1. Ch. 14 Gas Laws
Mrs. Fox

2. Ideal Gas Law
PV=nRT P= pressure (atm) V= volume (L) n= number of moles (moles) R= (mol*atm/L*K) T= temperature (kelvin)

3. Also note: Temperature must be in Kelvin
To convert from Celsius to Kelvin
“convert from C add 273”
23 degrees Celsius = ___________ Kelvin
= 296K

4. Try some:
Calculate the number of moles of gas contained in a 3.0 L vessel at 300K with a pressure of 1.50 atm
0.2 mole
If the pressure exerted by a gas at 25°C in a volume of 0.044L is 3.81 atm, how many moles of gas are present?
6.9x10-3 mol
Determine the Celsius temperature of 2.49 moles of gas contained in a 1.00L vessel at a pressure of 143 kPa.
-266°C

5. Boyle’s Law Boyle Potatoes Violently Look at the Ideal Gas Law: PV=nRT
only use the symbols P & V
Make sure both are on the left of the equal sign
Set it equal to itself
P1V1=P2V2

6. Charles’s Law Charles The Velociraptor
Look at the Ideal Gas Law: PV= nRT
only use the symbols T & V
Make sure both are on the left of the equal sign
Set it equal to itself
V1 = V2
T1 T2

7. Gay-Lussac’s Law Gay-Lussac’s Potty Training
Look at the Ideal Gas Law: PV=nRT
only use the symbols P & T
Make sure both are on the left of the equal sign
Set it equal to itself
P1 = P2
T1 T2

8. Steps to figure out which gas law to use for each question:
Underline each variable in the question
There will be 2 circumstances of each (before and after)
Ignore all constant variables
Write the initial circumstances under “before”
Write final circumstances under “after”
You are solving for the unknown variable
Move your variables BEFORE you insert numbers

9. A sample of helium gas in a balloon is compressed from 4. 0L to 2
A sample of helium gas in a balloon is compressed from 4.0L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0L volume is 210 kPa, what will the pressure be at 2.5 L?
BEFORE
AFTER
V1= 4.0 L P1= 210 kPa
V2= 2.5 L P2 = ??
Which gas law uses V & P?
Boyle’s Law!! P1V1=P2V2
Solve for P2 first:
P2= P1V1/V2
Now plug in the numbers:
P2 = (210 x 4.0) / (2.5)
Solve:
340 kPa

10. A gas sample at 40. 0°C occupies a volume of 2. 32 L
A gas sample at 40.0°C occupies a volume of 2.32 L. If the temperature is raised to 75°C , what will the volume be, assuming the pressure remains constant?
BEFORE
AFTER
T1= 40.0°C + 273= 313K V1= 2.32 L
T2= 75°C = 348K V2= ??
Which gas law uses T & V?
Charles’s Law!! V1/T1= V2/T2
Solve for V2 first:
V2= (V1/T1)*T2
Now plug in the numbers:
V2 = (2.32/313) * (348)
Solve:
2.58L  2.6L

11. Gay-Lussac’s Law!! P1/T1= P2/T2
The pressure of a gas in a tank is 3.20 atm at 22.0°C. If the temperature rises to 60.0°C what will the pressure in the tank be?
BEFORE
AFTER
P1= 3.20 atm T1= = 295K
P2=? T2= = 333K
Which gas law uses T & P?
Gay-Lussac’s Law!! P1/T1= P2/T2
Solve for P2 first:
P2= ( P1/T1)/T2
Now plug in the numbers:
P2 = (3.2/295) / (333)
Solve:
3.61 atm

12. Combined Gas Law P1V1 = P2V2 T1 T2
Combines/uses all of the variables with the exception of n & R in the Ideal Gas Law
PV=nRT

13. BEFORE AFTER Solve for V2:
A gas at 110 kPa and 30.0°C fills a flexible container with an initial volume of 2.00L. If the temperature is raised to 80.0°C and the pressure increased to 440kPa, what is the new volume?
BEFORE
P1=110 kPa
T1= 30.0°C
V1= 2.00 L
AFTER
P2=440 kPa
T2= 80.0°C
V2= ??
Which Gas Law should we use?
Combined Gas Law!!
P1V1/T1= P2V2/T2
Solve for V2:
V2= V1(P1/T1)(T2/P2)
V2= 0.58 L

14. Avogadro’s Principle
Equal volumes of gasses at the same temperature and pressure contains equal number of moles
STP(standard temperature & pressure)
When volume is 22.4L
When pressure is 1 atm
When temperature is 0 degrees Celsius
If at STP, the number of moles is always 1 mole
Remember that 1 mole = 6.02x1023 particles
Therefore, at STP:
1 mole=22.4L
1 mole= 1 atm
1 mole = 0.0°C

15. Calculate the volume that 0.881 mole of gas at STP will occupy
Since at STP: T=0.0°C V= 22.4L P= 1 atm n= moles
0.881 mole x 22.4L = 19.7L 1 mole

16. Challenge question!
If you are given grams instead of moles, can you solve a volume at STP??
Calculate the volume that a 2.0kg of CH4 will occupy at STP:
Convert kg into grams
Convert grams into moles
Convert moles into L (1 mole= 22.4L)
2.0kg= 2000g
2000g CH4 x 1 mole = 125 mole
16.05g
125 mole x L = 2800L
1 mole