Mixed Gas Laws Worksheet - Solutions 1) How many moles of gas occupy 98 L at a pressure of 2.8 atmospheres and a temperature of 292 K? n = PV = (2.8 atm)(98 L) = 11 moles of gas RT (0.0821 L.atm/mol.K)(292 K) 2) If 5.0 moles of O2 and 3.0 moles of N2 are placed in a 30.0 L tank at a temperature of 250 C, what will the pressure of the resulting mixture of gases be? 250 C = 298 K O2: P = nRT = (5.0 mol)(0.0821 L.atm/mol.K)(298 K) = 4.1 atm V (30.0 L) N2: P = nRT = (3.0 mol)(0.0821 L.atm/mol.K)(298 K) = 2.4 atm V (30.0 L) PTot = PO2 + PN2 = 4.1 atm + 2.4 atm = 6.5 atm 3) A balloon is filled with 35.0 L of helium in the morning when the temperature is 20.00 C. By noon the temperature has risen to 45.00 C. What is the new volume of the balloon? T1 = 20.00 C = 293 K, V1 = 35.0 L, T2 = 45.00 C = 318 K, V2 = ? V2 = V1 T2 = (35.0 L)(318 K) = 38.0 L T1 (293 K) 4) A 35 L tank of oxygen is at 315 K with an internal pressure of 190 atmospheres. How many moles of gas does the tank contain? n = PV = (190 atm)(35 L) = 260 moles of gas RT (0.0821 L.atm/mol.K)(315 K) 5) A balloon that can hold 85 L of air is inflated with 3.5 moles of gas at a pressure of 1.0 atmosphere. What is the temperature in 0C of the balloon? T = PV = (1 atm)(85 L) = 296 K = 230 C . . nR (3.5 mol)(0.0821 L atm/mol K) 6) CaCO3 decomposes at 12000 C to form CO2 gas and CaO. If 25 L of CO2 are collected at 12000 C, what will the volume of this gas be after it cools to 250 C? T1 = 12000 C = 1473 K, V1 = 25 L, T2 = 250 C = 298 K, V2 = ? V2 = V1 T2 = (25 L)(298 K) = 5.1 L T1 (1473 K) 7) A helium balloon with an internal pressure of 1.00 atm and a volume of 4.50 L at 20.00 C is released. What volume will the balloon occupy at an altitude where the pressure is 0.600 atm and the temperature is –20.00 C? P1 = 1.00 atm, V1 = 4.50 L, T1 = 20.00 C = 293 K, P2 = 0.600 atm, V2 = ?, T2 = -20.00 C = 253 K V2 = P1 V1 T2 = (1.00 atm)(4.50 L)(253 K) = 6.48 L T1P2 (293 K)(0.600 atm) 8) There are 135 L of gas in a container at a temperature of 2600 C. If the gas was cooled until the volume decreased to 75 L, what would the temperature of the gas be? T1 = 2600 C = 533 K, V1 = 135 L, T2 = ?, V2 = 75 L T2 = V2 T1 = (75 L)(533 K) = 296 K = 230 C V1 (135 L) 9) A 75 L container holds 62 moles of gas at a temperature of 2150 C. What is the pressure in atmospheres inside the container? 2150 C = 488 K P = nRT = (62 mol)(0.0821 L.atm/mol.K)(488 K) = 33 atm V (75 L) 10) 6.0 L of gas in a piston at a pressure of 1.0 atm are compressed until the volume is 3.5 L. What is the new pressure inside the piston? P1 = 1.0 atm, V1 = 6.0 L, P2 = ?, V2 = 3.5 L P2 = P1 V1 = (1.0 atm)(6.0 L) = 1.7 atm V2 (3.5 L) 11) A gas canister can tolerate internal pressures up to 210 atmospheres. If a 2.0 L canister holding 3.5 moles of gas is heated to 13500 C, will the canister explode? 13500 C = 1623 K P = nRT = (3.5 mol)(0.0821 L.atm/mol.K)(1623 K) = 230 atm V (2.0 L) Yes, the canister will explode. 12) The initial volume of a gas at a pressure of 3.2 atm is 2.9 L. What will the volume be if the pressure is increased to 4.0 atm? P1 = 3.2 atm, V1 = 2.9 L, P2 = 4.0 atm, V2 = ? V2 = P1 V1 = (3.2 atm)(2.9 L) = 2.3 L P2 (4.0 atm) 13) An airtight container with a volume of 4.25 x 104 L, an internal pressure of 1.00 atm, and an internal temperature of 15.00 C is washed off the deck of a ship and sinks to a depth where the pressure is 175 atm and the temperature is 3.000 C. What will the volume of the gas inside be when the container breaks under the pressure at this depth? P1 = 1.00 atm, V1 = 4.25 x 104 L, T1 = 15.00 C = 288 K, P2 = 175 atm, V2 = ?, T2 = 3.000 C = 276 K V2 = P1 V1 T2 = (1.00 atm)(4.25 x 104 L)(276 K) = 233 L T1P2 (288 K)(175 atm) 14) Two flasks are connected with a stopcock. Flask #1 has a volume of 2.5 L and contains oxygen gas at a pressure of 0.70 atm. Flask #2 has a volume of 3.8 L and contains hydrogen gas at a pressure of 1.25 atm. When the stopcock between the two flasks is opened and the gases are allowed to mix, what will the resulting pressure of the gas mixture be? (P’ & V’ are initial conditions before mixing) P’O2 = 0.70 atm, P’H2 = 1.25 atm, V’O2 = 2.5 L, V’H2 = 3.8 L, V = 6.3 L O2: P2 = P’O2 V’O2 = (0.70 atm)(2.5 L) = 0.28 atm V (6.3 L) H2: P2 = P’H2 V’H2 = (1.25 atm)(3.8 L) = 0.75 atm V (6.3 L) PTot = PO2 + PH2 = 0.28 atm + 0.75 atm = 1.0 atm 15) A weather balloon has a volume of 35 L at sea level (1.0 atm). After the balloon is released it rises to where the air pressure is 0.75 atm. What will the new volume of the weather balloon be? P1 = 1.0 atm, V1 = 35 L, P2 = 0.75 atm, V2 = ? V2 = P1 V1 = (1.0 atm)(35 L) = 47 L P2 (0.75 atm)