Gas Laws Combined and Ideal Worksheets Answer Key

Chemistry Gas Laws WS #1 Gay-Lussac's Law: Temperature & Pressure The relationship between temperature and pressure is: Direct The formula for converting °C to K is: 273 A gas with a pressure of 5.4 atm and at 25°C is raised to a new temperature of 78°C. What is the new pressure? P₁ = 5.4 atm T₁ = 25°C + 273 = 298K P₂ = ? T₂ = 78°C + 273 = 351K P₁/T₁ = P₂/T₂ P₂ = (P₁T₂)/T₁ P₂ = (5.4 atm * 351 K) / 298 K = 6.4 atm A gas with a pressure of 550 torr and at 110°C is raised to a new pressure of 760 torr. What is the new temperature? P₁ = 550 torr T₁ = 110°C + 273 = 383 K P₂ = 760 torr T₂ = ? P₁/T₁ = P₂/T₂ T₂ = (P₂T₁)/P₁ T₂ = (760 torr * 383 K) / 550 torr = 530 K A gas with a pressure of 780 mmHg and at 55°C is lowered to a new pressure of 640 mmHg. What is the new temperature? P₁ = 780 mmHg T₁ = 55°C + 273 = 328 K P₂ = 640 mmHg T₂ = ? P₁/T₁ = P₂/T₂ T₂ = (P₂T₁)/P₁ T₂ = (640 mmHg * 328 K) / 780 mmHg = 270 K A gas with a pressure of 5.6 atm and at -150°C is lowered to a new temperature of 20°C. What is the new pressure? P₁ = 5.6 atm T₁ = -150°C + 273 = 123 K P₂ = ? T₂ = 20°C + 273 = 293 K P₁/T₁ = P₂/T₂ P₂ = (P₁T₂)/T₁ P₂ = (5.6 atm * 293 K) / 123 K = 13 atm Gas Laws WS #2 Boyle's Law: Pressure and Volume What is the relationship between pressure and volume? Inverse A gas at 5.4 atm has a volume of 1.3 L. What volume would the gas have at 6.8 atm? P₁ = 5.4 atm P₂ = 6.8 atm V₁ = 1.3 L V₂ = ? P₁V₁ = P₂V₂ V₂ = (P₁V₁)/P₂ V₂ = (5.4 atm * 1.3 L) / 6.8 atm = 1.0 L A gas at 355 torr has a volume of 850 mL. What pressure would you need to decrease the volume to 550 mL? P₁ = 355 torr V₁ = 850 mL P₂ = ? V₂ = 550 mL P₁V₁ = P₂V₂ P₂ = (P₁V₁)/V₂ P₂ = (355 torr * 850 mL) / 550 mL = 550 torr A gas at 67 L is reduced to 44 L. If the original pressure was 330 torr, what is the new pressure? V₁ = 67 L V₂ = 44 L P₁ = 330 torr P₂ = ? P₁V₁ = P₂V₂ P₂ = (P₁V₁)/V₂ P₂ = (330 torr * 67 L) / 44 L = 500 torr A quick review: Temperature must be in KELVIN! 56°C 329 K 77°C 350 K -45°C 228 K -269°C 4 K Gas Laws WS #3 Charles Law: Temperature & Volume [\frac{V_1}{T_1} = \frac{V_2}{T_2}] The relationship between temperature and volume is: 1 Directly proportional (T↑V↑, T↓V↓)   1. cbsd.org cbsd.org A gas at 5°C occupies a volume of 7.5 liters. What volume will the gas occupy at 100°C? (V_1 = 7.5 L) (V_2 = ?) (T_1 = 5 + 273 = 278 K) (T_2 = 100 + 273 = 373 K) [\frac{V_1}{T_1} = \frac{V_2}{T_2} \implies V_2 = \frac{V_1 T_2}{T_1} = \frac{(7.5 L)(373 K)}{278 K} = 10.07 L] A gas at -20°C occupies a volume of 35.0 liters. What volume will the gas occupy at 20°C? (V_1 = 35.0 L) (V_2 = ?) (T_1 = -20 + 273 = 253 K) (T_2 = 20 + 273 = 293 K) [\frac{V_1}{T_1} = \frac{V_2}{T_2} \implies V_2 = \frac{V_1 T_2}{T_1} = \frac{(35.0 L)(293 K)}{253 K} = 40.5 L] A gas fills a balloon and occupies a volume of 22.4 L at a temperature of 27°C. What would the new volume of the balloon be if the gas were heated to 127°C? (V_1 = 22.4 L) (V_2 = ?) (T_1 = 27 + 273 = 300 K) (T_2 = 127 + 273 = 400 K) [\frac{V_1}{T_1} = \frac{V_2}{T_2} \implies V_2 = \frac{V_1 T_2}{T_1} = \frac{(22.4 L)(400 K)}{300 K} = 29.9 L] A gas occupies a volume of 30.0 cm³ at 73.5°C. If the pressure is held constant and the temperature is changed to 22.5°C, what will the new volume be? (V_1 = 30.0 cm^3) (V_2 = ?) (T_1 = 73.5 + 273 = 346.5 K) (T_2 = 22.5 + 273 = 295.5 K) [\frac{V_1}{T_1} = \frac{V_2}{T_2} \implies V_2 = \frac{V_1 T_2}{T_1} = \frac{(30.0 cm^3)(295.5 K)}{346.5 K} = 25.6 cm^3] A sample of argon gas is cooled and its volume went from 3.8 L to 2.3 L. If its final temperature was 45°C, what was the original temperature? (V_1 = 3.8 L) (V_2 = 2.3 L) (T_1 = ?) (T_2 = 45 + 273 = 318 K) [\frac{V_1}{T_1} = \frac{V_2}{T_2} \implies \frac{1}{T_1} = \frac{V_2}{T_2 V_1} \implies T_1 = \frac{T_2 V_1}{V_2} = \frac{(318 K)(3.8 L)}{2.3 L} = 526 K] Chemistry Gas Laws WS #4: Combined Gas Law [\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}] Combined Gas Law 4.5 L of Carbon dioxide at 23°C has a pressure of 3.2 atm. What is the pressure of the carbon dioxide at 95°C and 3.4 L? (V_1 = 4.5 L) (V_2 = 3.4 L) (T_1 = 296 K) (T_2 = 368 K) (P_1 = 3.2 atm) (P_2 = ?) [\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \implies P_2 = \frac{P_1 V_1 T_2}{T_1 V_2}] [P_2 = \frac{(3.2 atm)(4.5 L)(368 K)}{(296 K)(3.4 L)} = 5.3 atm] 7.6 L of nitrogen at 146°C has a pressure of 755 torr. What is the pressure of the nitrogen at 57°C and 10.5 L? (V_1 = 7.6 L) (V_2 = 10.5 L) (T_1 = 419 K) (T_2 = 330 K) (P_1 = 755 torr) (P_2 = ?) [P_2 = \frac{(755 torr)(7.6 L)(330 K)}{(419 K)(10.5 L)} = 430 torr] Oxygen at 25°C and 760 torr pressure occupies a volume of 21.2 L. What is the volume of oxygen gas at 133°C and 830 torr? (T_1 = 298 K) (T_2 = 406 K) (P_1 = 760 torr) (P_2 = 830 torr) (V_1 = 21.2 L) (V_2 = ?) [\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \implies V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}] [V_2 = \frac{(760 torr)(21.2 L)(406 K)}{(298 K)(830 torr)} = 26 L] 4.3 L of methane at 5.4 kPa has a temperature of 46°C. What is the temperature of methane at 5.4 L at 6.6 kPa? (V_1 = 4.3 L) (V_2 = 5.4 L) (P_1 = 5.4 kPa) (P_2 = 6.6 kPa) (T_1 = 319 K) (T_2 = ?) (Units added!) [\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \implies T_2 = \frac{P_2 V_2 T_1}{P_1 V_1}] [T_2 = \frac{(6.6 kPa)(5.4 L)(319 K)}{(5.4 kPa)(4.3 L)} = 490 K]