Chapter 5: Gases

Chapter 5: Gases

CHE 123: General Chemistry IDr. Jerome Williams, Ph.D.

Saint Leo University

Overview

• Gas Laws• Ideal Gas Law

– Example Problems

• General Gas Law– Example Problem

Boyle’s LawRobert Boyle (1627–1691)

3Tro: Chemistry: A Molecular Approach, 2/e

• Pressure of a gas is inversely proportional to its volume constant T and amount of gas graph P vs V is curve graph P vs 1/V is straight line

• As P increases, V decreases by the same factor

• P x V = constant

• P1 x V1 = P2 x V2

Boyle’s Experiment, P x V




Boyle’s Law: A Molecular View• Pressure is caused by the molecules striking the sides

of the container• When you decrease the volume of the container with

the same number of molecules in the container, more molecules will hit the wall at the same instant

• This results in increasing the pressure


P1 ∙ V1 = P2 ∙ V2

Example 5.2: A cylinder with a movable piston has a volume of 7.25 L at 4.52 atm. What is the volume at 1.21 atm?

because P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does

V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm

V2, L

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V1, P1, P2 V2


Practice – A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the

balloon is now 2.78 x 103 mL, what was it originally?


P1 ∙ V1 = P2 ∙ V2 , 1 atm = 760 torr (exactly)

A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon is now 2.78x 103

mL, what was it originally?

because P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does

V2 =2780 mL, P1 = 782 torr, P2 = 0.500 atm

V1, mL

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V2, P1, P2 V1


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Charles’s LawJacques Charles (1746–1823)


• Volume is directly proportional to temperatureconstant P and amount of gasgraph of V vs. T is straight line

• As T increases, V also increases

• Kelvin T = Celsius T + 273

• V = constant x Tif T measured in Kelvin

If the lines are extrapolated back to a volume of “0,” they all show the same temperature, −273.15 °C, called absolute zero

If you plot volume vs. temperature for any gas at constant pressure, the points will all fall on a straight line


Example 5.3: A gas has a volume of 2.57 L at 0.00 °C. What was the temperature in Celsius at 2.80 L?

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because T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does

V1 =2.57 L, V2 = 2.80 L, t2 = 0.00 °C

t1, K and °C

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V1, V2, T2 T1

Tro: Chemistry: A Molecular Approach, 2/e

Practice – The temperature inside a balloon is raised from 25.0 °C to 250.0 °C. If the volume of cold air was

10.0 L, what is the volume of hot air?


15

The temperature inside a balloon is raised from 25.0 °C to 250.0 °C. If the volume of cold air was 10.0 L, what is the

volume of hot air?

when the temperature increases, the volume should increase, and it does

V1 =10.0 L, t1 = 25.0 °C L, t2 = 250.0 °C

V2, L

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V1, T1, T2 V2

Tro: Chemistry: A Molecular Approach, 2/e

Avogadro’s LawAmedeo Avogadro (1776–1856)


• Volume directly proportional to the number of gas moleculesV = constant x nconstant P and Tmore gas molecules = larger

volume

• Count number of gas molecules by moles

• Equal volumes of gases contain equal numbers of moleculesthe gas doesn’t matter

Example 5.4:A 0.225 mol sample of He has a volume of 4.65 L. How many moles must be added to give 6.48 L?

because n and V are directly proportional, when the volume increases, the moles should increase, and they do

V1 = 4.65 L, V2 = 6.48 L, n1 = 0.225 mol

n2, and added moles

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V1, V2, n1 n2


Ideal Gas Law

• Ideal gases obey an equation incorporating the laws of Charles, Boyle, and Avogadro.

P V = n R T

• 1 mole of an ideal gas occupies 22.414 L at STP

• STP conditions are 273.15 K and 1 atm pressure

• The gas constant R = 0.08206 L·atm·K–1·mol–1

Example 5.6: How many moles of gas are in a basketball with total pressure 24.3 psi, volume of 3.24 L at 25°C?

1 mole at STP occupies 22.4 L, because there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas

V = 3.24 L, P = 24.3 psi, t = 25 °C n, mol

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

P, V, T, R n


Example Problems

• Sulfur hexafluoride (SF6) is a colorless, odorless, very

unreactive gas. Calculate the pressure (in atm)

exerted by 1.82 moles of the gas in a steel vessel of

volume 5.43 L at 69.5°C.

– Answer = 9.42 atm

Example Problems

• What is the volume (in liters) occupied by 7.40 g of

CO2 at STP?

– Answer = 3.77 L

Ideal Gas Law: Application

• Density and Molar Mass Calculations

• Given the following: Note: MM = molar mass

– Density d = mass/volume – No. moles n n = mass / MM

• Show that MM = dRT / P

Ideal Gas Law: Application Examples

• What is the molar mass of a gas with a density of

1.342 g/L–1 at STP?

• What is the density of uranium hexafluoride, UF6,

(MM = 352 g/mol) under conditions of STP?

• The density of a gaseous compound is 3.38 g/L–1 at 40°C and

1.97 atm. What is its molar mass?

Ideal Gas Law: Application Answers

• Molar mass = 30.08 g/mole

• Density of UF6 = 15.7 g/mL

• Molar mass = 44.09 g/mole

General Gas Law Equation

• (P1V1 / T1 )= (P2V2 / T2 )

• Only equation that uses two sets of data for V, P, and T.

Example Problem

• Oxygen gas is normally sold in 49.0 L steel containers at a pressure of 150.0 atm. What volume would the gas occupy if the pressure was reduced to 1.02 atm and the temperature raised from 20oC to 35oC?

Chapter 5: Gases

Documents

molecular approach

volume decreases

factorp x v

v decreases

molecular viewpressure

gasgraph of v

gasgraph p vs v

constant x tif t