Chemistry Gas Law <- Chemistry Gas Law Questions - Answers + Explanations Bundle Chemistry Gas Law Questions - Answers + Explanations Bundle Share Quiz Get Embed Code Copy the code below to embed this quiz on your website: <iframe src="https://tutorone.ca/practice-test/?embed=true" width="100%" height="800" style="border: none; max-width: 100%;" data-source="tutorone" allowfullscreen></iframe> Copy Code 123456789101112 Chemistry Gas Law Questions - Answers + Explanations Bundle 1 / 12 Ammonia gas occupies a volume of \(b = 5 \text{ L}\) at a pressure of 632.4 atm. If the pressure were lowered to 356.8 atm, what would the new volume be? Gas Law used: Boyle's Law 8.86 L 6.25 L 10.2 L Boyle's Law states that \(P_1V_1 = P_2V_2\). Substituting the given values: \(632.4 \times 5 = 356.8 \times V_2\). Solving for \(V_2\), \(V_2 = \frac{632.4 \times 5}{356.8} \approx 8.86 \text{ L}\). 2 / 12 A tank full of xenon gas has a volume of 596 L. When the gas is transferred to a new tank, the pressure is measured at 252.3 atm and the volume is 323 L. What was the original pressure? Gas Law used: Boyle's Law 136.63 atm 152.4 atm 175 atm Boyle's Law states that \(P_1V_1 = P_2V_2\). Substituting the given values: \(P_1 \times 596 = 252.3 \times 323\). Solving for \(P_1\), \(P_1 = \frac{252.3 \times 323}{596} \approx 136.63 \text{ atm}\). 3 / 12 A sample of bromine gas has a pressure of 86 atm and a temperature of 378 K. When the gas is heated to 584 K, what is the new pressure? Gas law used: Gay-Lussac's Law 132.87 atm 128.5 atm 140 atm Gay-Lussac's Law states that \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\). Substituting the given values: \(\frac{86}{378} = \frac{P_2}{584}\). Solving for \(P_2\), \(P_2 = \frac{86 \times 584}{378} \approx 132.87 \text{ atm}\). 4 / 12 A 22.5 L container of neon had a pressure change from 225 atm to 45 atm. What is the new volume? Gas Law used: Boyle's Law 112.5 L 90 L 150 L Boyle's Law states that \(P_1V_1 = P_2V_2\). Substituting the given values: \(225 \times 22.5 = 45 \times V_2\). Solving for \(V_2\), \(V_2 = \frac{225 \times 22.5}{45} = 112.5 \text{ L}\). 5 / 12 A container full of helium gas has a temperature of \(52^\circ\text{C}\). When the gas is transferred to a new container, it has a volume of 795.6 L and a temperature of \(58^\circ\text{C}\). What was the volume of the original container? Gas Law used: Charles's Law 781.42 L 800 L 770 L Charles's Law states that \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\). Temperatures must be converted to Kelvin: \(T_1 = 52 + 273 = 325 \text{ K}, T_2 = 58 + 273 = 331 \text{ K}\). Substituting: \(\frac{V_1}{325} = \frac{795.6}{331}\). Solving for \(V_1\), \(V_1 = \frac{325 \times 795.6}{331} \approx 781.42 \text{ L}\). 6 / 12 A gas has a volume of 240.0 mL at \(25.0^\circ\text{C}\) and 600.0 mmHg. Calculate its volume at STP. Gas Law used: Combined Gas Law 188.8 mL 190.0 mL 185.0 mL STP conditions are \(P_2 = 760 \text{ mmHg}\) and \(T_2 = 273 \text{ K}\). The Combined Gas Law is \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\). Convert temperature to Kelvin: \(T_1 = 25.0 + 273 = 298 \text{ K}\). Substituting the given values: \(\frac{600.0 \times 240.0}{298} = \frac{760 \times V_2}{273}\). Solving for \(V_2\): \(V_2 = \frac{600.0 \times 240.0 \times 273}{298 \times 760} \approx 188.8 \text{ mL}\). 7 / 12 A given sample of gas has a volume of 4.20 L at \(60.0^\circ\text{C}\) and 1.00 atm pressure. Calculate its pressure if the volume is changed to 5.00 L and the temperature to \(27^\circ\text{C}\). Gas Law used: Combined Gas Law 0.76 atm 0.80 atm 0.70 atm The Combined Gas Law is \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\). Convert temperatures to Kelvin: \(T_1 = 60.0 + 273 = 333 \text{ K}\), \(T_2 = 27 + 273 = 300 \text{ K}\). Substituting the given values: \(\frac{1.00 \times 4.20}{333} = \frac{P_2 \times 5.00}{300}\). Solving for \(P_2\): \(P_2 = \frac{1.00 \times 4.20 \times 300}{333 \times 5.00} \approx 0.76 \text{ atm}\). 8 / 12 A certain gas has a volume of 500.0 mL at \(77.0^\circ\text{C}\) and 600.0 torr. Calculate the temperature, \(^\circ\text{C}\), if the volume decreased to 400.0 mL and the pressure is increased to 1.00 atm. Gas Law used: Combined Gas Law 81.7°C 90°C 70°C The Combined Gas Law is \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\). Convert pressures to the same unit: \(P_2 = 1.00 \text{ atm} = 760 \text{ torr}\). Convert temperature to Kelvin: \(T_1 = 77.0 + 273 = 350 \text{ K}\). Substituting the given values: \(\frac{600.0 \times 500.0}{350} = \frac{760 \times 400.0}{T_2}\). Solving for \(T_2\): \(T_2 = \frac{760 \times 400.0 \times 350}{600.0 \times 500.0} \approx 354.7 \text{ K}\), which is \(354.7 - 273 \approx 81.7^\circ\text{C}\). 9 / 12 A 280.0 mL sample of neon exerts a pressure of 660 torr at \(26.0^\circ\text{C}\). At what temperature, \(^\circ\text{C}\), would it exert a pressure of 940 torr in a volume of 440.0 mL? Gas Law used: Combined Gas Law 420.6°C 400°C 450°C The Combined Gas Law is \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\). Temperatures must be in Kelvin: \(T_1 = 26.0 + 273 = 299.0 \text{ K}\). Substituting the given values: \(\frac{660 \times 280.0}{299.0} = \frac{940 \times 440.0}{T_2}\). Solving for \(T_2\): \(T_2 = \frac{940 \times 440.0 \times 299.0}{660 \times 280.0} \approx 693.6 \text{ K}\), which is \(693.6 - 273 \approx 420.6^\circ\text{C}\). 10 / 12 A sample of argon gas has a volume of 598.7 L. The temperature is decreased to \(34.8^\circ\text{C}\), and the volume also decreases to 399.1 L. What was the original temperature? Gas Law used: Charles's Law 461.8 K 450 K 400 K Charles's Law states that \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\). Temperatures must be in Kelvin: \(T_2 = 34.8 + 273 = 307.8 \text{ K}\). Substituting the given values: \(\frac{598.7}{T_1} = \frac{399.1}{307.8}\). Solving for \(T_1\), \(T_1 = \frac{598.7 \times 307.8}{399.1} \approx 461.8 \text{ K}\), which is \(461.8 - 273 \approx 188.8^\circ\text{C}\). 11 / 12 A sample of an unknown gas has a temperature of 478 K. When the temperature is increased to 346 K, the volume also increases to 2.047 L. What was the original volume? Gas Law used: Charles's Law 2.826 L 3.00 L 2.50 L Charles's Law states that \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\). Substituting the given values: \(\frac{V_1}{478} = \frac{2.047}{346}\). Solving for \(V_1\), \(V_1 = \frac{2.047 \times 478}{346} \approx 2.826 \text{ L}\). 12 / 12 A sample of hydrogen gas has an initial pressure of 3.86 atm and an initial volume of 747.2 mL. If the volume of the gas were decreased to 465.8 mL, what would the new pressure be? Gas Law used: Boyle's Law 6.19 atm 5.50 atm 7.00 atm Boyle's Law states that \(P_1V_1 = P_2V_2\). Substituting the given values: \(3.86 \times 747.2 = P_2 \times 465.8\). Solving for \(P_2\), \(P_2 = \frac{3.86 \times 747.2}{465.8} \approx 6.19 \text{ atm}\). Your score is Follow us on socials! LinkedIn Facebook Twitter 0% Restart quiz Send feedback About This Quiz Chemistry Gas Law Questions - Answers + Explanations Bundle Learn through doing the questions. Reach out to us if you run into any difficulty. 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