Mr. Shields
Regents Chemistry
U05 L05
1
Robert Boyle
PV=k
(constant T and n)
Robert Boyle (1627 – 1691)
-Boyle’s law (1662)
Expresses the relationship between
Pressure and Volume for gases
2
Developing Boyles Law
If volume is held constant how can we inc. Pressure?
Recall that the KMT assumes Pressure is created by the
Frequent collision of gas molecules with the wall of the
Container since …
Assumption 5 states that KE of a gas is directly
proportional to Temperature …
If T
then KE
Then as KE
and since P = F/A
(F = KE)
so does P
3
Developing Boyles Law
Since KE (1/2 mv2) increases with increasing T then so
Does molecular velocity
Molecules moving with a higher velocity (v) hit the
container walls more frequently and with higher force
than molecules with lower velocity and that’s what
creates Pressure (P=F/A).
Think of the force you would feel in a collision if
The car you’re in hits a wall while moving at 5 mph vs.
30 mph
4
Developing Boyles Law
So, The frequency of collision per unit area, and the force
of the collision will determine the final pressure.
So if P is increased by increasing the Temperature how
Can we explain Boyles Law which holds T constant as
Pressure is made to change by changing volume?
Let’s see how this works.
5
Developing Boyles Law
What do you think happens to the number of collisions
Per unit area of the container walls if we take the same
molecules from the larger cube and put them into the
smaller cube?
So what happens to P?
6
Developing Boyles Law
The relationship between P and V is known as
BOYLE’S LAW.
PV=k
P, V = variables
T, n = Constants
So Pressure x volume is always equal to the same value
To keep k constant P must increase the same number of
Times V decreases (and vice versa).
This is known as an inverse relationship
7
Developing Boyles Law
Let’s look at an example:
If P1 = 2 atm and the vol = 64cm3 then
PV=K
2atm x (64cm3)= k
so
k=128
What is P when the volume dec. to 8cm3 ?
PV =k
P(8) = 128
so P increases to 16 atm
(i.e V decreased 1/8th and P increases 8x)
8
What does a plot of this data looks like?
Below is a typical data table comparing P,V measurements:
Trial
P (atm)
V (cm3)
PV (atm*cm3)
1
4.0
2.5
10
2
2.0
5.0
10
3
1.0
10
10
4
0.5
20
10
A graph of this data looks
like this
An inverse relationship
P
V
Boyles law
9
Developing Boyles Law
If P1V1=K
and P2V2=k (i.e. different p and v values)
Then P1V1 = P2V2
Note: units on both sides of the
equation MUST BE the same!)
This is the form of Boyles law we will use to solve
Problems
If we know three of the variables we can calculate the
fourth:
For example if we wanted
to find the new volume
in the image to the right
it would be P1V1/P2 = V2
10
Boyles Law Problem
P 1 V 1 = P2 V 2
(P1 & V1)
1atm x 1L = 2atm x V2
(P2 & V2)
V2 = 0.5L
In the above figure what would P2 be if P1 is 1 atm & V1
is 5L (V2 is still 0.5L)?
P1V1=P2V2;
1atm(5L) = P2(0.5L);
P2 = (1x5)/0.5 = 10 atm
11
Problem 1:
A 250 ml sample of gas at 26 KPa is expanded to
1000 ml at Constant T.
What is the pressure exerted by the gas at the new
volume?
(note: n is not mentioned. This is common. If n is not
Mentioned assume it is constant)
P1v1 = P2V2
26KPa(250ml) = P2(1000ml)
P2 = 26KPa(250ml)/1000ml = 6.5KPa
12
Problem 2:
A 5.5L sample of gas at 700 torr is compressed to
250.0 ml at Constant T.
What is the pressure exerted by the gas at the new
volume?
P 1 V 1 = P2 V 2
700torr(5.5L) = P2(0.25L)
P2 = 700torr(5.5L)/0.25L = 15,400 torr
NOTE: Make sure you keep units the SAME!
How many atm is this?
20.26atm
13
PROBLEM 3:
A discarded spray paint can contains only a small
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 run
over by a garbage truck and flattened to a volume of
13.16 ml what is the pressure (in Pa & atm) in the can
Assuming there’s no leaks?
P 1 V1 = P2 V2
34,470Pa x 473.18ml = P2 x 13.16ml
P2 = 16310514.6 / 13.16 = 1,239,401 Pa = 12.23 atm
14