1. Introduction to Gases Ch 13 HW: 1, 31, 39, 41 Suggested: 42-46 10/21

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 Thus, each molecule behaves as if the others are not there

3. The Kinetic Theory of Gases Kinetic Theory tells us the following: 1.Gas molecules are ALWAYS in motion. They collide randomly with each other and with the walls of the container 2.All collisions involving gas particles are ELASTIC. This means they simply “bounce off” of one another. No energy is lost, and the speed of the particles does not change. 3.The distance between gas molecules is large. So, gas mainly consists of empty space. 4.Interactions between gas molecules are negligible (no attraction or repulsion). 5.The kinetic energy (speed) of gas molecules is proportional to temperature.

4. 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.

5. 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.

6. Atmospheric Pressure To calculate atmospheric pressure, we calculate the mass of a 1m 2 column of air extending through the entire atmosphere. That column would have an approximate mass of 10 4 kg. The force exerted on that 1m 2 area would be: F= ma = (10 4 kg)(9.8 ms -2 ) = 10 5 N Then, P = 10 5 N/1m 2 = 10 5 N/m 2 = 10 5 Pa SI unit of pressure is the Pascal (Pa). Related units are bar, mmHg, and atmospheres.

7. Units of Pressure 1 atm = 760 mm Hg

8. Atmospheric Pressure The image to the bottom left shows a barometer. The tube is inverted into a dish that contains mercury. Some mercury flows into the because the atmospheric pressure forces it upward (pressure in tube is lower than atmosphere). The flow stops when the atmospheric pressure is equal to the pressure inside the tube (weight of Hg/area). The height h is proportional to the atmospheric pressure. h Closed, under vacuum Open-end, exposed to atmosphere

9. Gas Laws: Boyle’s Law Large volume Small volume

10. Charles’s Law: A Demonstration a.Liquid nitrogen (-196 o C) poured on a balloon b.As air inside balloon is cooled, balloon compresses c.Balloon is allowed to warm back to room temperature d.Balloon re-inflates

11. Relationship between Volume and Temperature: Charles’s Law kinetic theory

12. 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. If V = mT (m is just a constant) Then V i /T i = m, where i stands for ‘initial’. And… V f /T f = m, where f stands for ‘final’. Thus… Charles’s Law

13. Example At 0 o C, the volume of the air trapped in a glass tube by a metal plug is 0.180 mL. Then, the water is raised to its boiling temperature and the air expands. What is the new volume of air? Here, we are looking for the final volume. We will use Charles’s Law and plug in the given information. USE KELVIN SCALE !!!

14. Avogadro’s Law Avogadro determined that equal volumes of gases at a given pressure and temperature must contain equal numbers of molecules, and thus, equal moles Volume is directly related to moles, and both are related to the stoichiometry 1L H 2 (g) 1L Cl 2 (g) 2L HCl(g) 2L H 2 (g) 1L O 2 (g) 2L H 2 O(g)

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

16. 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.

17. 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.

18. 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.

19. Example, continued. 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. 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)

20. Example 2 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 = 0.0821 Latmmol -1 K -1 PV = nRT

21. 2KClO 3 (s)  2KCl(s) + 3O 2 (g) 1.34 g.0109 mol Before we can find P, we must find n

22. Partial Pressures in Gas Mixtures To this point, we have not discusses mixtures of gases, yet gaseous mixtures are an important part of everyday life For example, the air we breathe is a gas mixture 78% N 2 (g) 21% O 2 (g) 1 % Ar and other gases In a gas mixture, each gas exerts its own pressure. If we had a sample of air in a container, the total pressure would be equal to the sum of the partial pressures (Dalton’s Law)… P total = P N 2 + P O 2 + P Ar …..

23. Partial Pressures The total pressure can further be defined using the Ideal gas law From this expression, we can determine the mole fraction (x i ) of each gas XN2XN2 XO2XO2 X Ar

24. Partial Pressures Therefore, we see that the partial pressure of a gas is dictated by its mole fraction. So, if we know the moles of a given gas in a mixture, and the total moles of gas, we can easily calculate the partial pressure: The sum of the mole fractions must equal 1. + + = 1

25. Example A 0.428 g sample of air at 2 atmospheres is contained with a vessel. Using the mass percentages previously mentioned, what are the partial pressures of N 2, O 2, and Ar ? Find moles of each gas 78% N 2.012 mol N 2 21% O 2.0028 mol O 2 1 % Ar.0001 mol Ar Calculate mole fractions, then partial pressures