vacuum.

Graham’s Law of DiffusionGraham’s Law of Diffusion

The rate at which gases diffuse is The rate at which gases diffuse is inversely proportional to the square inversely proportional to the square root of their densities, or …root of their densities, or …

Graham’s Law of DiffusionGraham’s Law of Diffusion

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

Graham’s Law of EffusionGraham’s Law of Effusion

The rate of The rate of effusioneffusion of a gas is of a gas is inversely proportional to the square inversely proportional to the square root of either the density or the root of either the density or the molar mass of the gas. molar mass of the gas.

Graham’s Law of Diffusion - Graham’s Law of Diffusion - restatedrestated

Graham’s Law of Diffusion - Graham’s Law of Diffusion - ExampleExample

Example 1. What is the ratio of the Example 1. What is the ratio of the velocity of helium atoms to the velocity velocity of helium atoms to the velocity of radon atoms when both gases are at of radon atoms when both gases are at the same temperature?the same temperature?

Solution: the ratio of VSolution: the ratio of V11 to V to V22 = = 7.45:1 7.45:1

Upcoming Dates to Keep in Upcoming Dates to Keep in MindMind

Graham’s Law Lab –Tuesday, April 10Graham’s Law Lab –Tuesday, April 10thth 20122012

Chapter 9 Test – Thursday, April 12Chapter 9 Test – Thursday, April 12thth 20122012– Assignment #2 is due that day!!Assignment #2 is due that day!!

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