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Gas Laws ch 13 Chem. Section 1 Laws involving changing conditions of a gas If one condition changes something else does If one condition changes something.

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Presentation on theme: "Gas Laws ch 13 Chem. Section 1 Laws involving changing conditions of a gas If one condition changes something else does If one condition changes something."— Presentation transcript:

1 Gas Laws ch 13 Chem

2 Section 1 Laws involving changing conditions of a gas If one condition changes something else does If one condition changes something else does Predict the effect changing pressure or temperature would have on volume of a gas. Predict the effect changing pressure or temperature would have on volume of a gas. Use the combined gas law to solve problems involving changes in pressure, temperature and/or volume. Use the combined gas law to solve problems involving changes in pressure, temperature and/or volume.

3 What are 4 variables that relate to gases? What are 4 variables that relate to gases? P pressure V volume T temperature n moles (number of molecules) P pressure V volume T temperature n moles (number of molecules) What are two types of containers? What are two types of containers? collapsible walls (balloons, syringes) rigid walls collapsible walls (balloons, syringes) rigid walls two apply to Boyles Law pressure and volume. What stays constant? What kind of container is needed? two apply to Boyles Law pressure and volume. What stays constant? What kind of container is needed?

4 Compare a gas to a liquid. Compare a gas to a liquid. Low density, particles do not take up space they occupy space, easily compressed, there are no attractive forces between particles, elastic collisions Low density, particles do not take up space they occupy space, easily compressed, there are no attractive forces between particles, elastic collisions Boyles law animated Boyles law animated Boyles law animated Boyles law animated Boyles law animated 2 Boyles law animated 2 Boyles law animated 2 Boyles law animated 2

5 P 1 V 1 = P 2 V 2 Read Example Problem 13.1 in your text. Read Example Problem 13.1 in your text. Problem Problem Helium gas in a balloon is compressed from 4.0 L to 2.5 L at constant temperature. The gass pressure at 4.0 L is 210 kPa. Determine the pressure at 2.5 L. Helium gas in a balloon is compressed from 4.0 L to 2.5 L at constant temperature. The gass pressure at 4.0 L is 210 kPa. Determine the pressure at 2.5 L. 1. Analyze the Problem 1. Analyze the Problem Known: Unknown: Known: Unknown: V1 = V1 = P2 = P2 = V2 = V2 = P1 = P1 = Use the equation for Boyles law to solve for P2. Use the equation for Boyles law to solve for P2. 2. Solve for the Unknown Write the equation for Boyles law: 2. Solve for the Unknown Write the equation for Boyles law: To solve for P2, divide both sides by V2. P2 To solve for P2, divide both sides by V2. P2 Substitute the known values. P2 Substitute the known values. P2 Solve for P2. P2 = Solve for P2. P2 = 3. Evaluate the Answer 3. Evaluate the Answer When the volume is, the pressure is. The answer is in ?, a unit of pressure. When the volume is, the pressure is. The answer is in ?, a unit of pressure.

6 units What are units of pressure? What are units of pressure? Atm kPa torr mmHg psi Atm kPa torr mmHg psi What are units of volume? What are units of volume? L mL cm 3 dm 3 m 3 L mL cm 3 dm 3 m 3

7 Do problems 1 and 2 on page 443 Do problems 1 and 2 on page 443 Check your answers in the back of the book page 999 Check your answers in the back of the book page 999 2) atm 2) atm

8 Charless Law What are 4 variables that relate to gases? What are 4 variables that relate to gases? What are two types of containers? What are two types of containers? two apply to Charless Law temperature and volume. What stays constant? What kind of container is needed? two apply to Charless Law temperature and volume. What stays constant? What kind of container is needed?

9 V1/T1 = V2/T2 What is absolute zero? What is its value? What is absolute zero? What is its value? Where no motion exists o C or 0 K Where no motion exists o C or 0 K Compare a direct and an indirect relationship. Compare a direct and an indirect relationship. Direct as one increases the other does as well Direct as one increases the other does as well Inverse as one increases the other decreases Inverse as one increases the other decreases

10 What are units of volume? What are units of volume? L mL cm 3 dm 3 m 3 L mL cm 3 dm 3 m 3 What are units of Temperature? What are units of Temperature? o C K o F o C K o F Which one must be used in Charless law? Which one must be used in Charless law? K

11 Charless Law animated Charless Law animated Charless Law animated Charless Law animated Liquid nitrogen demonstration Liquid nitrogen demonstration Liquid nitrogen demonstration Liquid nitrogen demonstration

12 Charless Law Charless Law Use with ExampleProblem 13.2, page 446. Use with ExampleProblem 13.2, page 446. Example Problem Example Problem Problem Problem A gas sample at 40.0°C occupies a volume of 2.32 L. Assuming the pressure is constant, A gas sample at 40.0°C occupies a volume of 2.32 L. Assuming the pressure is constant, if the temperature is raised to 75.0°C, what will the volume be? if the temperature is raised to 75.0°C, what will the volume be? 1. Analyze the Problem 1. Analyze the Problem Known: Unknown: Known: Unknown: T1 = V1 = V2= T2 = T1 = V1 = V2= T2 = Use Charless law and the known values for T1,V1, and T2 to Use Charless law and the known values for T1,V1, and T2 to solve for V2. solve for V2. 2. Solve for the Unknown 2. Solve for the Unknown Convert the T1 and T2 Celsius temperatures to kelvin: Convert the T1 and T2 Celsius temperatures to kelvin: T °C = K T °C = K T °C = K T °C = K Write the equation for Charless law: Write the equation for Charless law: To solve for V2, multiply both sides by T2: To solve for V2, multiply both sides by T2: Substitute known values: Substitute known values: Solve for V2. Solve for V2. V2 = V2 = 3. Evaluate the Answer 3. Evaluate the Answer

13 Do problems 4-6 page 446 Do problems 4-6 page 446 Check your answers page 999 Check your answers page 999 4) 3.1 L 6) 2.58 L 4) 3.1 L 6) 2.58 L

14 Gay-Lussacs Law What are 4 variables that relate to gases? What are 4 variables that relate to gases? What are two types of containers? What are two types of containers? two apply to Gay-Lussacs Law pressure and temperature. What stays constant? What kind of container is needed? two apply to Gay-Lussacs Law pressure and temperature. What stays constant? What kind of container is needed?

15 P 1 /T 1 = P 2 /T 2 What are units of pressure? What are units of pressure? kPa atm mmHg torr psi kPa atm mmHg torr psi What unit of temperature must be used in this law? What unit of temperature must be used in this law? K

16 The pressure of a gas stored in a refrigerated container is 4.0 atm at 22.0°C. Determine the gas pressure in the tank if the temperature is lowered to 0.0°C. 1. Analyze the Problem Known: Unknown: P1 4.0 atm P2 ? T1 = T2 = Use Gay-Lussacs law and the known values for T1,P1, and T2 to solve for P2. 2. Solve for the Unknown Convert the T1 and T2 Celsius figures to kelvin. T1 22.0°C = K T °C = K Write the equation for Gay-Lussacs law. To solve for P2, multiply both sides by T2. Substitute known values. Solve for P2. P2 = 3.7 atm 3. Evaluate the Answer

17 Do problems 8-10 page 448 check your answers Do problems 8-10 page 448 check your answers 8) 1.96 atm 10) 273 degrees C 8) 1.96 atm 10) 273 degrees C

18 To solve any problem with changing conditions use the combined gas law. If any variable is not used or is constant leave it out of the equation and you will have the correct equation to use To solve any problem with changing conditions use the combined gas law. If any variable is not used or is constant leave it out of the equation and you will have the correct equation to use

19 Combined Gas Law P1V1/ T1 = P2 V2/T2 P1V1/ T1 = P2 V2/T2

20 Problem Problem A gas at kPa and 25.0°C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 60.0° and the pressure increased to kPa,what is the new volume? A gas at kPa and 25.0°C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 60.0° and the pressure increased to kPa,what is the new volume? Step 1. Analyze the problem. Step 1. Analyze the problem. Known Variables Known Variables P1 = k Pa P2 =320.0 kPa P1 = k Pa P2 =320.0 kPa T1 = 25.0°C T2 = 60.0°C V1 = 2.00 L T1 = 25.0°C T2 = 60.0°C V1 = 2.00 L Unknown Variable Unknown Variable V2 =? L V2 =? L Step 2. Solve for the unknown.Add 273 to the Celsius temperature for T1 and T2 to obtain the kelvin temperature. Step 2. Solve for the unknown.Add 273 to the Celsius temperature for T1 and T2 to obtain the kelvin temperature. T °C = 298 K; T °C = 298 K; T °C =333 K T °C =333 K Multiply both sides of the equation for the combined law by T2 and divide by P2 to solve for V2. Multiply both sides of the equation for the combined law by T2 and divide by P2 to solve for V2. Substitute the known values into the rearranged equation; multiply and divide numbers and units to solve for V2. Substitute the known values into the rearranged equation; multiply and divide numbers and units to solve for V2. V2 =0.698 L V2 =0.698 L Step 3. Evaluate the answer. Step 3. Evaluate the answer.

21 Do problems page 450 and check your answers in the back of the book page 999 Do problems page 450 and check your answers in the back of the book page ) 72 mL 12) 72 mL

22 Do problems page 451 Do problems page torr 711 torr


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