1. Katie Hillman Michael Griffin
Ideal Gas Law
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Katie Hillman
Michael Griffin

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3. PV = nRT Ideal Gas Law equation
P – pressure
V – Volume
n – number of moles
R – ideal gas constant
T – temperature
Use the equation to find an unknown value

4. Pressure Pressure Conversions (all are equal/interchangeable)
3 units of pressure
Atm (Atmospheres)
kPa (Kilopascals)
Torrs /mm of Hg (millimeters of Mercury)
Pressure Conversions (all are equal/interchangeable)
101.3 kPa
760 mm of Hg/torrs
1 atm

5. Volume
3 Units of Volume
L (Liters)
mL (Milliliters)
Cm3 (Centimeters Cubed)
Volume Conversions (all are equal/interchangeable)
1 liter
1000 mL
1000 cm3
1 mL=1 cm3

6. Number of Moles (n) If given moles, use moles
If given grams, convert to moles
Click here for mole converter/factor label

7. Gas Constant (R)
8.31 kPa (Kilopascals)
for atm (atmospheres)

8. Temperature (T) 3 main units of Temperature Temperature Conversions
Fahrenheit (F)
Celsius (C)
Kelvin (K) (MUST USE for gas equations)
Temperature Conversions
Kelvin=C+273
Celsius=(5/9)(F-32)
Fahrenheit=(9/5)(C)+32

9. Factor Label Grams to Moles
Take value given in grams and multiply by 1 mole over atomic mass of element or compound
Example: value given is 10 grams of oxygen
(10 grams)(1 mole/16 grams)
Grams cancel out and the unit is moles
Answer: moles
Click here to get back to mole slide

10. Practice Problems
1) Given the following sets of values, calculate the unknown quantity.
P = 1.01 atm
V = ?
n = mol
T = 25°C
Click for answer
2) Given the following sets of values, calculate the unknown quantity. (Hint: Pressure will be in atm)
P = ?
V= L
n = mol
T = 311 K
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11. Practice Problems (cont.)
3) At what temperature would 2.10 moles of N2 gas have a pressure of 1.25 atm
and in a 25.0 L tank?
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4) What volume is occupied by 5.03 g of O2 at 28°C and a pressure of 0.998atm?
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12. Question #5 Click for answer
Calculate the pressure in a 212 Liter tank containing 23.3 kg of argon gas at 25°C?
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13. Answer to Question #1
Temperature is 25 degrees Celsius plus 273 to equal 298 degrees Kelvin
(1.01 atm) V= ( mol)(0.0821katm)(298 K)
Answer is 0.20 Liters
Click here to go to Question #2

14. Answer to Question #2 P (0.602 L)=(0.00801 mol)(0.0821 atm)(311 K)
Answer is atm
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15. Answer to Question #3 (1.25atm)(25.0L)=(2.10m)(0.0821 atm) T
Answer is 181 degrees Kelvin
Click here to go to Question #4

16. Answer to Question #4 (5.03 grams)(1mole/32 grams)=0.157 moles
32 grams came from atomic number of oxygen, 02 so multiply atomic mass by is 32.
28 degrees Celsius plus 273 equals 301 degrees Kelvin
(0.998atm)V= (0.157m)( atm)(301 K)
Final answer is 3.89 Liters
Click here to go to Question #5

17. Answer to Question #5 (23.3 grams) (1mole/40 grams) =582.5 moles
25 degrees Celsius plus 273 equals 298 degrees Kelvin
P (212 L)=(582.5m)(0.0821atm)(298 K)
The final answer is 67.3 atm

18. THE END!!!!! CONGRATS FOR MAKING IT THROUGH ALL THIS TROUBLE!!