CHE 123: General Chemistry I Chapter 5: Gases CHE 123: General Chemistry I Dr. Jerome Williams, Ph.D. Saint Leo University

Overview Gas Laws Ideal Gas Law General Gas Law Example Problems

Boyle’s Law Pressure of a gas is inversely proportional to its volume Robert Boyle (1627–1691) Pressure of a gas is inversely proportional to its volume constant T and amount of gas graph P vs V is curve graph P vs 1/V is straight line As P increases, V decreases by the same factor P x V = constant P1 x V1 = P2 x V2 Tro: Chemistry: A Molecular Approach, 2/e

Boyle’s Experiment, P x V Tro: Chemistry: A Molecular Approach, 2/e

Tro: Chemistry: A Molecular Approach, 2/e

Tro: Chemistry: A Molecular Approach, 2/e

Boyle’s Law: A Molecular View Pressure is caused by the molecules striking the sides of the container When you decrease the volume of the container with the same number of molecules in the container, more molecules will hit the wall at the same instant This results in increasing the pressure Tro: Chemistry: A Molecular Approach, 2/e

Example 5. 2: A cylinder with a movable piston has a volume of 7 Example 5.2: A cylinder with a movable piston has a volume of 7.25 L at 4.52 atm. What is the volume at 1.21 atm? Given: Find: V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm V2, L Conceptual Plan: Relationships: P1 ∙ V1 = P2 ∙ V2 V1, P1, P2 V2 Solution: Check: because P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does Tro: Chemistry: A Molecular Approach, 2/e

Practice – A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon is now 2.78 x 103 mL, what was it originally? Tro: Chemistry: A Molecular Approach, 2/e

A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon is now 2.78x 103 mL, what was it originally? Given: Find: V2 =2780 mL, P1 = 782 torr, P2 = 0.500 atm V1, mL Conceptual Plan: Relationships: P1 ∙ V1 = P2 ∙ V2 , 1 atm = 760 torr (exactly) V2, P1, P2 V1 Solution: Check: because P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does Tro: Chemistry: A Molecular Approach, 2/e

Charles’s Law Volume is directly proportional to temperature Jacques Charles (1746–1823) Volume is directly proportional to temperature constant P and amount of gas graph of V vs. T is straight line As T increases, V also increases Kelvin T = Celsius T + 273 V = constant x T if T measured in Kelvin Tro: Chemistry: A Molecular Approach, 2/e

If you plot volume vs. temperature for any gas at constant pressure, the points will all fall on a straight line If the lines are extrapolated back to a volume of “0,” they all show the same temperature, −273.15 °C, called absolute zero Tro: Chemistry: A Molecular Approach, 2/e

Example 5. 3: A gas has a volume of 2. 57 L at 0. 00 °C Example 5.3: A gas has a volume of 2.57 L at 0.00 °C. What was the temperature in Celsius at 2.80 L? Given: Find: V1 =2.57 L, V2 = 2.80 L, t2 = 0.00 °C t1, K and °C Conceptual Plan: Relationships: V1, V2, T2 T1 Solution: Check: because T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does Tro: Chemistry: A Molecular Approach, 2/e

Practice – The temperature inside a balloon is raised from 25 Practice – The temperature inside a balloon is raised from 25.0 °C to 250.0 °C. If the volume of cold air was 10.0 L, what is the volume of hot air? Tro: Chemistry: A Molecular Approach, 2/e

The temperature inside a balloon is raised from 25. 0 °C to 250. 0 °C The temperature inside a balloon is raised from 25.0 °C to 250.0 °C. If the volume of cold air was 10.0 L, what is the volume of hot air? Given: Find: V1 =10.0 L, t1 = 25.0 °C L, t2 = 250.0 °C V2, L Conceptual Plan: Relationships: V1, T1, T2 V2 Solution: Check: when the temperature increases, the volume should increase, and it does Tro: Chemistry: A Molecular Approach, 2/e

Avogadro’s Law Amedeo Avogadro (1776–1856) Volume directly proportional to the number of gas molecules V = constant x n constant P and T more gas molecules = larger volume Count number of gas molecules by moles Equal volumes of gases contain equal numbers of molecules the gas doesn’t matter Tro: Chemistry: A Molecular Approach, 2/e

Example 5. 4:A 0. 225 mol sample of He has a volume of 4. 65 L Example 5.4:A 0.225 mol sample of He has a volume of 4.65 L. How many moles must be added to give 6.48 L? Given: Find: V1 = 4.65 L, V2 = 6.48 L, n1 = 0.225 mol n2, and added moles Conceptual Plan: Relationships: V1, V2, n1 n2 Solution: Check: because n and V are directly proportional, when the volume increases, the moles should increase, and they do Tro: Chemistry: A Molecular Approach, 2/e

Ideal Gas Law Ideal gases obey an equation incorporating the laws of Charles, Boyle, and Avogadro. P V = n R T 1 mole of an ideal gas occupies 22.414 L at STP STP conditions are 273.15 K and 1 atm pressure The gas constant R = 0.08206 L·atm·K–1·mol–1

Example 5.6: How many moles of gas are in a basketball with total pressure 24.3 psi, volume of 3.24 L at 25°C? Given: Find: V = 3.24 L, P = 24.3 psi, t = 25 °C n, mol Conceptual Plan: Relationships: P, V, T, R n Solution: Check: 1 mole at STP occupies 22.4 L, because there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas Tro: Chemistry: A Molecular Approach, 2/e

Chapter 5, Unnumbered Figure 1, Page 191

Example Problems Sulfur hexafluoride (SF6) is a colorless, odorless, very unreactive gas. Calculate the pressure (in atm) exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 L at 69.5°C. Answer = 9.42 atm

Example Problems What is the volume (in liters) occupied by 7.40 g of CO2 at STP? Answer = 3.77 L

Ideal Gas Law: Application Density and Molar Mass Calculations Given the following: Note: MM = molar mass Density d = mass/volume No. moles n n = mass / MM Show that MM = dRT / P

Figure: 05-15-06UN Title: Molar Volume Caption: One mole of any gas occupies approximately 22.4 L at standard temperature (273 K) and pressure (1.0 atm).

Figure: 05-15-07UN Title: Molar density (1) Caption: The density in grams/liter can be obtained from molar density by multiplying by the molar mass (M).

Figure: 05-15-08UN Title: Molar density (2) Caption: The density in grams/liter can be obtained from the Ideal Gas Law.

Ideal Gas Law: Application Examples What is the molar mass of a gas with a density of 1.342 g/L–1 at STP? What is the density of uranium hexafluoride, UF6, (MM = 352 g/mol) under conditions of STP? The density of a gaseous compound is 3.38 g/L–1 at 40°C and 1.97 atm. What is its molar mass?

Ideal Gas Law: Application Answers Molar mass = 30.08 g/mole Density of UF6 = 15.7 g/mL Molar mass = 44.09 g/mole

General Gas Law Equation (P1V1 / T1 )= (P2V2 / T2 ) Only equation that uses two sets of data for V, P, and T.

Example Problem Oxygen gas is normally sold in 49.0 L steel containers at a pressure of 150.0 atm. What volume would the gas occupy if the pressure was reduced to 1.02 atm and the temperature raised from 20oC to 35oC?