Gas Laws ch 13 Chem Gas Laws ch 13 Chem

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

• Predict the effect changing pressure or Predict the effect changing pressure or temperature would have on volume of a gas.temperature would have on volume of a gas.

• Use the combined gas law to solve problems Use the combined gas law to solve problems involving changes in pressure, temperature involving changes in pressure, temperature and/or volume.and/or volume.

• What are 4 variables that relate to gases?What are 4 variables that relate to gases?

• P pressure V volume T temperature P pressure V volume T temperature n moles (number of molecules) n moles (number of molecules)

• What are two types of containers?What are two types of containers?

• collapsible walls (balloons, syringes) rigid collapsible walls (balloons, syringes) rigid wallswalls

• two apply to Boyle’s Law pressure and two apply to Boyle’s Law pressure and volume. What stays constant? What kind of volume. What stays constant? What kind of container is needed?container is needed?

• Compare a gas to a liquid.Compare a gas to a liquid.

• Low density, particles do not take up Low density, particles do not take up space they occupy space, easily space they occupy space, easily compressed, there are no attractive compressed, there are no attractive forces between particles, elastic forces between particles, elastic collisionscollisions

• Boyles law animatedBoyles law animated

• Boyles law animated 2Boyles law animated 2

PP11VV11 = P = P22VV22• Read Example Problem 13.1 in your text.Read Example Problem 13.1 in your text.• ProblemProblem• Helium gas in a balloon is compressed from 4.0 L to 2.5 L at constant 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 temperature. The gas’s pressure at 4.0 L is 210 kPa. Determine the pressure at 2.5 L.pressure at 2.5 L.

• 1. Analyze the Problem1. Analyze the Problem• Known: Unknown:Known: Unknown:• VV1 =1 =• PP2 =2 =• VV2 =2 =• PP1 =1 =• Use the equation for Boyle’s law to solve for Use the equation for Boyle’s law to solve for PP2.2.• 2. Solve for the Unknown Write the equation for Boyle’s law:2. Solve for the Unknown Write the equation for Boyle’s law:• To solve for To solve for PP2, divide both sides by 2, divide both sides by VV2. 2. PP2 2 • Substitute the known values. Substitute the known values. PP2 2 • Solve for Solve for PP2. 2. PP2 =2 =• 3. Evaluate the Answer3. Evaluate the Answer• When the volume is , the pressure is . The answer is in ?, a unit of When the volume is , the pressure is . The answer is in ?, a unit of

pressure.pressure.

unitsunits

• What are units of pressure?What are units of pressure?

• Atm kPa torr mmHg psiAtm kPa torr mmHg psi

• What are units of volume?What are units of volume?

• L mL cmL mL cm3 3 dmdm33 m m33

• Do problems 1 and 2 on page 443Do problems 1 and 2 on page 443

• Check your answers in the back of Check your answers in the back of the book page 999 the book page 999

• 2) 0.494 atm2) 0.494 atm

Charles’s LawCharles’s Law

• What are 4 variables that relate to What are 4 variables that relate to gases?gases?

• What are two types of containers?What are two types of containers?

• two apply to Charles’s Law temperature two apply to Charles’s Law temperature and volume. What stays constant? and volume. What stays constant? What kind of container is needed?What kind of container is needed?

V1/T1 = V2/T2V1/T1 = V2/T2

• What is absolute zero? What is its What is absolute zero? What is its value?value?

• Where no motion exists -273.15 Where no motion exists -273.15 ooC or 0 KC or 0 K• Compare a direct and an indirect Compare a direct and an indirect

relationship.relationship.• Direct as one increases the other does Direct as one increases the other does

as wellas well• Inverse as one increases the other Inverse as one increases the other

decreasesdecreases

• What are units of volume?What are units of volume?

• L mL cmL mL cm3 3 dmdm3 3 mm33

• What are units of Temperature? What are units of Temperature?

• ooC K C K ooFF

• Which one must be used in Charles’s Which one must be used in Charles’s law?law?

• KK

• Charles’s Law animatedCharles’s Law animated

• Liquid nitrogen demonstrationLiquid nitrogen demonstration

• Charles’s LawCharles’s Law• Use with ExampleProblem 13.2, page 446.Use with ExampleProblem 13.2, page 446.• Example Problem 13.2.Example Problem 13.2.• ProblemProblem• A gas sample at 40.0°C occupies a volume of 2.32 L. Assuming the A gas sample at 40.0°C occupies a volume of 2.32 L. Assuming the

pressure is constant, 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 Problem1. Analyze the Problem• Known: Unknown:Known: Unknown:• TT1 = 1 = VV1 = 1 = VV2= 2= TT2 =2 =• Use Charles’s law and the known values for Use Charles’s law and the known values for TT1,1,VV1, and 1, and TT2 to2 to• solve for solve for VV2.2.• 2. Solve for the Unknown2. Solve for the Unknown• Convert the Convert the TT1 and 1 and TT2 Celsius temperatures to kelvin:2 Celsius temperatures to kelvin:• TT1 273 +40.0°C = K 1 273 +40.0°C = K TT2 273 +75.0°C = K2 273 +75.0°C = K• Write the equation for Charles’s law:Write the equation for Charles’s law:• To solve for To solve for VV2, multiply both sides by 2, multiply both sides by TT2:2:• Substitute known values:Substitute known values:• Solve for Solve for VV2.2.• VV2 =2 =• 3. Evaluate the Answer3. Evaluate the Answer

• Do problems 4-6 page 446Do problems 4-6 page 446

• Check your answers page 999Check your answers page 999

• 4) 3.1 L 6) 2.58 L4) 3.1 L 6) 2.58 L

Gay-Lussac’s LawGay-Lussac’s Law

• What are 4 variables that relate to What are 4 variables that relate to gases?gases?

• What are two types of containers?What are two types of containers?

• two apply to Gay-Lussac’s Law two apply to Gay-Lussac’s Law pressure and temperature. What pressure and temperature. What stays constant? What kind of container stays constant? What kind of container is needed?is needed?

PP11/T/T11 = P = P22/T/T22

• What are units of pressure?What are units of pressure?

• kPa atm mmHg torr psikPa atm mmHg torr psi

• What unit of temperature must be What unit of temperature must be used in this law?used in this law?

• KK

• 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 Do problems 8-10 page 448 check your answersyour answers

• 8) 1.96 atm 10) 273 degrees C8) 1.96 atm 10) 273 degrees C

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

Combined Gas LawCombined Gas Law

• P1V1/ T1 = P2 V2/T2P1V1/ T1 = P2 V2/T2

• ProblemProblem

• A gas at 100.0 kPa and 25.0°C fills a flexible container with 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 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 60.0° and the pressure increased to 320.0 kPa,what is the new volume?new volume?

• Step 1. Step 1. Analyze the problem.Analyze the problem.• Known VariablesKnown Variables• P1 P1 == 100.0 k Pa 100.0 k Pa P2 P2 ==320.0 kPa320.0 kPa• T1 T1 = = 25.0°C 25.0°C T2 T2 = = 60.0°C 60.0°C V1 V1 = = 2.00 L2.00 L• Unknown VariableUnknown Variable• V2 V2 ==? L? L• Step 2. Step 2. Solve for the unknown.Add 273 to the Celsius Solve for the unknown.Add 273 to the Celsius

temperature for temperature for T1 T1 and and T2 T2 to obtain the kelvin temperature.to obtain the kelvin temperature.• T1 T1 273 273 ++25.0°C 25.0°C = = 298 K; 298 K; • T2 T2 273 273 + + 60.0°C 60.0°C ==333 K333 K• Multiply both sides of the equation for the combined law by Multiply both sides of the equation for the combined law by T2 T2 and and

divide by divide by P2 P2 to solve for to solve for V2V2..• Substitute the known values into the rearranged equation; Substitute the known values into the rearranged equation;

multiply and divide numbers and units to solve for multiply and divide numbers and units to solve for V2V2..• V2 V2 ==0.698 L0.698 L• Step 3. Step 3. Evaluate the answer.Evaluate the answer.

• Do problems 11 - 13 page 450 and Do problems 11 - 13 page 450 and check your answers in the back of check your answers in the back of the book page 999the book page 999

• 12) 72 mL12) 72 mL

• Do problems 15-18 page 451Do problems 15-18 page 451

• 711 torr711 torr