Katie Hillman Michael Griffin Ideal Gas Law Click this arrow to go to next slide Katie Hillman Michael Griffin

Navigation Click on the leftward arrow to go back one slide Click on the rightward arrow to go forward one slide If you get lost at anytime, and feel as if you’ll explode with confusion, just click on the explode button below to go back to this slide

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

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

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

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

Gas Constant (R) 8.31 kPa (Kilopascals) 0.0821 for atm (atmospheres)

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

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: 0.625 moles Click here to get back to mole slide

Practice Problems 1) Given the following sets of values, calculate the unknown quantity. P = 1.01 atm V = ? n = 0.00831 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= 0.602 L n = 0.00801 mol T = 311 K Click for answer

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? Click for answer 4) What volume is occupied by 5.03 g of O2 at 28°C and a pressure of 0.998atm? Click for answer

Question #5 Click for answer Calculate the pressure in a 212 Liter tank containing 23.3 kg of argon gas at 25°C? Click for answer

Answer to Question #1 Temperature is 25 degrees Celsius plus 273 to equal 298 degrees Kelvin (1.01 atm) V= (0.00831 mol)(0.0821katm)(298 K) Answer is 0.20 Liters Click here to go to Question #2

Answer to Question #2 P (0.602 L)=(0.00801 mol)(0.0821 atm)(311 K) Answer is 0.340 atm Click here to go to Question #3

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

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 2. 16+16 is 32. 28 degrees Celsius plus 273 equals 301 degrees Kelvin (0.998atm)V= (0.157m)(0.0821 atm)(301 K) Final answer is 3.89 Liters Click here to go to Question #5

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

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