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Good Morning!

Good Morning!. Today we are going to… Hand back work and tests. Show the answers for the Review problems. Turn in 12.4 & 12.5 FYI: you will be given the equations on the test You do not need to know what name goes with which law.

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Good Morning!

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  1. Good Morning! • Today we are going to… • Hand back work and tests. • Show the answers for the Review problems. • Turn in 12.4 & 12.5 • FYI: • you will be given the equations on the test • You do not need to know what name goes with which law. • There are several multiple choice that ask you about the how P,V, & T relate to each other. • And you will also need to complete calculations using the gas laws.

  2. What is the volume of a sample of CO2 at STP that has a volume of 75.0 mL at 30.0 °C and 91 kPa? • First you have to figure out what you have and what you are being asked to find , then choose the right law. • So what do we have, what do we need to find, and which law do we need to use.

  3. Make sure they are in the right units • V1 =0.075 L • P1 = 91 kPa • T1= 303 K • V2 = ? • P2 = 101.3 kPa • T2= 273 K • Now which law will take all of this into account? Now divide by P2 Now we can put the numbers in : ) To do this we multiply by T2 First solve for V2

  4. Chapter 12 Problems with Gas Law • Let’s get started.

  5. State the equation for Boyles Law. • P1V1 = P2V2

  6. 2. Identify the quantity held constant in Boyles Law. • Temperature, it’s not present in the equation.

  7. 3. Draw the general shape of a Boyles Law graph.

  8. A quantity of gas under a pressure of 50 kPa has a volume of 565 cm3. The pressure is increased to 450 kPa, while the temperature is kept constant. What is the new volume? • P1 =50 kPa • V1 = 0.565 L • P2 = 450 kPa • V2 = ? • Which Law do we need to use?

  9. A quantity of gas under a pressure of 50 kPa has a volume of 565 cm3. The pressure is increased to 450 kPa, while the temperature is kept constant. What is the new volume? • P1V1 = P2V2 • To solve for V2 divide both sides by P2

  10. State the equation for Charles’s Law.

  11. 6. What is the relationship between temperature and volume? • Inverse relationship or direct relationship

  12. A gas sample has a volume of 25.0 L at a temperature of 30.0 C = 303 K. The temperature is raised to 227 C = 500 K while the pressure remains unchanged. What is the new volume of the gas? • T1 = 303 K • V1 = 25.0 L • T2 = 500 K • V2 = ? • Which Law do we need?

  13. A gas sample has a volume of 25.0 L at a temperature of 30.0 C = 303 K. The temperature is raised to 227 C = 500 K while the pressure remains unchanged. What is the new volume of the gas? ÷ by T2

  14. What volume will 0.375 mol of Oxygen take up at STP?

  15. Determine the volume occupied by 0.582 mol of a gas at 15C if the pressure is 81.8 kPa. • V = ? • n= 0.582 mols • T=288 K • P = 81.8 kPa • R = 8.31

  16. Determine the volume occupied by 0.582 mol of a gas at 15C if the pressure is 81.8 kPa.

  17. A mixture of gases at a total pressure of 97 kPa contains N2, CO2, and O2. The partial pressure of the CO2 is 24 kPa and the partial pressure of the N2 is 48 kPa. What is the partial pressure of the O2? • Ptotal = P1 + P2 + P3 • Ptotal - P1 - P2 = P3 • 97kPa - 24kPa - 48 kPa = P3 • 25 kPa = P3

  18. Using Graham's Law of Diffusion calculate the rate of diffusion between methane CH4 (mass = 16) and hydrogen sulfide H2S (mass = 34)If 2 gases are at the same temp. they have the same kinetic energy (KE)

  19. First where does that equation come from? No, you do not have to do this on the test, I just wanted you to know that there is a reason why the velocity is “opposite” of the mass. Now you can express this in 2 different ways. You could say that H2S is 68% slower than CH4 or you could say CH4 is 146% faster than H2S Finally, take the square root of both sides to get the equation that you need to use. If 2 gases are at the same temperature they have the same KE. And the formula for KE is ½ mv2. So we can sub in and get the equation above. Divide both sides by the ½ and get rid of it. Now divide both sides by the m1 so it can cancel out on the left Divide both sides by the V22 so it can cancel out on the right

  20. 12. Calculate the relative rates of diffusion of gaseous UF6 containing these isotopes Formula mass of UF6 containing uranium - 235 = 349 amu. Formula mass of UF6 containing uranium - 238 = 352 amu. Uranium 235 will reach you 1.0043 times as fast.

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