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## Gas Laws ch 13 Chem

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**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.**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?**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**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.**units**• What are units of pressure? • Atm kPa torr mmHg psi • What are units of volume? • L mL cm3 dm3 m3**Do problems 1 and 2 on page 443**• Check your answers in the back of the book page 999 • 2) 0.494 atm**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?**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**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**Charles’s Law animated**• Liquid nitrogen demonstration**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**Do problems 4-6 page 446**• Check your answers page 999 • 4) 3.1 L 6) 2.58 L**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?**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**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**Do problems 8-10 page 448 check your answers**• 8) 1.96 atm 10) 273 degrees C**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**Combined Gas Law**• P1V1/ T1 = P2 V2/T2**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.**Do problems 11 - 13 page 450 and check your answers in the**back of the book page 999 • 12) 72 mL**Do problems 15-18 page 451**• 711 torr