Good Morning!

1
Good Morning!

• Today we are going to
• Hand back work and tests.
• Show the answers for the Review problems.
• Turn in 12.4 12.5
• FYI
• you will be given the equations on the test
• You do not need to know what name goes with which
  law.
• There are several multiple choice that ask you
  about the how P,V, T relate to each other.
• And you will also need to complete calculations
  using the gas laws.

2
What is the volume of a sample of CO2 at STP that
has a volume of 75.0 mL at 30.0 C and 91 kPa?

• First you have to figure out what you have and
  what you are being asked to find , then choose
  the right law.
• So what do we have, what do we need to find, and
  which law do we need to use.

3
Make sure they are in the right units

• V1 0.075 L
• P1 91 kPa
• T1 303 K
• V2 ?
• P2 101.3 kPa
• T2 273 K
• Now which law will take all of this into account?

T2 P1 V1 P2 V2 T2
T1 T2
T2 P1 V1 P2 V2
T1 P2 V2
T2 P1 V1 P2 V2
T1P2 P2
T2 P1 V1 V2
T1 P2 V2
P1 V1 P2 V2
T1 T2
Now divide by P2
Now we can put the numbers in )
To do this we multiply by T2
First solve for V2
273K 91kPa 0.075 L V2
303 K 101.3 kPa V2
61 mL or 0.061 L V2
4
Chapter 12 Problems with Gas Law

• Lets get started.

5
State the equation for Boyles Law.

• P1V1 P2V2

6
2. Identify the quantity held constant in Boyles
Law.

• Temperature, its not present in the equation.

7
3. Draw the general shape of a Boyles Law graph.
8
A quantity of gas under a pressure of 50 kPa has
a volume of 565 cm3. The pressure is increased
to 450 kPa, while the temperature is kept
constant. What is the new volume?

• P1 50 kPa
• V1 0.565 L
• P2 450 kPa
• V2 ?
• Which Law do we need to use?

9
A quantity of gas under a pressure of 50 kPa has
a volume of 565 cm3. The pressure is increased
to 450 kPa, while the temperature is kept
constant. What is the new volume?

• P1V1 P2V2
• To solve for V2 divide both sides by P2

P1V1 P2V2
P2 P2
50kPa?565 cm3
450 kPa
V2 P1V1
V2 P2
62.7 cm3
10
State the equation for Charless Law.
V1 V2
T1 T2
11
6. What is the relationship between temperature
and volume?

• Inverse relationship or direct relationship

12
A gas sample has a volume of 25.0 L at a
temperature of 30.0 ?C 303 K. The temperature
is raised to 227 ?C 500 K while the pressure
remains unchanged. What is the new volume of the
gas?

• T1 303 K
• V1 25.0 L
• T2 500 K
• V2 ?
• Which Law do we need?

13
A gas sample has a volume of 25.0 L at a
temperature of 30.0 ?C 303 K. The temperature
is raised to 227 ?C 500 K while the pressure
remains unchanged. What is the new volume of the
gas?
T2 V1 T2 V2
T1 T2
V1 V2
T1 T2
by T2
T2V1 V2
T1 V2
500K?25.0 L
303 K
41.3 L
14
What volume will 0.375 mol of Oxygen take up at
STP?
0.375 moles O2
x 22.4 Liters O2
x 1 moles O2
8.4 liters O2
15
Determine the volume occupied by 0.582 mol of a
gas at 15?C if the pressure is 81.8 kPa.

• V ?
• n 0.582 mols
• T288 K
• P 81.8 kPa
• R 8.31

kPa L
mol K
16
Determine the volume occupied by 0.582 mol of a
gas at 15?C if the pressure is 81.8 kPa.
V P nRT
P P
V P nRT
kPa L
mol K
0.582 mol 8.31 288 K
81.8 kPa
nRT V
P V
V 17.0 L
17
A mixture of gases at a total pressure of 97
kPa contains N2, CO2, and O2. The partial
pressure of the CO2 is 24 kPa and the partial
pressure of the N2 is 48 kPa. What is the
partial pressure of the O2?

• Ptotal P1 P2 P3
• Ptotal - P1 - P2 P3
• 97kPa - 24kPa - 48 kPa P3
• 25 kPa P3

18
Using Graham's Law of Diffusion calculate the
rate of diffusion between methane CH4 (mass 16)
and hydrogen sulfide H2S (mass 34)If 2 gases
are at the same temp. they have the same kinetic
energy (KE)
19
First where does that equation come from?
No, you do not have to do this on the test, I
just wanted you to know that there is a reason
why the velocity is opposite of the mass.
Now you can express this in 2 different ways.
You could say that H2S is 68 slower than CH4 or
you could say CH4 is 146 faster than H2S
Finally, take the square root of both sides to
get the equation that you need to use.
If 2 gases are at the same temperature they have
the same KE. And the formula for KE is ½ mv2. So
we can sub in and get the equation above.
Divide both sides by the ½ and get rid of it.
Now divide both sides by the m1 so it can cancel
out on the left
Divide both sides by the V22 so it can cancel out
on the right
20
12. Calculate the relative rates of diffusion of
gaseous UF6 containing these isotopes Formula
mass of UF6 containing uranium - 235 349 amu.
Formula mass of UF6 containing uranium - 238
352 amu.

• Uranium 235 will reach you 1.0043 times as fast.