1. Basic Gas Laws (Combined and Partial Pressures Laws)

2. Steps for Solving ANY Gas Law Problem: 1.Write out a column of information down the left-hand side. Make sure all of your variable’s units match (i.e. if P 1 is in kPa, then P 2 must be in kPa as well). If one doesn’t match, convert it to match the other, using a conversion table. Put a question mark in the space for the variable you are trying to solve for (what you DON’T have). 2.Write the original equation for the gas law you will be using. 3.Rearrange the equation to solve for the variable you need. 4.Plug in the values and units you have in to the rearranged equation, and make sure all your units will cancel except for one. This will be the unit for your answer. 5.Calculate, then box your answer!

3. Part 5: The Combined Gas Law (1802) the Combined Gas Law takes Boyle’s, Charles’s, and Gay-Lussac’s Law and combines them into one. Ex4: A sample of gas took up 35.0L of space at 124  C and 84.56 kPa. If the temperature increased by 16  C and the pressure changed to 4.56 atm, what would the resulting volume be? Ex5: The volume of a balloon is 3.70L at 15.8  C and 1.3 atm pressure. What volume will the balloon have at STP? Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 UnitAbbr.STP value atmospheresatm1 atm millimeters of mercurymmHg760 mmHg pounds per square inchpsi14.7 psi kilopascalskPa101.325 kPa

4. Ex4: A sample of gas took up 35.0L of space at 124  C and 84.56 kPa. If the temperature increased by 16  C and the pressure changed to 4.56 atm, what would the resulting volume be? P 1 = __________ V 1 = __________ T 1 = ____  C  ____K P 2 = __________ V 2 = __________ T 2 = ____  C  ____K Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 84.56 kPa 35.0 L 124 4.56 atm ? 140 _____ P 2 T 1 P 2 T 1 397 413 P 1 V 1 T 2 = P 2 V 2 T 1 V 2 = P 1 V 1 T 2 P 2 T 1 V 2 = (0.835 atm)(35.0 L)(413 K) = (4.56 atm)(397 K) 6.67 L V 2 = 0.835  35.0  413 ÷ (4.56  397) = 84.56 kPa 1 atm = 101.325 kPa 0.835 atm = 0.835 atm

5. Ex5: The volume of a balloon is 3.70L at 15.8  C and 1.3 atm pressure. What volume will the balloon have at STP? P 1 = __________ V 1 = __________ T 1 = ___  C  _____K P 2 = __________ V 2 = __________ T 2 = ____  C  ____K Part 6: The Partial Pressures Law (1803) the Partial Pressures Law states that the total pressure of a mixture of gases equals the sum of the partial pressure of each gas. partial pressures are written as P[formula of gas] P 1, P 2, and P 3 represent the partial pressures in the formula Ex6: A gas mixture contains oxygen, nitrogen, and carbon dioxide @ 72.9 kPa. If P[O 2 ] = 26.60 kPa, and P[N 2 ] = 0.25 atm, what is P[CO 2 ] ? Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 1.3 atm 3.70 L 15.8 1 atm ? 0 273 288.8 V 2 = P 1 V 1 T 2 P 2 T 1 V 2 = (1.3 atm)(3.70 L)(273 K) = (1 atm)(288.8 K) V 2 = 4.55 L V 2 = 1.3  3.70  273 ÷ (1  288.8) = Partial Pressures Law P T = P 1 + P 2 + P 3 Partial Pressures Law P T = P 1 + P 2 + P 3

6. Ex6: A gas mixture contains oxygen, nitrogen, and carbon dioxide @ 72.9 kPa. If P[O 2 ] = 26.60 kPa, and P[N 2 ] = 0.25 atm, what is P[CO 2 ] ? P T = __________ P 1 = __________ P 2 = __________ P 3 = __________ Partial Pressures Law P T = P 1 + P 2 + P 3 Partial Pressures Law P T = P 1 + P 2 + P 3 72.9 kPa 26.60 kPa 0.25 atm ? P T = P 1 + P 2 + P 3 - P 1 - P 2 P 3 = P T - P 1 - P 2 0.25 atm P 3 = 72.9 kPa – 26.60 kPa – 25.33 kPa = 1 atm 101.325 kPa = 25.33 kPa = 25.33 kPa 20.97 kPa