The document provides information about a chemistry lesson on gas laws and the relationship between temperature and pressure. It includes objectives, examples, and questions for students. The lesson explains that temperature and pressure are directly related, as temperature increases so does molecular motion and collisions, thus increasing pressure. Examples show how to use the direct relationship equation to calculate new pressure or temperature given one variable.

1. Launch: 2/10
 Grab your binder and immediately take a seat!
 Place launch paper on your desk.
 Yesterday’s Benchmark Data:
96%!
 Today’s Objectives: I can apply the gas laws to relations
between temperature and pressure.

2. Launch 2/10
1. What is the equation for converting
between the Celcius and Kelvin scales?
Label the variables.

3. Launch 2/10
2. Molecules in a substance at 50°C
a. move slower than molecules in a substance at 25°C
because they have less kinetic energy
b. move slower than molecules in a substance at 25°C
because they have more kinetic energy
c. move faster than molecules in a substance at 25°C
because they have less kinetic energy
d. move faster than molecules in a substance at 25°C
because they have more kinetic energy

4. Launch 2/10
3. What is the equivalent of 423K in degrees
Celsius?
a. -223°C
b. -23°C
c. 150°C
d. 696°C

5. Launch 2/10
4. Convert -50°C to Kelvin.
a. -223 K
b. -50 K
c. 223 K
d. 323 K

6. Announcements
 Great job yesterday!
 No quiz on Friday
 Class points end on Friday
 Only 3 days until break…
 Let’s finish strong

7. February Calendar
Quiz!
Today No quiz!
Exam!

8. Opening
 Today’s Objective: I can apply the gas laws to
relations between temperature and pressure.
 Standard – CH.4.c
 Today’s Questions: How are P and T related?

9. How are P & T related?
Mr. Heffner
2/10/10

10. How are P & T related?
 Temperature is…
 a measure of movement
 °C & K
 Pressure is…
 a measure of random collisions
 atm & mmHg

11. How are P & T related?
temperature = collisions = pressure
temperature = pressure

12. How are P & T related?
 Pressure & Temperature are…
 directly related change in the
same direction
 In math terms…
P
= constant
T
“some number”

13. How are P & T related?
 By rearranging the relationship, we can get an equation
P1 P2
starting
conditions T1 = constant = T ending
conditions
2
P1 P2
=
T1 T2

14. How are P & T related?
 There is a 3 step process for solving gas equations:
1. Write down the equation and given conditions
2. Plug in for the variables
3. Solve for the unknown

15. Example #1
At 298K (room temp), the pressure inside an empty Coke
can is 1 atm. It is then thrown into a bucket of ice at
273K. What is the new gas pressure?
 Step 1: Write down the equation and given conditions
P1 = 1 atm
P1 P2
= T1 = 298K
T1 T2
P2 = ?
T2 = 273K

16. Example #1
At 298K (room temp), the pressure inside an empty Coke
can is 1 atm. It is then thrown into a bucket of ice at
273K. What is the new gas pressure?
 Step 2: Plug in for the variables
P1 = 1 atm
1 P1
atm P2
= T1 = 298K
T1 T2
298K 273K
P2 = ?
T2 = 273K

17. Example #1
At 298K (room temp), the pressure inside an empty Coke
can is 1 atm. It is then thrown into a bucket of ice at
273K. What is the new gas pressure?
 Step 3: Solve for the unknown cross-
1 atm P2
multiply
=
✕
298K 273K
(P2)(298K) = (1 atm)(273K)
298K 298K
P2 = .92 atm

18. Example #2
The pressure of oxygen gas at 300K is reduced to from
760mmHg to 380mmHg. What is the final temperature?
 Step 1: Write down the equation and given conditions
P1 = 760 mmHg
P1 P2
= T1 = 300K
T1 T2
P2 = 380 mmHg
T2 = ?

19. Example #2
The pressure of oxygen gas at 300K is reduced to from
760mmHg to 380mmHg. What is the final temperature?
 Step 2: Plug in for the variables
P1 = 760 mmHg
P1 P2P2
760mmHg 380mmHg
= T1 = 300K
T
300K 1 T2T?
2 P2 = 380mmHg
T2 = ?

20. Example #2
The pressure of oxygen gas at 300K is reduced to from
760mmHg to 380mmHg. What is the final temperature?
 Step 3: Solve for the unknown
760mmHg 380mmHg
300K
✕= T2
1
(380mmHg)(300K) = (760mmHg)(T2)
760mmHg 760mmHg
2 150K = T2

21. Example #3
The pressure of Cl2 gas is 1 atm at 50K. What would the
final pressure be if the temperature was raised to 100K?
 Step 1: Write down the equation and given conditions
P1 = 1 atm
P1 P2
= T1 = 50K
T1 T2
P2 = ?
T2 = 100K

22. Whiteboards
 Work in pairs
 Trade-off marker every question
 You have 60 seconds to calculate the answer
 Show all of your work!
 Don’t forget units!
 Lift board only when prompted

23. The temperature of a gas is
raised by 50K. Draw an
arrow (up/down) to show
how pressure would be
affected.

24. The pressure of a gas is
lowered to .5 atm from STP .
Draw an arrow showing
what would happen to the
temperature.

25. A gas is sealed in a bottle at
400K and 2 atm. If put
into a freezer at 200K, what
would the new pressure be?

26. The pressure of CH4
(methane) gas at 10K is
increased from 2 atm to
10 atm. What is the new
pressure?

27. Closing
 Today’s Objective: I can describe how the random
collisions between gas molecules creates pressure
on a surface.
 Standard – CH.4.a
Molecular Motion Applet

28. Exit Slip
1. What is pressure?
a. a force that comes from the interaction of molecules in
a chemical bond
b. the amount of kinetic energy in a substance
c. a quantity that can be measured in mm
d. the force that a gas exerts on an object

29. Exit Slip
2. Which of the following factors effect the
force that is exerted on an object?
# of gas molecular
collisions temperature speed
I II III
a. I only
b. II only
c. II & III
d. I, II and III

30. Exit Slip
3. Which of the following relationships is
correct?
a. an increase in temperature  an increase in
pressure
b. a decrease in temperature  an increase in kinetic
energy
c. an increase in temperature  a decrease in pressure
d. a decrease in temperature  a increase in the
number and speed of collisions

31. Exit Slip
4. A soccer ball is inflated to a certain pressure
and is then heated in an oven. As it warms
up the pressure increases because
a. air molecules hit the walls of the ball less frequently
b. rubber in the soccer ball reacts with the air inside the
ball
c. air molecules speed up and collide with the wall of the
ball more often
d. air molecules diffuse quickly out of the ball

32. Exit Slip
5. Would you expect a gas at 100°C to have a
higher pressure than a gas at 0°C?
a. No, because it has less kinetic energy
b. No, because it has a higher temperature
c. Yes, because it has more random collisions
d. Yes, because it has less kinetic energy

33. Homework
 Finish practice questions
 Study for the quiz!