

11.31 Gas Laws (Combined Reteach) Notes
Presentation
•
Science
•
9th - 12th Grade
•
Medium
+1
Standards-aligned
Stacy King
Used 4+ times
FREE Resource
16 Slides • 7 Questions
1
Gas Laws
Class Announcements:
8 days to makeup or retake exams!
School Announcements:
May 15th Early Release
Boyle's Law
Charles’s Law
Combined, Avogadro's, Ideal
Dalton's Law Partial Pressures
2
Standard Temperature and Pressure
STP
temperature of 0.00°C or 273 K and a pressure of 1.00 atm.
3
Multiple Choice
Standard temperature is defined as
0.0 K
273 K
273°C
4
Multiple Choice
Standard pressure is defined as
1.00 atm
870 mmHg
0.00 atm
5
Combined Gas law
The combination of Boyle's Law and Charles's Law is called the Combined Gas Law
6
Combined Gas law Variables
| Symbol | Units |
|---|---|---|
Pressure | P | atm, Pa (kPa), torr, mmHg, Bar, psi, at |
Temperature | T | C converts to K |
Volume | V | L, mL, m3, cm3 |
Temperatures in oC +273.15 = K
ALWAYS check that your units match!
If P1 is in atm P2 also needs to be in atm
7
Key step Identifying Variables Combined Gas Law
A helium balloon has a volume of 48.7 L at 37°C and 700 Pa . What Volume will it have at 970 Pa and 87oC?
Step one Identify all the variables
V1=
P1=
T1=
V2=
P2=
T2=
8
Labelling
Identify the variables in the question.
T1
T2
V2
V1
P2
P1
9
Don't Forget
Convert oC!!!
C + 273.15 = K
10
Combined Gas Law
A helium balloon has a volume of 48.7 L at 37°C and 700 Pa . What Volume will it have at 970 Pa and 87oC?
Step one Identify all the variables
V1= 48.7 L
P1= 700 Pa
T1= 37 oC +273.15 = 310.15 K
V2= X
P2= 970 Pa
T2= 87 oC + 273.15 = 360.15 K
11
Manipulate the original equation and plug in your values.
Step 1
Step 2
Step 3
Final EQ
12
plug in your variables.
Final EQ
V1= 48.7 L
P1= 700 Pa
T1= 37 oC +273.15 = 310.15 K
V2= X
P2= 970 Pa
T2= 87 oC + 273.15 = 360.15 K
13
Multiple Choice
A helium balloon has a volume of 48.7 L at 37°C and 700 Pa . What Volume will it have at 970 Pa and 87oC?
40.81
392664
1
14
Fill in the Blanks
15
Combined Gas Law
The volume of a gas at 45°C and 0.678 atm is 180 mL. What Volume will it have at standard Temperature and a new volume of 79 mL?
Step one Identify all the information we have
V1=
P1=
T1=
V2=
P2=
T2=
16
Combined Gas Law
The volume of a gas at 45°C and 0.678 atm is 180 mL. What Pressure will it have at standard Temperature and a new volume of 79 mL?
Step one: Identify all the information we have
V1= 180 mL
P1= 0.678 atm
T1= 45 C + 273.15 = 318.15 K
V2= 79 mL
P2=
T2= 273 K
17
Manipulate the original equation and plug in your values.
Step 1
Step 2
Step 3
Final EQ
18
plug in your values.
Final EQ
V1= 180 mL
P1= 0.678 atm
T1= 45 C + 273.15 = 318.15 K
V2= 79 mL
P2=
T2= 273 K
19
Fill in the Blanks
20
Solving for T2
21
Manipulate the original equation and plug in your values.
Step 1
Step 2
Step 3
Final EQ
Step 4
Step 5
22
Manipulate the original equation and plug in your values.
Final EQ
23
Open Ended
The initial temperature of a 1.00 L sample of argon is 20.0° C. The pressure is decreased from 720.0 mm Hg to 360.0 mm Hg, and the volume increases to 2.14 liters. What was the change in temperature of the argon?
Gas Laws
Class Announcements:
8 days to makeup or retake exams!
School Announcements:
May 15th Early Release
Boyle's Law
Charles’s Law
Combined, Avogadro's, Ideal
Dalton's Law Partial Pressures
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