States and Properties of Matter - The Gas Laws Your feedback on these self-help problems is appreciated. Click here to send an e-mail. For on-line help with this topic, see the chemcal module "Behaviour of Gases" Shortcut to Questions Q: 1 2 3 4 5 6 7 8 9 10 11 1 What will be the volume of a sample of carbon dioxide at 98.0 kPa if its volume at 96.4 kPa is 223 mL , the temperature being unchanged? 2 What volume will be occupied by a sample of hydrogen at 153 kPa if the volume at 98.6 kPa is 325 mL, the temperature being held constant? 3 A sample of hydrogen sulfide was collected in a 250 mL flask at a pressure of 98.6 kPa and 310 K. What volume would the gas occupy at 369 kPa and 313 K? 4 Calculate the molar weight of a gas a sample of which (0.638 g) occupies a volume of 223 mL at 300 K and 750 mmHg. 5 A mixture of Ar (3.28 x 10-3 mole) and N2 (1.92 x 10-2 mole) was collected in a vessel whose capacity was 62.5 mL. What would be the total gas pressure at 298 K? 6 A spark is passed through a mixture formed by adding hydrogen (3.00 litre), oxygen (1.00 litre), and neon (2.00 litre), all the volumes being measured at 273 K and 101 kPa. What is the final volume of the reaction mixture at these conditions? 7 (a) If air is 80% nitrogen by mass and 20% oxygen by mass, how many molecules would be present in a sample (2.17 g) of air? (b) What volume would this occupy at 273 K and 101 kPa? 8 (Advanced question) The mass of an evacuated glass bulb is 35.194 g. When filled with O2 at 101 kPa pressure and 298 K its mass is 36.406 g. When filled with a mixture of CH4 and C2H6 under the same physical conditions its mass is 36.033 g. What is the % by volume of CH4 in the mixture? 9 (Advanced question) An equilibrium mixture of gaseous NO2 and N2O4 exerts a total pressure of 101 kPa in a container at 298 K, and the mole fraction of NO2 in the mixture is 0.200. What is the total pressure in the container at 473 K if all the N2O4 has been converted into NO2? 10 (Advanced question) Pure phosphine gas (PH3) was prepared by complete decomposition of Ca3P2 (1.00 g) with excess acid. The (dry) PH3 gas was heated to 673 K to decompose it into its gaseous elements, Pz and H2 . The final volume was 1.060 litre at 673 K and 101 kPa pressure. What is the molecular formula of gaseous phosphorus under these conditions? 11 (Advanced question) Chlorine gas(Cl2) can be prepared by adding concentrated hydrochloric acid to MnO2. The equation for the reaction is MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O Calculate (a) the mass of MnO2 and, (b) the volume of hydrochloric acid of density 1.12 g ml-1 and containing 40.0% HCl by mass needed to produce 2.50 litre of Cl2 gas at 293 K and 103 kPa. The Gas Laws (Answers) 1 As temperature is constant, Boyle's Law can be used. PV = constant or P1V1 = P2V2 P1 = 96.4 kPa and V1 = 223 mL P2 = 98.0 kPa and V2 = ? So 96.4 x 223 = 98.0 x V2 V2 = (96.4 x 223) / 98.0 = 219 mL [Note: the units of P and V must be the same on both sides of the equation.] 2 As temperature is constant, Boyle's Law can be used. PV = constant or P1V1 = P2V2 P1 = 98.6 kPa and V1 = 325 mL P2 = 153 kPa and V2 = ? So 98.6 x 325 = 153 x V2 V2 = (98.6 x 325) / 153 = 209 mL 3 The temperature and pressure both change so the combined Boyle's and Charles's Law expression must be used. P1V1 / T1 = P2V2 / T2 P1 = 98.6 kPa, V1 = 250 mL and T1 = 310 K (Note that temperature must be expressed in Kelvin) P2 = 369 kPa, V2 = ? and T2 = 313 K So (98.6 x 250) / 310 = (369 x V2) / 313 V2 = (98.6 x 250 x 313) / (310 x 369) = 67.4 mL 4 The Ideal Gas Equation is required to solve this problem. PV = nRT The constant R has a value which depends on the units for P and V. (Units for n are always moles). If P is expressed in kPa and V in litres, R = 8.314 J K-1 mol-1. Temperature is always in Kelvin. Let the molar mass of the gas = m P = 750 mmHg Converting this to kPa: P = (750 / 760) x 101.3 kPa = 100.0 kPa V 223 mL = 0.223 L n = mass / molar mass = 0.638 / m mol R = 8.314 J K-1 mol. T = 300 K (Note: T must be in Kelvin) PV = nRT So 100.0 x 0.223 = (0.638 / m) x 8.314 x 300 m = (0.638 x 8.314 x 300) / (100.0 x 0.223) = 71.4 g mol-1 5 The Ideal Gas Equation is required to solve this problem. PV = nRT The constant R has a value which depends on the units for P and V. (Units for n are always moles). If P is expressed in kPa and V in litres, R = 8.314 J K-1 mol-1. Temperature is in Kelvin. P = ? kPa, V 62.5 mL = 0.0625 L and n = total moles of gas present = 3.28 x 10-3 + 1.92 x 10-2 = 2.248 x 10-2 mol R = 8.314 J K-1 mol-1 and T = 298 K (Note: T must be in Kelvin) P x 0.0625 = 2.248 x 10-2 x 8.314 x 298 P = (2.248 x 10-2 x 8.314 x 298) / 0.0625 = 891 kPa 6 The hydrogen and oxygen gases combine to give water which is a liquid at the prevailing conditions of temperature and pressure. The neon gas does not react. The equation for the reaction is 2H2(g) + O2(g) → 2H2O(l) From the stoichiometry of the equation, 2.00 mole of hydrogen reacts exactly with 1.00 mole of oxygen. Because one mole of any gas occupies the same volume as one mole of any other gas at the same conditions of temperature and pressure, then these gases also combine in the same ratio by volume as they do by moles i.e. 2.00 volume of hydrogen gas reacts exactly with 1.00 volume of oxygen gas. If all the oxygen (1.00 L) is used in the reaction, it would require 2.00 L of hydrogen, leaving 1.00 L of hydrogen gas unreacted. The volume of the water liquid formed is negligible, so the final volume is just 1.00 L of hydrogen gas + the non-reacting neon gas (2.00 L) = 3.00 L Note: Gases combine in simple ratio by volume, the same ratio as for moles, because gases expand to uniformly fill their container. However, the same is not true for liquids or solids which have a specified volume for a given mass. 7 (a) Mass of nitrogen, N2, = 0.80 x 2.17 = 1.74 g Mass of oxygen, O2, = 0.20 x 2.17 = 0.434 g Moles of N2 = mass / molar mass = 1.736 / (2 x 14.01) = 0.0620 mol Moles of O2 = mass / molar mass = 0.434 / (2 x 16.00) = 0.0136 mol Total moles of gas present = 0.0620 + 0.0136 = 0.0756 mol 1 mole of any gas contains an Avogadro number of molecules, 6.022 x 1023. Therefore 0.0756 mole contains 0.0756 x 6.022 x 1023 molecules = 4.6 x 1022 molecules. (b) Two methods can be used. Either: from the ideal gas equation PV = nRT, V = nRT / P For V in L and P in kPa, R = 8.324 J K-1 mol-1 V = (0.0756 x 8.314 x 273) / 101 = 1.7 L or: one mole of any gas at 273 K and 101 kPa occupies a volume of 22.4 L 0.0756 mole of gas occupies 0.0756 x 22.4 L = 1.7 L 8 The essential concept required in this question is that at a given temperature and pressure, a fixed volume of any gas or mixture of gases will always have the same number of moles present regardless of which gases are present. This follows from the property of all gases that they expand to completely and uniformly occupy any fixed-volume container because the attractive forces between gas phase molecules are very small. From the data given for the system when oxygen is present, the moles of O2(g) can be deduced: Mass of oxygen = 36.406 - 35.194 = 1.212 g So, moles of O2 = mass/molar mass = 1.212 / 32.00 = 0.03788 mol Thus moles of (methane + ethane) = 0.03788 mol as the temperature, pressure and volume are unchanged. Let moles of methane present = m and moles of ethane present = y. As total moles = 0.03788 mol, 0.03788 = m + y .............(1) A second equation in m and y can be obtained from the mass of the mixture of methane + ethane. Mass of methane + ethane = 36.033 - 35.194 = 0.839 g This mass is made up of m moles of methane + y moles of ethane. Molar mass of CH4 = 16.04 g mol-1 and molar mass of C2H6 = 30.08 g mol-1 Therefore m x 16.04 + y x 30.08 = 0.839 ....................(2) From (1), y = (0.03788 - m) Substituting in (2) m x 16.04 + (0.03788 - m) x 30.08 = 0.839 whence 14.04 m = 0.300 m = 0.0214 mol The fraction of methane by volume will be the same as its fraction by moles as these are gases. Fraction of CH4 in the mixture = m / (m + y) = 0.0214 / 0.03788 = 0.565 and % methane = 100 x 0.565 = 56.6 % 9 The volume is fixed but not P or T. Therefore any convenient volume can be taken as long as the same volume is used throughout the calculation. Take V = 1.00 L Given P = 101 kPa and T = 298 K, then: At 298 K, total moles of gas present in 1.00 L = PV / RT = (101 x 1.00) / (8.314 x 298) = 0.04077 mol 0.200 of this total number of moles is NO2(g) and 0.800 of the total moles is N2O4(g). Therefore, moles of NO2 = 0.200 x 0.04077 = 0.00815 mol and moles of N2O4 = 0.800 x 0.04077 = 0.03261 mol At 473 K, all the N2O4 has been converted to NO2, N2O4(g) → 2NO2(g) 1 mole → 2 moles Thus the total moles of gas present = moles of NO2 originally present at 298 K + 2 x moles of N2O4 originally present at 298 K. i.e. moles of gas at 473 K = 0.00815 + 2 x 0.03261 = 0.0734 mol Substituting in the ideal gas equation, P = nRT / V = (0.0734 x 8.314 x 473) / 1.00 = 289 kPa 10 Each mole of Ca3P2 would produce 2 moles of PH3(g) as there are 2 phosphorous atoms in each formula unit of Ca3P2. The PH3 then undergoes decomposition according to the equation PH3 → (1/z)Pz + (3/2)H2. From this equation, 1 mole of PH3 results in the formation of 1/z moles of gaseous Pz molecules and 3/2 moles of hydrogen gas, H2. Therefore from 2 moles of PH3 there would be formed 2 x 1/z moles of Pz and 2 x 3/2 moles of H2. i.e. 1 mole of Ca3P2 would result in the formation of 2/z moles of Pz and 3 moles of H2. Moles of Ca3P2 = mass / molar mass = 1.00 / 182.2 = 0.005488 mol Therefore moles of Pz resulting from the decomposition of the PH3 formed = (2/z) x 0.005488 mol = 0.01098 / z mol. Moles of H2 resulting from the decomposition of the PH3 = 3 x 0.005488 mol = 0.01646 mol. Total moles of gas present = moles of Pz + moles of H2 = (0.01098 / z + 0.01646) mol In the Ideal Gas Equation, n = PV / RT P = 101 kPa V = 1.060 L T = 673 K R = 8.314 J K-1 mol-1 i.e. 0.01098 / z + 0.01646 = (101 x 1.060) / (8.314 x 673) 0.01913 = 0.01098 / z + 0.01646 z = 0.01098 / 0.00267 = 4.1 As z must be an integer, the formula Pz is P4 11 MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O From the equation, 1 mole of Cl2 requires 1 mole of MnO2 and 4 moles of HCl. Moles of Cl2 formed can be calculated from the Ideal Gas Equation. n = PV / RT = (103 x 2.50) / (8.314 x 293) mol = 0.1057 mol. (a) Moles of MnO2 = moles of Cl2 = 0.1057 mol. Therefore mass of MnO2 = 0.1057 x molar mass = 0.1057 x 86.94 = 9.19 g (b) Moles of HCl = 4 x moles of Cl2 = 4 x 0.1057 = 0.4228 mol. [Note: while only 2 moles of HCl are needed to form 1 mole of Cl2, the additional 2 moles of HCl are needed to provide the required total moles of H+ to form the 2 moles of H2O.] So, mass of HCl = mol HCl x molar mass HCl = 0.4228 x 36.46 = 15.42 g The hydrochloric acid solution contains 40.0 % HCl by mass and has a density = 1.12 g mL. i.e 1.00 mL of solution contains 0.400 x 1.12 g of HCl Therefore 1.00 / (0.400 x 1.12) mL contains 1.00 g of HCl and (15.42 x 1.00) / (0.400 x 1.12) mL contains 15.42 g of HCl = 34.4 mL