Chapter 5 Gases and the Gas Laws

University:
Kentucky State University
Course:
CHE 101 | General Chemistry I
Academic year:

2021
Views:

232

Pages:

2
Author:

argatikzu1

Chapter 5: Gases and the Gas Laws GAS LAWS AND RELATED PROBLEMS 𝑉 Charles’s Law (𝑇 = 𝐾) - Jacques Charles discovered that when a gas under constant pressure is heated from 0∘ 𝐶to 1∘ 𝐶, it expands 1/273 of its volume. - If a gas at 0∘ 𝐶was cooled to −273∘ 𝐶, its volume would be zero. - If the pressure remains constant, the volume of a gas varies directly as the absolute temperature. Then: 𝑉 𝑉 𝑉 - Initial𝑇1 =Final 𝑇2at constant pressure or 𝑇 = 𝐾 1 2 Example:The volume of a gas at 20∘ 𝐶is 500 mL. Find its volume at standard temperature if pressure is held constant. Constant Temperatures: 20∘ 𝐶 = 20∘ + 273 = 293 𝐾 0∘ 𝐶 = 0∘ + 273 = 273 𝐾 Use the formula and substitute known values. 𝑉1 𝑉 = 𝑇2 𝑇 1 500.𝑚𝐿 2 𝑥 𝑚𝐿 = 273 𝑥 𝑚𝐿 = 465 𝑚𝐿 293 Boyle’s Law (PV = K) Robert Boyle, found that the volume of a gas decreases when the pressure on it is increased, and vice versa, when the temperature is held constant. If the temperature remains constant, the volume of a gas varies inversely as the pressure changes. Then: 𝑃1 𝑉1 = 𝑃2 𝑉2 at a constant temperature or PV = K Ex: Given the volume of a gas as 200. mL at 1.05 atm pressure, calculate the volume of the same gas at 1.01 atm. Temperature Is held constant. 𝑃1 𝑉1 = 𝑃2 𝑉2 𝑃 - 𝑉2 = 𝑉1 × 𝑃1 - = 200. 𝑚𝐿 × 1.01 𝑎𝑡𝑚 = 208 𝑚𝐿 2 1.05 𝑎𝑡𝑚 Combined Gas Law - A combination of the 2 previous gas laws: 𝑃1 𝑉1 𝑃 𝑉 = 𝑇2 2 𝑇 1 2 1 2 Ex: The volume of a gas at 780. mm pressure and 30∘ 𝐶is 200. mL. What volume would the gas occupy at STP? 𝑃1 𝑉1 𝑃 𝑉 = 𝑇2 2 𝑇 𝑃1 𝑇2 - 𝑉2 = 𝑉1 × - 𝑉2 = 500. 𝑚𝐿 × 𝑃2 × 𝑇1 780.𝑚𝑚 𝐻𝑔 760.𝑚𝑚 𝐻𝑔 × 273 𝐾 303 𝐾 = 462 𝑚𝐿 Gay-Lussac’s Law - At constant volume, the pressure of a given mass of gas varies directly w/ the absolute temperature. Then: 𝑃 𝑃 𝑃 - 𝑇1 = 𝑇2 at constant temperature or 𝑇1 = 𝑘 1 2 1 2 1 Ex: A steel tank contains a gas at 27∘ 𝐶and a pressure of 12.0 atms. Determine the gas pressure when the tank is heated to 100∘ 𝐶. 𝑃 𝑃 - 𝑇1 = 𝑇2 𝑇 - 𝑃2 = 𝑃1 × 𝑇2 - 𝑃2 = 12.0 𝑎𝑡𝑚 × 300.𝐾 = 14.9 𝑎𝑡𝑚 1 373 𝐾