A Directed Learning Activity for Hartnell College Chemistry 1

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A Directed Learning Activity for Hartnell College Chemistry 1
Using the Gas Laws A Directed Learning Activity for Hartnell College Chemistry 1 Funded by the Title V – STEM Grant #P031S through Hartnell College Start For information contact

Student Learning Objectives
This tutorial will help you to: Manipulate the ideal gas laws to calculate pressure, volume, temperature and amount of gas present & Use balanced equations and the ideal gas laws to predict the amounts of reagents and/or products for gaseous reactions Next

Getting Started This set of Power Point slides will lead you through a series of short lessons and quizzes on the topics covered by this Directed Learning Activity tutorial. Move through the slideshow at your own pace. There are several hyperlinks you can click on to take you to additional information, take quizzes, get answers to quizzes, and to skip to other lessons. You can end this slide show at any time by hitting the “ESC” key on your computer keyboard. Next

Table of Topics What You Should Already Know Boyle’s Law Charles Law
Gay-Lussac’s Law Combined Gas Law Avogadro’s Law Ideal Gas Law Putting it All Together (Gas Stoichiometry) Dalton’s Law of Partial Pressures Next

How to manipulate algebraic equations Conversions between different measurement units The basics of the Kinetic Molecular Theory of ideal gases Understand the “mole” Write balanced chemical equations for reactions If you are uncertain about these skills, please refer to your lecture text for help. Next

Boyle’s Law At constant temperature (T), the volume (V) of a fixed mass of gas is inversely proportional to the Pressure (P). 𝑃 ∝ 1 𝑉 If we have two sets of conditions (1 and 2) for the same gas, we can then write this equation, which is known as Boyle’s Law: 𝑃1𝑉1=𝑃2𝑉2 which is true when moles of gas and T are constant. Next

Example 1 for Boyle’s Law
1. A sample of an ideal gas occupies 2.00 L at 760 torr. What volume will this amount of gas occupy if the temperature remains constant but the pressure changes to 1.25 atm? Remember that 1 atm = 760 torr. Solution: use Boyle’s law and substitute in the values for conditions 1 and 2. Property 1 2 P 760 torr = 1 atm 1.25 atm V 2.00 L ? Next

Example 1 for Boyle’s Law (cont’d)
𝑃1𝑉1=𝑃2𝑉2 Solve for the new volume 𝑉2= 𝑃1𝑉1 𝑃2 = 1.00 𝑎𝑡𝑚 ∗ 2.00𝐿 1.25 𝑎𝑡𝑚 𝑉2= 1.60 L Next

Example 2 for Boyle’s Law
A 1.00 L sample of an ideal gas at 760 torr is compressed to L at constant temperature. Calculate the final pressure of the gas. Solution: use Boyle’s law and substitute in the values for conditions 1 and 2. Property 1 2 P 760 torr = 1 atm ? V 1.00 L 0.800 L Next

Example 2 for Boyle’s Law (cont’d)
𝑃1𝑉1=𝑃2𝑉2 Solve for the new pressure 𝑃2= 𝑃1𝑉1 𝑉2 = 760 𝑡𝑜𝑟𝑟 ∗1.00𝐿 0.800𝐿 𝑃2= 950 torr Next

Quiz Question 1 A mass of oxygen occupies 7.00 L under a pressure of 740 torr. Determine the volume of the same mass of gas at the standard pressure of 760 torr, the temperature remaining constant. Click to review Check answer

Solution: 𝑉2= 6.82 L Explanation: 𝑃1𝑉1=𝑃2𝑉2 Solve for the new volume: 𝑉2= 𝑃1𝑉1 𝑃2 = 740 𝑡𝑜𝑟𝑟 ∗7.00𝐿 760 𝑡𝑜𝑟𝑟 𝑉2= 6.82 L Click to review Next question

Quiz Question 2 Ten (10.0) liters of hydrogen under 7.0 atm pressure is slowly compressed until it occupies only 4.0 L of volume. Assume that the temperature of the gas remains constant. What pressure is needed for the gas to remain compressed? Click to review Check answer

Solution to Quiz Question 2
Solution: 𝑃2= 18 atm Explanation: 𝑃1𝑉1=𝑃2𝑉2 Solve for the new pressure 𝑃2= 𝑃1𝑉1 𝑉2 = 7.0 𝑎𝑡𝑚 ∗10.0𝐿 4.00𝐿 𝑃2= 18 atm Click to review Next lesson

Charles’ Law At constant pressure the volume (V) of a fixed mass of an ideal gas is directly proportional to the Kelvin temperature (T). 𝑉 ∝𝑇 If we have two sets of conditions (1 and 2) for the same gas, we can then write this equation, which is known as Charles’ Law: 𝑉1 𝑇1 = 𝑉2 𝑇2 which is true when moles and P are constant. Next

Example for Charles’ Law
A given mass of chlorine gas occupies 25.0 L at 20 °C. What is the new volume at 45 °C, assuming that the pressure remains constant? Solution: Use Charles’ Law and substitute the values for conditions 1 and 2. Remember temperature must be in Kelvin. Property 1 2 V 25.0 L ? T K K Next

Example for Charles’ Law (cont’d)
𝑉1 𝑇1 = 𝑉2 𝑇2 Solve for V2 and substitute in values: 𝑉2=𝑉1∗ 𝑇2 𝑇1 = 25.0 𝐿 𝐾 ( 𝐾) 𝑉2 = 27.1 L Next

Quiz Question 3 A sample of gaseous argon is maintained at a constant pressure. The sample has an initial volume of 10.5 L at 25 °C. What will be volume be if the same sample is kept at the same pressure, but heated to 250 °C? Click to review Check answer

Solution: V2 = 18.4 L Explanation: Use Charles’ Law. Use temperature in Kelvin. 𝑉1 𝑇1 = 𝑉2 𝑇2 Solve for V2 and substitute in values: 𝑉2=𝑉1∗ 𝑇2 𝑇1 = 10.5 𝐿 𝐾 ( 𝐾) 𝑉2 = 18.4 L Click to review Next question

Quiz Question 4 A certain amount of gas occupies of volume of 100. mL at a temperature of 20 °C. What will the new volume be at 10 °C, if the pressure remains constant? Click to review Check answer

Solution: V2 = 96.6 mL Explanation: Use Charles’ Law. Temperature in Kelvin. 𝑉1 𝑇1 = 𝑉2 𝑇2 Solve for V2 and substitute in values: 𝑉2=𝑉1∗ 𝑇2 𝑇1 = 100 𝑚𝐿 𝐾 ( 𝐾) 𝑉2 = 96.6 mL Click to review Next lesson

Gay-Lussac’s Law The pressure (P)of a fixed mass of an ideal gas, at constant volume, is directly proportional to the Kelvin temperature. 𝑃 ∝𝑇 If we have two sets of conditions (1 and 2) for the same gas, we can then write this equation, which is known as Gay-Lussac’s Law: 𝑃1 𝑇1 = 𝑃2 𝑇2 which is true when moles and V are constant. Next

Example for Gay-Lussac’s Law
The air in a cylindrical tank has a pressure of 640 torr at 23 °C. When the tank was placed in the sun, the temperature rose to 48 °C. What was the final pressure in the tank if the mass and volume of the gas does not change? Solution : Use Gay-Lussac’s Law. Remember that temperature must be in Kelvin. Property 1 2 P 640 torr ? T K K Next

Example for Gay-Lussac’s Law (cont’d)
𝑃1 𝑇1 = 𝑃2 𝑇2 Solving for 𝑃2=𝑃1∗ 𝑇2 𝑇1 =640 𝑡𝑜𝑟𝑟∗ 321 𝐾 296 𝐾 = 694 torr Next

Quiz Question 5 A sealed glass bulb contains a sample of He gas at a pressure of 750 torr and 27 °C. The bulb was cooled down to -73 °C. What was the new gas pressure inside the bulb? Click to review Check answer

Solution: P2 = 500 torr Explanation: Using Gay-Lussac’s Law and substituting in the values, solve for P2, Remember to convert the temperatures to Kelvin. 𝑃2=𝑃1∗ 𝑇2 𝑇1 =750 𝑡𝑜𝑟𝑟∗ 200 𝐾 300 𝐾 = 500 torr Click to review Next question

Quiz Question 6 A steel tank contains carbon dioxide gas at 27 °C and at a pressure of 11.0 atm. Determine the internal pressure when the gas and its contents are heated to 100 °C. Assume that the amount of carbon dioxide and the volume of the tank are constant. Click to review Check answer

Solution: P2 = 13.7 atm Explanation: Using Gay-Lussac’s Law and substituting in the values, solve for P2. Remember to convert the temperatures to Kelvin. 𝑃2=𝑃1∗ 𝑇2 𝑇1 =11.0 𝑎𝑡𝑚∗ 373 𝐾 300 𝐾 = 13.7 atm Click to review Next lesson

The Combined Gas Law The Combined Gas Law is a combination of Boyle’s and Charles’ Laws. It may be used whenever the mass of an ideal gas remains constant and the Kelvin temperature (T) is used. 𝑃1𝑉1 𝑇1 = 𝑃2𝑉2 𝑇2 as long as the moles are constant. You can use this equation for two different conditions (1 and 2) and solve for any one of the six variables. Next

Example for the Combined Gas Law
What would be the new pressure for a 2.00 L sample of gas at 1.00 atm and -20 °C that is compressed to a new volume of L at 40 °C? Solution: Use the combined gas law. Temperatures must be converted to Kelvin. 1 2 P 1.00 atm ? V 2.00 L K T 0.500 L K Next

Example for the Combined Gas Law (cont’d)
𝑃2=𝑃1 𝑉1 𝑉2 𝑇2 𝑇1 = 1.00 𝑎𝑡𝑚 𝐿 0.500𝐿 𝐾 253𝐾 =4.95 𝑎𝑡𝑚 Next

Quiz Question 7 A 2.50 L sample of gas is at 0 °C and 1.00 atm pressure. What will the temperature of the gas be if it is placed in a 2.00 L container at 1.50 atm pressure? Click to review Check answer

Solution: T2 = 328 K Explanation: Use the combined gas law and solve for T2. Remember to convert T to Kelvin. 𝑇2=𝑇1 𝑃2 𝑃1 𝑉2 𝑉1 =273 𝐾 1.50 𝑎𝑡𝑚 1.00 𝑎𝑡𝑚 2.00 𝐿 2.50 𝐿 =328 𝐾 Click to review Next lesson

Avogadro’s Gas Law Under Standard Temperature and Pressure (STP) conditions, the volume of one mole of an ideal gas will occupy Liters. 1 mole ideal gas = L, at STP STP conditions are 273K and 1 atmosphere (760 mm Hg) of pressure. Next lesson

The Ideal Gas Law To adequately describe an ideal gas under a particular set of physical conditions, you need to know: the temperature, pressure and volume of the gas; and the amount of gas. This is summarized in the following equation: Next

The Ideal Gas Law cont’d
𝑃𝑉=𝑛𝑅𝑇 where P is the pressure of the gas V is the volume of the gas n is the number of moles of the gas in the sample T is the temperature in Kelvin R is the Universal Gas Constant, which varies depending on the units of the other variables R = 𝐿∙𝑎𝑡𝑚 𝑀𝑜𝑙∙𝐾 or R = 𝐿ˑ𝑘𝑃𝑎 𝑀𝑜𝑙ˑ𝐾 Next

Example for the Ideal Gas Law
What is the pressure in atmospheres of 3.4x10-3 moles of argon gas in a 75-mL glass bulb at 20 °C? Solution: The problem gives three out of the four properties of an ideal gas (moles, volume, and temperature) and asks for the fourth (pressure). Use the Ideal Gas Law. Next

Example for Ideal Gas Law cont’d
Rearrange to solve for P. Convert the temperature to K. Convert the volume to L. Choose R = 𝐿∙𝑎𝑡𝑚 𝑀𝑜𝑙∙𝐾 . 𝑃= 𝑛𝑅𝑇 𝑉 𝑃= 3.4𝑥10_3𝑚𝑜𝑙𝑒𝑠 𝐿ˑ𝑎𝑡𝑚/𝑚𝑜𝑙ˑ𝐾 293𝐾 0.075𝐿 P = 1.1 atm Next

Quiz Question 8 An incandescent light bulb contains g of Ar in a 23.0-mL volume. The pressure inside the light bulb under these conditions is 952 torr. What is the temperature of the Ar gas under these conditions? Click to review Check answer

Solution: T = 333 K. Explanation: We can use the Ideal Gas Law if we convert grams of Ar to moles, convert 23.0 mL to Liters, and 952 torrs to atm. 𝑇= 𝑃𝑉 𝑛𝑅 = 𝑃𝑉𝑀 𝑔𝑅 = 952 𝑡𝑜𝑟𝑟 23.0 𝑚𝐿 𝑔 𝑚𝑜𝑙 1 𝑎𝑡𝑚 760 𝑡𝑜𝑟𝑟 1 𝐿 1000 𝑚𝐿 𝑔 𝐿ˑ𝑎𝑡𝑚/𝑚𝑜𝑙ˑ𝐾 = 333 K Click to review Next lesson

Putting it All Together – Gas Stoichiometry
If we are given a chemical reaction where one or more of the reactants is a gas, we can use the balanced chemical equation to determine the volume of gas that is obtained if we know the amounts of the reagents. If we know the experimental conditions of temperature and pressure, we can use the Ideal Gas Law. Often problems like this are written to be solved under STP conditions, so the results of Avogadro’s Law can also be used. This kind of problem is sometimes called gas stoichiometry. Let’s look at an example. Next

Example Problem How many liters of carbon dioxide at STP will be formed from the complete combustion of g of ethanol, C2H5OH(l)? What would this volume be if we then changed the conditions of the gas to 23 °C and 0.95 atm to expand the gas after formation? Solution: First we need to write the balanced equation for the reaction. C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 2 H2O(l) Next, we need to know how many moles of ethanol we have as starting material. This will allow us to use this procedure: g C2H5OH → mol C2H5OH → mol CO2 → L STP For the last step, since the conditions are at STP, we know that each mole of CO2 = L. If the conditions are different than STP, we have to use the Ideal Gas Law to determine the volume. Next

Example Problem cont’d
?𝐿 𝐶𝑂2=82.60 𝑔 𝐶2𝐻5𝑂𝐻 1𝑚𝑜𝑙 𝐶2𝐻5𝑂𝐻 𝑔 𝐶2𝐻5𝑂𝐻 2𝑚𝑜𝑙 𝐶𝑂2 1𝑚𝑜𝑙 𝐶2𝐻5𝑂𝐻 𝐿 𝐶𝑂2 1𝑚𝑜𝑙 𝐶𝑂2 = L STP For the second part of the problem, we can calculate for the new volume of CO2 under the new conditions: 𝑉2=𝑉1 𝑃1 𝑃2 𝑇2 𝑇1 =80.37 𝐿 1.00𝑎𝑡𝑚 0.95𝑎𝑡𝑚 V2 = 91.7 L (Note that you could also solve by calculating for moles of CO2 at STP and substituting into the Ideal Gas Law.) Next

Quiz Question 9 Calculate the volume of O2 that can be prepared at 60 °C and 760 torr by the decomposition of 20.0 g H2O2 to H2O and O2. The reaction is: 2 H2O2 → 2 H2O + O2. Click to review Check answer

Solution: V = 8.03 L Explanation: Since the conditions are not at STP, you must use the Ideal Gas Law. This means that you must determine how many moles of O2 is produced from 20.0 g H2O2. ? 𝑚𝑜𝑙 𝑂2= 20.0𝑔 𝐻2𝑂2 1𝑚𝑜𝑙 𝐻2𝑂2 34.0𝑔 𝐻2𝑂2 1𝑚𝑜𝑙 𝑂2 2𝑚𝑜𝑙 𝐻2𝑂2 =0.294𝑚𝑜𝑙 𝑂2 Next

Answer to Quiz Question 9 cont’d
𝑉= 𝑛𝑅𝑇 𝑃 = 0.294𝑚𝑜𝑙 𝑂 𝐿ˑ𝑎𝑡𝑚 𝑚𝑜𝑙ˑ𝐾 333𝐾 1.00𝑎𝑡𝑚 V = 8.03 L Click to review Next lesson

Dalton’s Law of Partial Pressures
The partial pressure of a gas in a mixture of gases is the portion of the total pressure that one gas contributes. Dalton’s Law of Partial Pressures is generally written as: 𝑃𝑡𝑜𝑡𝑎𝑙=∑𝑛 𝑒𝑎𝑐ℎ 𝑔𝑎𝑠 𝑅𝑇 𝑉 or 𝑃𝑡𝑜𝑡𝑎𝑙=∑𝑃𝑝𝑎𝑟𝑡𝑖𝑎𝑙 If you have a problem involving only gases and no chemical reaction involved, you may be able to use this formula to calculate either the total pressure or the pressure of one of the gases. Next

Example Using Dalton’s Law
3.00L oxygen gas is collected over water at 27 °C and when the barometric pressure is 787 torr. The vapor pressure of water at 27 C is 27 torr. What is the partial pressure of the dry oxygen gas under these conditions? Solution: PO2 = Ptotal – PH2O = 787torr – 27torr = 760torr Next

Quiz Question 10 Exactly 100 mL of oxygen gas is collected over water at 23 °C and 800 torr. Calculate the standard volume of the dry oxygen if the vapor pressure of water at 23 °C is 21.1 torr. Click to review Check answer

Solution: V2 = 94.5 mL Explanation: We first use Dalton’s Law to find the partial pressure of oxygen PO2 = Ptotal – PH20 = 800torr – 21.1torr = 779torr Next

Solution to Quiz Question 10 cont’d
You must then use the combined gas law to determine the volume of the oxygen at STP. Here is what you know. 1 2 P 779 torr 760 torr V 100. mL ? T 273 Next

Solution to Quiz Question 10 cont’d
𝑉2=𝑉1 𝑃1 𝑃2 𝑇2 𝑇1 =100. 𝑚𝐿 779 𝑡𝑜𝑟𝑟 760 𝑡𝑜𝑟𝑟 273 𝐾 296 𝐾 V2 = 94.5 mL Click to review Next

Congratulations! You have successfully completed this Directed Learning Activity tutorial. We hope that this has helped you to better understand this topic. Click here to end. Click here to repeat this activity.

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