Gas Laws ch 13 Chem

# Gas Laws ch 13 Chem

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## Gas Laws ch 13 Chem

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1. Gas Laws ch 13 Chem

2. Section 1 Laws involving changing conditions of a gas • If one condition changes something else does • 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.

3. What are 4 variables that relate to gases? • P pressure V volume T temperature n moles (number of molecules) • What are two types of containers? • collapsible walls (balloons, syringes) rigid walls • two apply to Boyle’s Law pressure and volume. What stays constant? What kind of container is needed?

4. 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 • Boyles law animated • Boyles law animated 2

5. P1V1 = P2V2 • Read Example Problem 13.1 in your text. • Problem • Helium gas in a balloon is compressed from 4.0 L to 2.5 L at constant temperature. The gas’s pressure at 4.0 L is 210 kPa. Determine the pressure at 2.5 L. • 1. Analyze the Problem • Known: Unknown: • V1 = • P2 = • V2 = • P1 = • Use the equation for Boyle’s law to solve for P2. • 2. Solve for the Unknown Write the equation for Boyle’s law: • To solve for P2, divide both sides by V2. P2 • Substitute the known values. P2 • Solve for P2. P2 = • 3. Evaluate the Answer • When the volume is , the pressure is . The answer is in ?, a unit of pressure.

6. units • What are units of pressure? • Atm kPa torr mmHg psi • What are units of volume? • L mL cm3 dm3 m3

7. Do problems 1 and 2 on page 443 • Check your answers in the back of the book page 999 • 2) 0.494 atm

8. Charles’s Law • What are 4 variables that relate to gases? • What are two types of containers? • two apply to Charles’s 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? • Where no motion exists -273.15 oC or 0 K • Compare a direct and an indirect relationship. • Direct as one increases the other does as well • Inverse as one increases the other decreases

10. What are units of volume? • L mL cm3 dm3 m3 • What are units of Temperature? • oC K oF • Which one must be used in Charles’s law? • K

11. Charles’s Law animated • Liquid nitrogen demonstration

12. Charles’s Law • Use with ExampleProblem 13.2, page 446. • Example Problem 13.2. • Problem • 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? • 1. Analyze the Problem • Known: Unknown: • T1 = V1 = V2= T2 = • Use Charles’s law and the known values for T1,V1, and T2 to • solve for V2. • 2. Solve for the Unknown • Convert the T1 and T2 Celsius temperatures to kelvin: • T1 273 +40.0°C = K T2 273 +75.0°C = K • Write the equation for Charles’s law: • To solve for V2, multiply both sides by T2: • Substitute known values: • Solve for V2. • V2 = • 3. Evaluate the Answer

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

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

15. P1/T1 = P2/T2 • What are units of pressure? • kPa atm mmHg torr psi • 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-Lussac’s 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 T2 0 + 273 °C = K • Write the equation for Gay-Lussac’s 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 • 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

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

20. Problem • A gas at 100.0 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 320.0 kPa,what is the new volume? • Step 1. Analyze the problem. • Known Variables • P1 =100.0 k Pa P2 =320.0 kPa • T1 = 25.0°C T2 = 60.0°C V1 = 2.00 L • Unknown Variable • V2 =? L • Step 2. Solve for the unknown.Add 273 to the Celsius temperature for T1 and T2 to obtain the kelvin temperature. • T1 273 +25.0°C = 298 K; • T2 273 + 60.0°C =333 K • 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. • V2 =0.698 L • Step 3. Evaluate the answer.

21. Do problems 11 - 13 page 450 and check your answers in the back of the book page 999 • 12) 72 mL

22. Do problems 15-18 page 451 • 711 torr