<p>Sample Problems from Standard 5.2 (Ideal Gas Law) Detailed solutions below</p><p>1. A gas cylinder contains .722 L of hydrogen gas at a pressure of 10.6 atm. If the gas is used to fill a balloon at a pressure of 0.96 atm, what is the volume of the filled balloon?</p><p>2. A bubble of carbon dioxide gas in some unbaked bread dough has a volume of 1.15 mL at a temperature of 22.0oC. What volume will the bubble have when the bread is baked and reaches a temperature of 99.0oC?</p><p>3. A tank with a volume of 658 mL contains 1.50 g of neon gas. The maximum safe pressure that the tank can withstand is 450 kPa. At what temperature in degrees Celsius will the tank have that pressure?</p><p>4. 2 BaO2(s)  2 BaO(s) + O2(g)</p><p>What mass of BaO2 reacted if 265 mL of O2 is collected by water displacement at 0.975 atm and 10.0oC?</p><p>1. Since the pressure and volume are changing, we start with the relationship</p><p>P1V1 / n1RT1 = P2V2 / n2RT2</p><p>Since nothing is mentioned about temperature or the number of moles, we can assume they stay </p><p> the same. This will cancel them out of the equation along with R to leave us with P1V1 = P2V2.</p><p>(10.6 atm)(.722 L) = (0.96 atm)(V2)</p><p>Divide both sides by 0.96 atm to isolate the variable.</p><p>V2 = (10.6 atm)(.722 L) / (0.96 atm) = 7.97 L</p><p>2. In this example, volume and temperature are changing. Everything else is assumed to remain </p><p> constant. This time, we are left with V1 /T1 = V2 /T2. Be sure to convert temperature to Kelvins. (1.15 mL) / (295 K) = V2 / 371 K</p><p>Cross multiply and divide to determine the new volume.</p><p>V2 = (1.15 mL)(371 K) / 295 K = 1.45 mL</p><p>3. None of the variables are changing in this problem, so we will use the ideal gas law in its regular form, PV = nRT. Keep in mind, pressure must be in atmospheres, volume in Liters, and the grams will have to be converted to moles by dividing by the molar mass of neon.</p><p>P = 450 kPa / 101.3 = 4.44 atm V = .658 L n = 1.50 g / 20.18 = 0.0743 moles</p><p>(4.44 atm)(.658 L) = (0.0743 mol)(.0821)(T)</p><p>Divide both sides by (0.0743 mol)(.0821)to isolate the variable.</p><p>T = (4.44 atm)(.658 L) / (0.0743 mol)(.0821) = 479 K</p><p>Convert from K to oC.</p><p>479 K – 273 = 206oC</p><p>4. From the volume, pressure, and temperature of the gas, O2, we can calculate the number of moles</p><p> of O2. Then we can use stoichiometry to calculate moles of BaO2 which can then be converted to grams.</p><p>Be sure to convert the measurements to the necessary units required by the ideal gas law.</p><p>(0.975 atm)(0.265 L) = n(0.0821)(283 K)</p><p>Divide both sides by (0.0821)(283 K) to isolate the variable.</p><p> n = (0.975 atm)(0.265 L) / (0.0821)(283 K) = 0.0111 mol O2</p><p>Next, we multiply by B / A from the balanced equation to determine moles of BaO2.</p><p>0.0111 mol O2 x (2 / 1) = 0.0222 mol BaO2</p><p>Multiply by the molar mass of BaO2 to get to grams.</p><p>(0.0222 mol)(137 + 16 + 16) = 3.76 g BaO2</p>