Unit 7 Gas Laws and Thermodynamics Chapters 11 & 16

Unit 7Gas Laws and Thermodynamics

Chapters 11 & 16

CHAPTER 11Gases

Pressure and Force

• Pressure is the force per unit area on a surface.

Pressure =Force

Area

Chapter 11 – Section 1: Gases and Pressure

Gases in the Atmosphere

• The atmosphere of Earth is a layer of gases surrounding the planet that is retained by Earth's gravity.

• By volume, dry airis 78% nitrogen, 21% oxygen, 0.9% argon, 0.04% CO2, and small amounts of other gases.


Atmospheric Pressure

• Atmospheric pressure is the force per unit area exerted on a surface by the weight of the gases that make up the atmosphere above it.


Measuring Pressure

• A common unit of pressure is millimeters of mercury (mm Hg).

• 1 mm Hg is also called 1 torr in honor of Evangelista Torricelli whoinvented the barometer (used tomeasure atmospheric pressure).

• The average atmospheric pressure at sea level at 0°C is 760 mm Hg, so one atmosphere (atm) of pressure is 760 mm Hg.


Measuring Pressure (continued)

•Pressure can also be measured in pascals (Pa): 1 Pa = 1 N/m2.

•One pascal is verysmall, so usually kilopascals (kPa) are used instead.

•One atm is equal to 101.3 kPa.

1 atm = 760 mm Hg (Torr) = 101.3 kPa


Units of PressureChapter 11 – Section 1: Gases and Pressure

Converting PressureSample Problem

The average atmospheric pressure in Denver, CO is 0.830 atm. Express this pressure in:

a. millimeters of mercury (mm Hg)

b. kilopascals (kPa)


0.830 atmatm

mm Hg1

760x =

0.830 atmatm

kPa1

101.3x =

631 mm Hg

84.1 kPa

Dalton’s Law of Partial Pressures

• Dalton’s law of partial pressures - the total pressure of a gas mixture is the sum of the partial pressures of the component gases.


PT = P1 + P2 + P3 …

Dalton’s Law of Partial PressuresSample Problem

A container holds a mixture of gases A, B & C. Gas A has a pressure of 0.5 atm, Gas B has a pressure of 0.7 atm, and Gas C has a pressure of 1.2 atm.

a.What is the total pressure of this system?

b. What is the total pressure in mm Hg?


PT = P1 + P2 + P3 …PT = 0.5 atm+ 0.7 atm + 1.2 atm = 2.4 atm

2.4 atmatm

mm Hg1

760x = 1800 mm Hg

Gases and Pressure• Gas pressure is caused by collisions of the gas

molecules with each other and with the walls of their container.

• The greater the number of collisions, the higher the pressure will be.

Chapter 11 – Section 2: The Gas Laws

Pressure – Volume Relationship

• When the volume of a gas is decreased, more collisions will occur.

• Pressure is caused by collisions.

• Therefore, pressure will increase.

• This relationshipbetween pressure andvolume is inversely proportional.


Boyle’s Law

• Boyle’s Law – The volume of a fixed mass of gas varies inversely with the pressure at a constant temperature.

• P1 and V1 representinitial conditions, andP2 and V2 representanother set of conditions.


P1V1 = P2V2

Boyle’s LawSample Problem

A sample of oxygen gas has a volume of 150.0 mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant?Solution:


P1V1 = P2V2(0.947 atm)(150.0 mL) = (0.987 atm)V2

V2 =(0.947 atm)(150.0 mL)

(0.987 atm)= 144 mL

Volume – Temperature Relationship

• the pressure of gas inside and outside the balloon are the same.

• at low temperatures, the gas molecules don’t move as much – therefore the volume is small.

• at high temperatures, the gas molecules move more – causing the volume to become larger.


• Charles’s Law – The volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature.

Charles’s LawChapter 11 – Section 2: The Gas Laws

V1 V2=T1 T2

• V1 and T1 represent initial conditions, and V2 and T2 represent another set of conditions.

The Kelvin Temperature Scale

• Absolute zero – The theoretical lowest possible temperature where all molecular motion stops.

• The Kelvin temperature scale starts at absolute zero (-273oC.)

• This gives the followingrelationship between the two temperature scales:


K = oC + 273

Charles’s LawSample Problem

A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant?Solution:


V1 V2=T1 T2

752 mL V2=298 K 323 K

K = oC + 273T1 = 25 + 273 = 298T2 = 50 + 273 = 323

752 mLV2 =298 K

323 Kx = 815 mL

Pressure – Temperature Relationship

• Increasing temperature means increasing kinetic energy of the particles.

• The energy and frequency of collisions depend on the average kinetic energy of the molecules.

• Therefore, if volume is kept constant, the pressure of a gas increases with increasing temperature.


Gay-Lussac’s Law

• Gay-Lussac’s Law – The pressure of a fixed mass of gas varies directly with the Kelvin temperature.

• P1 and T1 representinitial conditions.P2 and T2 representanother set of conditions.

P1 P2=T1 T2


The Combined Gas Law

• The combined gas law is written as follows:

• Each of the other gas laws can be obtained from the combined gas law when the proper variable is kept constant.


P1 P2=T1 T2

V1 V2

The Combined Gas LawSample Problem

A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C?Solution:


(1.08 atm) (0.855 atm)=

298 K 283 K

K = oC + 273T1 = 25 + 273 = 298T2 = 10 + 273 = 283

(1.08 atm)V2 =(298 K)

(283 K)= 60.0 L

P1 P2=T1 T2

V1 V2

(50.0 L) V2

(50.0 L)(0.855 atm)

Avogadro’s Law

• In 1811, Amedeo Avogadro discovered that the volume of a gas is proportional to the number of molecules (or number of moles.)

• Avogadro’s Law - equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, or:

V1 V2=n1 n2

Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law

Standard Molar Volume

• Standard Temperature and Pressure (STP) is 0oC and 1 atm.

• The Standard Molar Volume of a gas is the volume occupied by one mole of a gas at STP. It has been found to be 22.4 L.


Molar Volume Conversion Factor

• Standard Molar Volume can be used as a conversion factor to convert from the number of moles of a gas at STP to volume (L), or vice versa.


Molar Volume ConversionSample Problem

a. What quantity of gas, in moles, is contained in 5.00 L at STP?

b. What volume does 0.768 moles of a gas occupy at STP?


5.00 LL

mol22.41x = 0.223 mol

0.768 molmol

L1

22.4x = 17.2 L

Volume Ratios

• You can use the volume ratios as conversion factors just like you would use mole ratios.

2CO(g) + O2(g) → 2CO2(g)2 molecules 1 molecule 2 molecules2 mole 1 mole 2 mol2 volumes 1 volume 2 volumes

• Example: What volume of O2 is needed to react completely with 0.626 L of CO to form CO2?


0.626 L COL COL O2

21x = 0.313 L O2

The Mole Map

• You can now convert between number of particles, mass (g), and volume (L) by going through moles.


Gas StoichiometrySample Problem

Assume that 5.61 L H2 at STP reacts with excess CuO according to the following equation:

CuO(s) + H2(g) → Cu(s) + H2O(g)

a.How many moles of H2 react?

b.How many grams of Cu are produced?


5.61 L H2L H2

mol H2

22.41x = 0.250 mol H2

5.61 L H2L H2

mol H2

22.41x

mol H2

mol Cu1

1xmol Cu

g Cu1

63.5x = 15.9 g Cu

The Ideal Gas Law

• All of the gas laws you have learned so far can be combined into a single equation, the ideal gas law:

• R represents the ideal gas constant which has a value of 0.0821 (L•atm)/(mol•K).


PV = nRT

The Ideal Gas LawSample Problem

What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298 K?Solution:


PV = nRTP (10.0 L) = (0.500 mol)(0.0821 L•atm/mol•K)

P =(0.500 mol) (298 K)

(10.0 L)= 1.22

atm

(298 K)

(0.0821 L•atm/mol•K)

Diffusion and Effusion

• Diffusion is the gradual mixing of two or more gases due to their spontaneous, random motion.

• Effusion is the process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in the container.

Chapter 11 – Section 4: Diffusion and Effusion

Graham’s Law of Effusion

• Light molecules move faster than heavy ones. • Graham’s

law of effusion saysthe greaterthe molarmass of a gas,the slower itwill effuse.

Chapter 11 – Section 4: Diffusion and Effusion

CHAPTER 16Thermochemistry

Heat and Temperature

• Temperature – a measure of the average kinetic energy of the particles in a sample of matter.

• The greater the kinetic energy of the particles in a sample, the hotter it feels.

• Heat – energy transferred between samples of matter due to a difference in their temperatures.

• Heat always moves spontaneously from matter at a higher temperature to matter at a lower temperature.

Chapter 16 – Section 1: Thermochemistry

Measuring Heat

• Heat energy is measured in joules (or calories – food only)

• Chemical reactions usually either absorb or release energyas heat.

• The energy absorbed or released as heat in a chemical or physical change is measured in a calorimeter.


Specific Heat

• A quantity called specificheat can be used to compare heat absorptioncapacities for different materials.

• Specific heat – the amount of energy required to raise the temperature of one gram of a substance by 1°C or 1 K.

• Specific heat can be measured in units of J/(g•°C), J/(g•K), cal/(g•°C), or cal/(g•K).


Heat Transfer Equation

• Specific heat can be used to find the quantity of heat energy gained or lost with a change in temperature according to the following equation:

• Where the variables stand for the following:Q = heat transferred (joules or calories)m = mass (g)cp = specific heat∆T = change in temperature (oC or K)


Q = m x cp x ∆T

Heat Transfer EquationSample Problem

A 4.0 g sample of glass was heated from 274 K to 314 K, a temperature increase of 40. K, and was found to have absorbed 32 J of energy as heat.

a.What is the specific heat of this type of glass?

b. How much energy will the same glass sample gain when it is heated from 314 K to 344 K?


Q = m x cp x ∆T 32 J = (4.0 g)(cp) (40. K)

cp = (4.0 g)(40. K)

32 J = 0.20 J/(g•K)

Q = m x cp x ∆T Q = (4.0 g)(0.20 J/(g•K)) (30 K) = 24 J

Enthalpy of Reaction

• The enthalpy of reaction (H) is the quantity of energy transferred as heat during a chemical reaction.

• The change in enthalpy (∆H) of a reaction is always the difference between the enthalpies of the products and the reactants.


∆H = Hproducts - Hreactants

Exothermic Reactions

• In an exothermic reaction, energy is released. Therefore, the energy of the products must be less than the energy of the reactants, and ∆H is negative.

• The great majority ofchemical reactions innature are exothermic.


Endothermic Reactions

• In an endothermic reaction, energy is absorbed. Therefore, the energy of the products must be greater than the energy of the reactants, and ∆H is positive.


Entropy and Reaction Tendency

• Entropy – a measure of the degree of randomness in a system.

• Processes in nature are driven in two directions: towards decreasing enthalpy and towards increasing entropy.

• The combined enthalpy-entropy function is called the Gibbs free energy (G).


Hess’s Law

• Hess’s Law – the overall enthalpy change in a reaction is equal to the sum of enthalpy changes for the individual steps in the process.

• Possible steps in Hess’s Law:1. Reverse equation/

change sign on ΔH.2. Multiply or Divide

coefficients/multiply or divide ΔH.


Hess’s LawSample Problem

Calculate the enthalpy of formation for CH4:C(s) + 2H2(g) → CH4(g) ∆Hf = ?The component reactions are:C(s) + O2(g) → CO2(g) ∆Hc = -393.5 kJH2(g) + ½O2(g) → H2O(l) ∆Hc = -285.8 kJCH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ∆Hc = -890.8 kJCO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ∆Hc = +890.8 kJ- 2 moles of H2 are used to make CH4,

so multiply the 2nd equation by 2 (including ∆H.)- CH4 is on the products side, not the reactants side, so

reverse the 3rd reaction and change the sign on ∆H.- Cancel unwanted terms and add the ∆H’s.


2 2 -571.6 kJ

-74.3 kJ

-74.3 kJ

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Unit 7 Gas Laws and Thermodynamics Chapters 11 & 16