2Section 1 Laws involving changing conditions of a gas If one condition changes something else doesPredict 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.
3What 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 wallstwo apply to Boyle’s Law pressure and volume. What stays constant? What kind of container is needed?
4Compare 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 collisionsBoyles law animatedBoyles law animated 2
5P1V1 = 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 ProblemKnown: 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. P2Substitute the known values. P2Solve for P2. P2 =3. Evaluate the AnswerWhen the volume is , the pressure is . The answer is in ?, a unit of pressure.
6units What are units of pressure? Atm kPa torr mmHg psi What are units of volume?L mL cm3 dm3 m3
7Do problems 1 and 2 on page 443Check your answers in the back of the book page 9992) atm
8Charles’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?
9V1/T1 = V2/T2 What is absolute zero? What is its value? Where no motion exists oC or 0 KCompare a direct and an indirect relationship.Direct as one increases the other does as wellInverse as one increases the other decreases
10What are units of volume? L mL cm3 dm3 m3What are units of Temperature?oC K oFWhich one must be used in Charles’s law?K
11Charles’s Law animated Liquid nitrogen demonstration
12Charles’s LawUse with ExampleProblem 13.2, page 446.Example Problem 13.2.ProblemA 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 ProblemKnown: Unknown:T1 = V1 = V2= T2 =Use Charles’s law and the known values for T1,V1, and T2 tosolve for V2.2. Solve for the UnknownConvert the T1 and T2 Celsius temperatures to kelvin:T °C = K T °C = KWrite 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
13Do problems 4-6 page 446Check your answers page 9994) 3.1 L 6) 2.58 L
14Gay-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?
15P1/T1 = P2/T2 What are units of pressure? kPa atm mmHg torr psi What unit of temperature must be used in this law?K
16The pressure of a gas stored in a refrigerated container is 4 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 ProblemKnown: Unknown:P atm P2 ?T1 = T2 =Use Gay-Lussac’s law and the known values for T1,P1, and T2 tosolve for P2.2. Solve for the UnknownConvert the T1 and T2 Celsius figures to kelvin.T1 22.0°C = K T °C = KWrite 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 atm3. Evaluate the Answer
17Do problems 8-10 page 448 check your answers 8) 1.96 atm ) 273 degrees C
18To solve any problem with changing conditions use the combined gas law 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
20ProblemA 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.Known VariablesP1 = k Pa P2 =320.0 kPaT1 = 25.0°C T2 = 60.0°C V1 = 2.00 LUnknown VariableV2 =? LStep 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 =333 KMultiply 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 LStep 3. Evaluate the answer.
21Do problems 11 - 13 page 450 and check your answers in the back of the book page 999 12) 72 mL