Boyle's Law (105kPa)(2.50L) = (40.5kPa)(V2) 6.48 = V2 #7 Nitrous oxide (N2O) is used as an anesthetic. The pressure on 2.50 L of N2O changes from 105 kPa to 40.5 kPa. If the temperature does not change, What will the new volume be? Step one: Ask which formula is needed here? This is a problem where the number of moles is not needed and the temperature is not changing. So the only things That change in this problem are PRESSURE and VOLUME. The formula that compares pressure and volume is Boyle’s Law. Boyle's Law P1V1 = P2V2 (105kPa)(2.50L) = (40.5kPa)(V2) 6.48 = V2

Boyle's Law #8 A gas with a volume of 4.00 L at a pressure of 205 kPa is allowed to expand To a volume of 12.0 L. What is the pressure in the container if the temperature Remains constant? This is a pressure vs volume problem. Note that we don’t add or take away Molecules or change the temperature. Therefore we need to use Boyle’s Law Boyle's Law P1V1 = P2V2 (205kPa)(4.00L) = (P2)(12.0L) 68.3kPa = P2

Charles's Law 3.39 L = V2 This is a problem where the pressure and #9 If a sample of gas occupies 6.80 L at 325°C, what will its volume be at 25°C if the pressure does not change? This is a problem where the pressure and number of moles of gas do not change. The two variables that change are volume and temperature. Therefore we need Charles’s Law to solve this problem. Charles's Law V1 = V2 T1 T2 6.8L = V2 598°K 298°K 3.39 L = V2

Charles's Law 8.36 L = V2 5.00L = V2 223°K 373°K #10 Exactly 5.00 L of air at -50°C is warmed to 100°C. What is the new volume if the pressure remains constant? This is a problem where the pressure and number of moles of gas do not change. The two variables that change are temperature and pressure. Therefore we need Charles’s Law to solve this problem. Charles's Law V1 = V2 T1 T2 5.00L = V2 223°K 373°K 8.36 L = V2

#11 A sample of nitrogen gas has a pressure of 6.58 kPa at 539 K. If the volume does not change, what will the pressure be at 211 K? This is a problem where the volume and number of moles of gas do not change. The two variables that change are temperature and pressure. Therefore we need Gay-Lussac’s Law to solve this problem. Gay-Lussac's Law P1 = P2 T1 T2 6.58kPa = P2 539°K 211°K 2.58kPa = P2

Gay-Lussac's Law 341°K = T2 This is a problem where the volume and #12 The pressure in a car tire is 198 kPa at 27°C After a long drive, the pressure is 225 kPa. What is the temperature of the air in the tire? Assume that the volume is constant. This is a problem where the volume and number of moles of gas do not change. The two variables that change are temperature and pressure. Therefore we need Gay-Lussac’s Law to solve this problem. Gay-Lussac's Law P1 = P2 T1 T2 198kPa = 225 kPa 300°K T2 341°K = T2

#13 A gas at 155 kPa and 25°C has an initial volume of 1.00L The pressure of the gas increases to 605 kPa as the temperature is raised to 125°C. What is the new volume? This is a problem where the ONLY thing that doesn’t change is the number of moles of gas.The three variables that change are temperature, pressure, and volume. Therefore we need the Combined Gas Law to solve this problem. Combined Gas Law P1V1 = P2V2 T1 T2 (155kPa)(1.00L) = (605 kPa)(V2) 298°K 398°K .342 L = V2

#14 A 5.00 L air sample has a pressure of 107 kPa at a temperature of -50°C. If the temperature is raised to 102°C and the volume expands to 7.00 L, what will the new pressure be? This is a problem where the ONLY thing that doesn’t change is the number of moles of gas.The three variables that change are temperature, pressure, and volume. Therefore we need the Combined Gas Law to solve this problem. Combined Gas Law P1V1 = P2V2 T1 T2 (107kPa)(5.00L) = (P2)(7.00 L) 223°K 375°K 128.5 kPa = P2