1Mr. ShieldsMr. Shields Regents Chemistry Regents Chemistry U05 L05 U05 L05

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Robert BoyleRobert Boyle

Robert Boyle (1627 – 1691)Robert Boyle (1627 – 1691)

-Boyle’s law (1662)Boyle’s law (1662)

Expresses the relationship between Expresses the relationship between Pressure and VolumePressure and Volume for gases for gases

PV=kPV=k (constant T and n)(constant T and n)

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Developing Boyles LawDeveloping Boyles Law

If volume is held constant how can we inc. Pressure?If volume is held constant how can we inc. Pressure?Recall that the KMT assumes Recall that the KMT assumes PressurePressure is created by the is created by theFrequent Frequent collisioncollision of gas molecules with the wall of the of gas molecules with the wall of theContainer since …Container since …

Assumption 5 states that KE of a gas is directlyAssumption 5 states that KE of a gas is directlyproportional to proportional to TemperatureTemperature … …

If T then KE and since P = F/A (F = KE)If T then KE and since P = F/A (F = KE)

Then as KE so does P Then as KE so does P

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Developing Boyles LawDeveloping Boyles Law

Molecules moving with a Molecules moving with a higher velocityhigher velocity (v) hit the (v) hit the container walls container walls more frequentlymore frequently and with and with higher forcehigher force than molecules with lower velocity and that’s whatthan molecules with lower velocity and that’s whatcreates Pressure (P=F/A).creates Pressure (P=F/A).

Think of the force you would feel in a collision ifThink of the force you would feel in a collision ifThe car you’re in hits a wall while moving at 5 mph vs.The car you’re in hits a wall while moving at 5 mph vs.30 mph30 mph

Since KE (1/2 mvSince KE (1/2 mv22) increases with increasing T then so) increases with increasing T then soDoes molecular velocity Does molecular velocity

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Developing Boyles LawDeveloping Boyles Law

So, So, The frequency of collision per unit areaThe frequency of collision per unit area, and the , and the forceforceof the collision will determine the final pressure. of the collision will determine the final pressure.

So if P is increased by increasing the Temperature how So if P is increased by increasing the Temperature how Can we Can we explainexplain Boyles Law which holds T constant as Boyles Law which holds T constant asPressure is made to change by changing volume? Pressure is made to change by changing volume?

Let’s see how this works.Let’s see how this works.

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Developing Boyles LawDeveloping Boyles LawWhat do you think happens to the What do you think happens to the number of collisionsnumber of collisions Per unit area of the container walls if we take the samePer unit area of the container walls if we take the samemolecules from the molecules from the larger cubelarger cube and put them into the and put them into thesmaller cubesmaller cube??

So what happens to P?So what happens to P?

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Developing Boyles LawDeveloping Boyles LawThe relationship between P and V is known as The relationship between P and V is known as BOYLE’S LAW.BOYLE’S LAW.

PV=kPV=k

So So Pressure x volumePressure x volume is always equal to the same value is always equal to the same value

To keep k constant P To keep k constant P mustmust increase the same number of increase the same number ofTimes V decreases (and vice versa). Times V decreases (and vice versa).

This is known as an This is known as an inverse relationshipinverse relationship

P, V = variables T, n = Constants

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Developing Boyles LawDeveloping Boyles Law

Let’s look at an example:Let’s look at an example:

If P1 = 2 atm and the vol = 64cmIf P1 = 2 atm and the vol = 64cm33 then then

PV=KPV=K 2atm x (64cm2atm x (64cm33)= k )= k soso k=128k=128

What is P when the volume dec. to 8cmWhat is P when the volume dec. to 8cm33 ? ?

PV =k P(8) = 128PV =k P(8) = 128 soso P increases to 16 atmP increases to 16 atm

(i.e V decreased 1/8(i.e V decreased 1/8thth and P increases 8x) and P increases 8x)

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Below is a typical data table comparing P,V measurements:

Trial Trial P (atm) P (atm) V (cmV (cm33) ) PV PV

(atm*cm(atm*cm33

) )

1 1 4.0 4.0 2.5 2.5 10 10

2 2 2.0 2.0 5.0 5.0 10 10

3 3 1.0 1.0 10 10 10 10

4 4 0.5 0.5 20 20 10 10

A graph of this data looks like this

An inverse relationshipAn inverse relationship

P V P V

Boyles lawBoyles law

What does a plot of this data looks like?What does a plot of this data looks like?

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Developing Boyles LawDeveloping Boyles LawIf PIf P11VV11=K and P=K and P22VV22=k (i.e. different p and v values)=k (i.e. different p and v values)

Then PThen P11VV11 = P = P22VV22

This is the form of Boyles law we will use to solveThis is the form of Boyles law we will use to solveProblemsProblems

If we know three of the variables we can calculate theIf we know three of the variables we can calculate thefourth:fourth:

For example if we wantedFor example if we wantedto find the new volumeto find the new volumein the image to the rightin the image to the rightit would be Pit would be P11VV11/P/P22 = V = V22

Note:Note: units on both sides of the units on both sides of theequation MUST BE the same!)equation MUST BE the same!)

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Boyles Law ProblemBoyles Law Problem

In the above figure what would PIn the above figure what would P22 be if P be if P11 is 1 atm & V is 1 atm & V11 is is 5L (V5L (V22 is still 0.5L)? is still 0.5L)?

PP11VV11=P=P22VV22; 1atm(5L) = P; 1atm(5L) = P22(0.5L); P(0.5L); P22 = (1x5)/0.5 = = (1x5)/0.5 = 10 atm10 atm

(P(P11 & V & V11))

(P(P22 & V & V22))

PP11VV11 = P = P22VV22

1atm x 1L = 2atm x V21atm x 1L = 2atm x V2

V2 = 0.5LV2 = 0.5L

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Problem 1:Problem 1:

A 250 ml sample of gas at 26 KPa is expanded toA 250 ml sample of gas at 26 KPa is expanded to1000 ml at Constant T.1000 ml at Constant T.

What is the pressure exerted by the gas at the newWhat is the pressure exerted by the gas at the newvolume?volume?

(note: n is not mentioned. This is common. If n is not(note: n is not mentioned. This is common. If n is notMentioned assume it is constant)Mentioned assume it is constant)

PP11vv11 = P = P22VV22 26KPa(250ml) = P26KPa(250ml) = P22(1000ml)(1000ml)

PP22 = 26KPa(250ml)/1000ml = 6.5KPa = 26KPa(250ml)/1000ml = 6.5KPa

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Problem 2:Problem 2:

A A 5.5L5.5L sample of gas at 700 torr is compressed to sample of gas at 700 torr is compressed to250.0 ml250.0 ml at Constant T. at Constant T.

What is the pressure exerted by the gas at the newWhat is the pressure exerted by the gas at the newvolume?volume?

PP11VV11 = P = P22VV22 700torr(5.5L) = P700torr(5.5L) = P22(0.25L)(0.25L)

PP22 = 700torr( = 700torr(5.5L5.5L)/)/0.25L0.25L = 15,400 torr = 15,400 torr

NOTE: Make sure you keep units the SAME! NOTE: Make sure you keep units the SAME!

How many atm is this?How many atm is this? 20.26atm20.26atm

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PROBLEM 3:PROBLEM 3: A discarded spray paint can contains only a smallA discarded spray paint can contains only a smallvolume of the propellant gas at a pressure of 34,470 Pa.volume of the propellant gas at a pressure of 34,470 Pa.The volume of the can is 473.18 ml. If the can is runThe volume of the can is 473.18 ml. If the can is runover by a garbage truck and flattened to a volume ofover by a garbage truck and flattened to a volume of13.16 ml what is the pressure (13.16 ml what is the pressure (in Pa & atmin Pa & atm) in the can) in the canAssuming there’s no leaks?Assuming there’s no leaks?

PP11VV11 = P = P22VV22

34,470Pa x 473.18ml = P34,470Pa x 473.18ml = P22 x 13.16ml x 13.16ml

PP22 = 16310514.6 / 13.16 = = 16310514.6 / 13.16 = 1,239,4011,239,401 Pa = Pa = 12.2312.23 atm atm