Basic Gas Laws (Combined and Partial Pressures Laws) 4/27/11

Steps for Solving ANY Gas Law Problem: Write out a column of information down the left-hand side. Make sure all of your variable’s units match (i.e. if P1 is in kPa, then P2 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). Write the original equation for the gas law you will be using. Rearrange the equation to solve for the variable you need. 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. Calculate, then box your answer!

millimeters of mercury 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 P1V1 = P2V2 T1 T2 Unit Abbr. STP value atmospheres atm 1 atm millimeters of mercury mmHg 760 mmHg pounds per square inch psi 14.7 psi kilopascals kPa 101.325 kPa

P1V1T2 = P2V2T1 V2 = P1V1T2 _____ _____ P2T1 P2T1 P2T1 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? P1 = __________ V1 = __________ T1 = ____C____K P2 = __________ V2 = __________ T2 = ____C____K Combined Gas Law P1V1 = P2V2 T1 T2 P1V1T2 = P2V2T1 _____ _____ P2T1 P2T1 V2 = P1V1T2 P2T1 84.56 kPa 35.0 L 124 4.56 atm ? 140 = 0.835 atm 397 413 84.56 kPa 1 atm = 0.835 atm 101.325 kPa V2 = (0.835 atm)(35.0 L)(413 K) = (4.56 atm)(397 K) V2 = 0.835 35.0  413 ÷ (4.56  397) = 6.67 L

V2 = P1V1T2 P2T1 V2 = (1.3 atm)(3.70 L)(273 K) = (1 atm)(288.8 K) 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? P1 = __________ V1 = __________ T1 = ___C_____K P2 = __________ V2 = __________ T2 = ____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] P1, P2, and P3 represent the partial pressures in the formula Ex6: A gas mixture contains oxygen, nitrogen, and carbon dioxide @ 72.9 kPa. If P[O2] = 26.60 kPa, and P[N2] = 0.25 atm, what is P[CO2] ? Combined Gas Law P1V1 = P2V2 T1 T2 V2 = P1V1T2 P2T1 1.3 atm 3.70 L 15.8 1 atm ? 0 273 V2 = (1.3 atm)(3.70 L)(273 K) = (1 atm)(288.8 K) 288.8 V2 = 1.3  3.70  273 ÷ (1  288.8) = V2 = 4.55 L Partial Pressures Law PT = P1 + P2 + P3

PT = P1 + P2 + P3 - P1 - P2 - P1 - P2 P3 = PT - P1 - P2 Ex6: A gas mixture contains oxygen, nitrogen, and carbon dioxide @ 72.9 kPa. If P[O2] = 26.60 kPa, and P[N2] = 0.25 atm, what is P[CO2] ? PT = __________ P1 = __________ P2 = __________ P3 = __________ Partial Pressures Law PT = P1 + P2 + P3 72.9 kPa 26.60 kPa 0.25 atm ? PT = P1 + P2 + P3 - P1 - P2 - P1 - P2 P3 = PT - P1 - P2 = 25.33 kPa 0.25 atm 101.325 kPa = 25.33 kPa 1 atm P3 = 72.9 kPa – 26.60 kPa – 25.33 kPa = 20.97 kPa