Measurements of Pressure

ì Pressurecanhaveavarietyofmethodsofmeasurement

ì Theseincludetheatmosphere(atm),thekilopascal(kPa),mmofMercury(mmofHg),milibars(mb),andpoundspersquareinch(psi)

ì YoumustknowhowtoconvertbetweenkPa/psi,mmHg/atm,andpsi/atm

Conversion Factors

ì Herearetheconversionfactorsforthevariouspressureunits:

ì 1atm=760mmofHg=14.696psi=1.103bar=101325Pascal=101.3Kpa=760Torr

ì BasicallymmofHgisthesameasTorr

ì AndKpaisjust1000xlargerthanPa.

Try the following

ì Convertthefollowingì 1atmtommofHgì 500mmofHgtoatmì 400TorrtommofHgì 300kpatoatmì 3atmtommofHg

CHAPTER 14 –GASES &

ATMOSPHEREMr.Yeung

Objective

■ 3 basic laws

– Boyle’s law– Charles’ law– Gay-Lussac’s law

Combining the above laws– Combined gas law– Ideal gas law

Intro

■ Gases– There are more

spaces between the particles in a gas than liquids and solids

Thinking in gases■ In gases, think of the kinetic theory– The movement of the molecules has the ability to ■ Expand (heat up a balloon)

■ Take shape (use gas to fill up any thing)■ Change in volume (change the pressure of a balloon)

Four variables used to describe a gas■ Pressure (P) in kilopascals, or others

■ Volume (V) in liters

■ Temperature (T) in kelvin

■ Number of moles (n)

Gases at STP (Standard Temperature Pressure)■ When gases are said to be at STP,– it means it is at 0°C (273K), 101.3kPa or 1atm.

Overall gas laws

■ Thereare4gaslaws:– Boyles’ law– Charles’ law– Gay-lussac’slaw– Combinedgaslaw– Idealgaslaw(involvesmoles)

Boyle’s Law

Variables: Pressure & VolumeConstant variable: Temperature

P1 V1 = P2 V2

Charles’ Law

Variables: Temperature & VolumeConstant variable: PressureTemperature must be in Kelvin!

𝑉1𝑇1= 𝑉2

𝑇2

GayLussac’s Law

Variables: Pressure & TemperatureConstant variable: VolumeTemperature in Kelvin𝑃1𝑇1= 𝑃2

𝑇2

Boyle’s law■ Hestates“foragivenmassofgasat

constanttemperature,thevolumeofthegasvariesinverselywithpressure.”

P1 V1 =P2 V2

– P=Pressure– V=Volume■ Assumingthesameamountofmols and

atthesametemperature■ Inverselyproportionalmeansthehigher

thepressure,thelowerthevolume

Note: There is still the same number of gas molecules from the first one to the other. The number of gas molecules do not

change, it is the space that the molecules can move around decreased (with added pressure!

Increase in pressure = low volume

Low pressure = high volume Knowthisgraphicalrelationship

inBoyle’slaw!!!

Graphing pressure vs 1/volume gives a linear relationshipWhy is it a linear relationship? (straight

line?)

Thegraphispressureandtheinverseofvolume.Sincetherelationshipbetweenpressureandvolumeisan“INVERSE” relationship,graphingtheinverseofvolumewillgiveyoua

straightline.

Example■ A balloon contains 30.0L of helium gas at 103

kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa?

– P1 = 103 kPa– V1 = 30.0 L– P2 = 25.0 kPa

■ Unknown: V2 = ?L

Answer

■ V2 = (V1 * P1)/P2

■ V2 = (30.0L * 103kPa)/25.0kPa

= 1.24 X 102 L

Charles’ Lawü You buy a helium balloon for your lovely girlfriend

for Valentines day. You take the balloon outside into the -30C temperature, and the balloon shrunk half its size. You walk back to the store and see the balloon in it’s regular size. What just happened?

Charles lawüAsthetemperatureoftheenclosedgasincreases,thevolumeincreases,ifthepressureisconstant (same).■ Sothehotteritgets,themore volume(morespacethegasmoleculeshastomovearound)youget…andviceversa■ Orthecolder itgets,themoleculesdistancewithineachotherdecreases,thesizeoftheballoonshrinks.

Know this graphical

representation!

Charles Law■ Symbolic

– V1 = V2T1 T2

V = volume (L)T = Temperature (K)

Why does the temperature have to be in Kelvin?

■ Whenhecollectedhisresults,heextrapolated (extended)andfoundthatwhenthevolumewasequaledto0L,itwasat-273.15degreesCelsius.

■ Converting Celsius to Kelvin 273K = 0C

So basically just add 273 to get Kelvin from Celsiusa) 15C b) 45Cc) -23C d) 0C

288K 318K 230K 273K

DO NOT USE Unit ANALYSIS for this because Celsius and Kelvin are not

proportional to each other. If you increase 5 degrees C, it does not

increase 5 degrees Kelvin!

Problem■ A balloon inflated in a room at 24°C has a volume of

4.00L. The balloon is then heated to a temperature of 58C.

■ Whatisthenewvolumeifthepressureremainsconstant?

Answer■ V1 = 4.0L

■ T1 = 24C

■ T2 = 58C V2 = ?

■ ConverttoKelvin–– T1=24+273=297K– T2=58+273=331K■ Use Charles’ equation■ V2 = (V1 * T2)/T1■ V2 = (4.0L * 331K)/297K

=4.46L

Gay-Lussac’s Law■ Gay-Lussac's Law states: that the pressure of a

sample of gas at constant volume, is directly proportional to its temperature in Kelvin.

■ Increased temperature = increased pressure■ Constant variable: Volume (doesn’t change), so

think about a enclosed can, the volume doesn’t change when heated.

Gay-Lussac’s law

■ Equation:

■ Examples:■ Exploding beer kegs, aerosol cans, or soda

cans in the heat. €

P1T1

=P2T2

Gay-Lussac law example

■ Example: 10.0 L of a gas is found to exert 97.0 kPa at 25.0°C. What would be the required temperature (in Celsius) to change the pressure to standard pressure keeping a constant volume?

P1: 97.0 kPa T1: 25C + 273 = 298

P2: 101.3 kPa T2:

€

P1T1

=P2T2

Change your temperatures to

Kelvin!

€

97.0298

=101.3T2

T2=311.21K

Combined Gas Law

■ Consider the following: If the top diagram is similar to a see-saw…

■ What happens when P is held constant? T and V move together (Proportional)

■ What happens when T is held constant? P moves up and V moves down and vice versa (Inverse relationship)

■ The combined gas law can be expressed as:

■ P = Pressure■ V = in Litres■ T = in KelvinHere’s an example:

“Ifagasoccupiesavolumeof25Lat25ºCand1.25atm,calculatethevolumeat128ºCand0.75atm”

€

P1V1T1

=P2V2T2

Step1:ChangetemptoKelvin25ºC+273=298K128ºC+273=401KStep2:SolveforV2 usingtheCGLV2 =P1V1T2 =(1.25atm)(25L)(401K)

P2T1 (0.75atm)(298K)

V2 =56.1L