1. Introduction to Gases Ch 13 Suggested HW: 1, 3, 15, 16, 31, 39, 40, 49, 53, 55

2. About Gases Gases are the most understood form of matter. Even though different gases have different chemical properties, they tend to exhibit similar physical properties This situation arises because gas molecules expand to fill a given space, and are relatively far apart from one another. A volume of gas consists mostly of empty space. Thus, each gas atom/molecule behaves as if the others are not there.

3. Pressure The most readily measured properties of a gas are its temperature, volume, and pressure Pressure describes the force that a gas exerts on an area, A. P = F/A The image below shows gas molecules inside of a cubic container. The gas molecules strike against the walls of the container. These collisions are the source of the pressure.

4. Atmospheric Pressure You and I are currently experiencing an attractive force that pulls us toward the center of the earth (gravity). Gas molecules in the atmosphere also experience gravity. Because of their small masses and thermal energies, gas molecules can somewhat counteract gravity, which is why gases don’t just sit on the surface Nonetheless, gravity causes the gases in the atmosphere to “press down” on the surface. This is atmospheric pressure.

5. Atmospheric Pressure The mass of a 1m 2 column of air extending through the entire atmosphere would be approximately 10 4 kg. The force exerted on the surface would be: F= ma = (10 4 kg)(9.8 ms -2 ) = 10 5 N Then, P = F/A = 10 5 N/m 2 = 10 5 Pa SI unit of pressure is the Pascal (Pa). Related units are bar, mmHg, and atmospheres.

6. Atmospheric Pressure A barometer is shown to the left. It is composed of a glass tube that is open on one end and closed on the other. The tube is under vacuum (no air in tube). The tube is inverted into a dish that contains mercury. Some mercury flows into the tube because the atmospheric pressure forces it upward (initial pressure in tube is zero). The flow stops when the atmospheric pressure is equal to the pressure inside the tube. Pressure inside the tube is caused by the weight of mercury. The height h is proportional to the atmospheric pressure. h Closed, under vacuum Open

7. Units of Pressure 1 atm = 760 mm Hg

8. Gas Laws: Boyle’s Law

9. High Pressure (P 2 ) Low Pressure (P 1 ) Constant Temperature Boyle’s Law

10. Example

11. C.I.R.L.: Boyle’s Law Everyday, without thinking about it, you move nearly 8500 L of air in and out of your lungs, equating to about 25 lbs. The ability of the lungs to create pressure gradients (differences in pressure between two regions) is what allows us to breathe. When you inhale, your lungs expand (increased volume). This expansion causes the molarity of air in the lungs to decrease, yielding low pressure (LP).

12. The pressure outside of the lung remains high (HP). Now, there is a pressure gradient. LP HP Immediately, air flows from the HP region (atmosphere) to the LP region (lungs) until the pressures are equal. Inhalation C.I.R.L.: Boyle’s Law

13. When you exhale, the lungs shrink, increasing the air molarity, and the air pressure. Now, the pressure inside the lung is higher than the pressure outside. The gradient is reversed. HP LP Air flows out C.I.R.L.: Boyle’s Law

14. C.I.R.L. Boyle’s Law Applied to the Rules of Deep Sea Diving 1.Never hold your breath! If you ascend even a few feet with air in your lungs, the decrease in pressure will cause the air to expand in your lungs! POP!! 2.Ascend slowly! The solubility of gas in fluid increases with pressure. Air is 79% N 2, and N 2 is quite soluble in water. As you ascend, all of that N 2 comes out of your blood. If this happens too fast…..POP!!

15. Relationship between Volume and Temperature: Charles’s Law

16. Charles’s Law So how can we predict the change in volume with temperature? Charles’s Law tells us that volume is directly proportional to temperature at constant pressure. Constant Pressure We can express this as:

17. Charles’ Law Constant Pressure

18. Example At 25 o C, the volume of the air in a balloon is 1.3L. Then, the balloon is cooled to -73 o C using N 2 (L). What is the new volume of air?

19. C.I.R.L.: Charles’ Law, Boyle’s Law, and The Internal Combustion Engine

20. Gas Laws: Avogadro’s Law Avogadro determined that equal volumes of gases at the same pressure and temperature must contain equal numbers of molecules, and thus, equal moles constant T, P

21. The Ideal Gas Law PV = nRT Depending on the units of Pressure, we can use multiple values of R

22. Ideal Gases Any gas that follows the ideal gas law is considered an ideal gas. One mole of an ideal gas at 0 o C and 1 atmosphere of pressure occupies 22.4 L of space. The value of R is based on these values of n, T, P, and V. The ideal gas law is valid only at low pressures The conditions listed above (0 o C, 1 atm) are referred to as standard temperature and pressure (STP) Note: The Ideal gas law is a theoretical approximation, and no gas follows this law exactly, but most gases are within a few percent of this approximation, so it is very useful.

23. Comparisons of Real Gases to the Ideal Gas Law Approximation at STP Ideal Gas 22.4 L 22.06 L 22.31 L 22.41 L 22.42 L Cl 2 CO 2 He H2H2 Molar Volume at STP (L) Deviations from the ideal gas law exist, but those deviations are reasonably small.

24. Example The pressure in a 10.0 L gas cylinder containing N 2 (g) is 4.15 atmospheres at 20.0 o C. How many moles of N 2 (g) are there in the cylinder? *When dealing with ideal gas law questions, follow these steps: 1) Determine what it is you are solving for. 2) List the given information. Pay attention to the units of each parameter. Convert as needed, and MAKE SURE THAT THE TEMPERATURE IS IN KELVIN ! 3) Rearrange the ideal gas law equation accordingly to solve for the desired parameter.

25. We are solving for moles. V = 10.0 L P = 4.15 atm. T = 20. 0 o C 293 o K R =.0821 Latmmol -1 K -1 Rearrange the equation to solve for n. Example, continued. The pressure in a 10.0 L gas cylinder containing N 2 (g) is 4.15 atmospheres at 20.0 o C. How many moles of N 2 (g) are there in the cylinder? PV = nRT = 1.72 moles N 2 (g)

26. Example 3 2KClO 3 (s)  2KCl(s) + 3O 2 (g) In the reaction above, 1.34 g of potassium chlorate is heated inside a container to yield oxygen gas and potassium chloride. The oxygen occupies 250 mL at 20.0 o C. What will the pressure of the gas be, in atmospheres? We are solving for pressure in atm. V =.250 L T = 20 o C  293 o K n = ? R = 0.821 Latmmol -1 K -1 PV = nRT

27. 2KClO 3 (s)  2KCl(s) + 3O 2 (g) 1.34 g.0109 mol Before we can find P, we must find n.0164 mol

28. Group Example (KNOW HOW TO DO THIS!!) 18.7g of solid manganese (IV) oxide is added to 600 mL of an aqueous solution of 1.33M hydrochloric acid in a sealed container at standard temperature and pressure. The reaction produces aqueous manganese (II) chloride, water, and chlorine gas. a. Write a balanced reaction b. Given that reaction generates 1.85L of chlorine gas, calculate the % yield.