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Chapter 14: Gases Objectives: The student will be able to:

Chapter 14: Gases Objectives: The student will be able to:. state Boyle’s, Charles’s, and Gay-Lussac’s Laws; apply three gas laws to problems involving P,T and V of gases; combine P, T, and V concepts into a combined gas law; relate n (moles) of particle to Avogadro’s principle;

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Chapter 14: Gases Objectives: The student will be able to:

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  1. Chapter 14: Gases Objectives: The student will be able to: state Boyle’s, Charles’s, and Gay-Lussac’s Laws; apply three gas laws to problems involving P,T and V of gases; combine P, T, and V concepts into a combined gas law; relate n (moles) of particle to Avogadro’s principle; relate amount of gas to P, T, and V using the ideal gas law; compare and contrast real and ideal gas laws using kinetic molecular theory assumptions; determine volume ratios for gaseous reactants and products by using coefficients from a chemical equation; apply gas laws to stoichiometry problems.

  2. From experience, what happens to volume when we push down on the plunger? Pushing on the plunger increases the pressure on the gas which in turn decreases the volume.

  3. Robert Boyle: 1670 Investigated the relationship between pressure (P) and volume (V) at constant temperature and constant amount of gas. As mercury is added, the pressure on the gas in the tube increases, and its volume decreases. Having one thing increase while the other decreases is known as an inverse relationship. What is the mathematical relationship between volume and pressure?

  4. Data from Boyle’s experiments. Notice that the initial volume is a maximum at the start of the experiment when the pressure lowest! The volume steadily decreases as the pressure is increased. Remember that the pressure is increasing because mercury is being added to the open end of the “J” tube. Is there a pattern in the data?

  5. Patterns in data are often recognized by graphing the data in some way. What do the graphs tell us? Plot of P versus V Question: why would we have predicted that b = 0? For a line: y = m(x) + b from the graph, y = “V”, x = “1/P” and b = 0. m is slope and b is y-intercept

  6. Therefore, m = y/x = (V)/(1/P) = (V)(P) and the slope of the line is a constant. Since the graphs indicate an inverse relationship between P and V, (P)(V) = a constant and the slope of the line is equal to the value of the constant. We usually call a generic constant “k”.

  7. (P1) or V2 = (V1) (P2) Boyle’s Law: The volume of a constant amount of gas held at constant temperature varies inversely with the pressure. Said another way: PV = constant How does this mathematical statement correctly express Boyle’s Law? However, under condition 2, P2V2 = k Under condition 1, P1V1 = k Since k is a constant, it must have the same value under both sets of conditions. So, P1V1 = P2V2 Now if we solve for V2, we get V2 = (V1)(P1/P2) This statement says that if we increase the pressure (P2 > P1), the volume (V2) must be less than V1 because P1/P2 will be a number smaller than 1.

  8. (P1) V2 = (V1) (P2) What will the volume be at condition 2? Condition 1 Condition 2 First identify the things that are given: P1 = 5.6X103 Pa V1 = 1.53 L P2 = 1.5X104 Pa V2 = ? What equation uses these letters? P1V1 = P2V2 Rearrange the equation to get the letter you need by itself:

  9. Complete any three practice problems on page 422. The key equation from Boyle’s Law is: P1V1 = P2V2

  10. (V1) (V2) = k = k (T1) (T2) (V1) (V2) So, = (T1) (T2) (T2) V2 = (V1) (T1) Jacques Charles: 1800 Investigated the relationship between volume and temperature for a constant amount of gas at constant pressure. Using experiments similar to those described for Boyle, Charles discovered that volume and temperature have a direct relationship (in contrast to the inverse relationship discovered by Boyle). Mathematical expression of Charles’ Law Now if we solve for V2, we get This statement says that if we increase the temperature (T2 > T1), the volume (V2) must be larger than V1 because T2/T1 will be a number greater than 1. In other words, volume increases when temperature increases.

  11. (V1) (V2) = (T1) (T2) (-25oC) V2 = (1.0 L) (25oC) (T2) V2 = (V1) (T1) Charles’s Law indicates that as the temperature of a gas increases, the volume of the gas will also increase (assuming a constant pressure and a constant amount of gas). This is a direct relationship. As one thing increases, the other also increases. What would happen if we started with 1.0 L of gas at 25oC and lowered the temperature to -25oC? From the equation: = -1.0 L What is wrong with this? It is impossible to have a negative volume!

  12. (248 K) V2 = (1.0 L) (298 K) Plot of V versus T for several gases. Since it is impossible to have a negative volume we must use the Kelvin temperature scale for all problems involving gases. What would happen if we started with 1.0 L of gas at 25oC and lowered the temperature to -25oC? Remember: K = oC + 273.15 = 0.83 L

  13. (V1) (V2) = (T1) (T2) V1 = 0.67 L V2 = 1.12 L T2 = ? (V2) T2 = (T1) (V1) (1.12 L) T2 = (362K) (0.67 L) Practice problem 6 on page 425. A gas at 89oC occupies a volume of 0.67 L. At what Celsius temperature will the volume increase to 1.12 L? Attack strategy: Identify what is given, what is wanted, and what equation connects these ideas. T1 = 89oC = 362 K because (89 + 273 = 362) The key equation that connects these ideas is from Charles’ Law. Rearranging the equation gives: Plugging in: = 605 K = 332 oC = 330 oC

  14. (V1) (V2) = (T1) (T2) Complete practice problems 7 and 8 on page 425. The key equation from Charles’ Law is: Algebra Practice: solve for each of the four variables in Charles’ law.

  15. (P1) (P2) (P1) (P2) = = k = k (T1) (T2) (T1) (T2) (T2) P2 = (P1) (T1) Joseph Gay-Lussac: Investigated the relationship between pressure and temperature for a constant amount of gas at constant volume. Using experiments similar to those described for Boyle and Charles, Gay-Lussac discovered that pressure and temperature have a direct relationship. This equation confirms that the pressure will increase if the temperature increases. After solving for P2

  16. (P1) (P2) = (T1) (T2) P1 = 125 kPa P2 = 201 kPa T2 = ? (P2) T2 = (T1) (P1) (201 kPa) T2 = (303.2K) (125 kPa) Practice problem 9 on page 427. A gas in a sealed container has a pressure of 125 kPa at a temperature of 30.0oC. If the pressure in the container is increased to 201 kPa, what is the new temperature? Attack strategy: Identify what is given, what is wanted, and what equation connects these ideas. T1 = 30.0oC = 303.2 K because (30.0 + 273.15 = 303.2) The key equation that connects these ideas is from Gay-Lussac’s Law. Rearranging the equation gives: Plugging in: = 488 K or 214 oC

  17. (P1) (P2) = (T1) (T2) Complete any two practice problems from 10-13 on page 427. The key equation from Gay-Lussac’s Law is: Algebra Practice: solve for each of the four variables in Gay-Lussac’s law.

  18. (V1)(P1) (V2)(P2) (V1)(P1) (V2)(P2) = k = k = (T1) (T2) (T1) (T2) (P1) (T2) V2 = (V1) (P2) (T1) (T2) (P1) V2 = (V1) = 1 (T1) (P2) Combined Gas Law: Combine all of the previous three gas laws into one equation. The only thing held constant is the amount of gas. Algebra Practice: solve for each of the six variables in the combined gas law. Notice that this law can be used to solve any of the previous problems if you understand what happens when something is held constant. Example: solve for V2 For a Boyle’s Law problem, temperature is constant, so and which is the same as before for Boyle’s Law problems.

  19. (V1)(P1) (V2)(P2) = (T1) (T2) Complete any three practice problems on page 430. The key equation from the Combined Gas Law is: Again, the attack strategy is to first assign letters to all of the numbers given in the question. Then rearrange the combined gas law equation until the letter you need to find is by itself. Then plug the values in for the letters and complete the math.

  20. (V1)(P1) (V2)(P2) = (T1) (T2) (P1) (V1) (P2) (V2) = = (T1) (T1) (T2) (T2) Summary: What is held constant in each of the gas laws? Moles of gas (n) is held constant in all four gas laws. Boyle’s Law also holds Temperature constant (Constant T) P1V1 = P2V2 Charles’ Law also holds Pressure constant (Constant P) Gay-Lussac’s Law also holds Volume Constant (Constant V) The Combined Gas Law only holds moles of gas constant (Constant n)

  21. What is the only concept that remains constant throughout all of the individual gas laws and the combined gas law? The amount (moles) of gas remains constant!!!! Avogadro’s Principle: Equal volumes of gases at the same temperature and pressure must contain the same number of particles-no matter what the identity of the gases is. If the four balloons have the same volume and are in the same room (i.e. they have the same temperature and pressure), then they must contain the same number of gas particles even though they contain different gases.

  22. Avogadro’s principle has a practical application. If we pick standard conditions for temperature and pressure (STP), we can measure the volume of one mole of gas. Standard Pressure is 1.00 atm (or any equivalent amount like 760 mm Hg). Standard temperature is 0.00 oC or more correctly 273.15 K. What is the volume of 1.00 mole of any gas at STP? 1.00 mole = 22.4 L at STP 22.4 L/mol of gas is known as the molar volume for a gas at STP Now, if we know a gas is at STP and we measure its volume, we can calculate the number of moles of gas that are present. This means we can use 1 mole = 22.4 L to convert from mole to L (or from L to mole)

  23. ( ) ( ) 1 mol 3.45 L 22.4 L How many moles of oxygen are contained in a balloon that has a volume of 3.45 L at STP? What is given? V = 3.45 L T = 273.15 K P = 1.00 atm n = ? Remember that at STP 1.00 mole = 22.4 L = 0.154 mol O2 How many moles of methane are contained in a container that has a volume of 38.17 L at STP?

  24. ( ) ( ) 22.4 L 0.750 mol 1 mol What will the volume of a gas be if there are 0.750 moles of the gas at 0.00oC and a pressure of 760.0 torr? What is given? n = 0.750 moles T = 273.15 K P = 760.0 torr V = ? Remember that at STP 1.00 mole = 22.4 L = 16.8 L What volume will 3.54 moles of methane gas occupy at STP?

  25. Complete any two practice problems on page 432. Complete any two practice problems on page 433. Please remember that 22.4 L of gas = 1 mole of gas only at STP You may not use 22.4 L = 1 mole if the pressure is not equal to 1 atm or if the temperature is not equal to 0oC.

  26. (V1)(P1) (V1)(P1) (V2)(P2) (V1)(P1) (V2)(P2) = constant = R = constant = (T1)(n1) (T1)(n1) (T2)(n2) (T1)(n1) (T2)(n2) Ideal Gas Law: incorporates the amount of gas (n in moles) into the combined gas law. The constant is called the ideal gas constant and is given the symbol “R”. The value of R depends upon the units for P since V is always in L, T is always in K, and n is always in moles. Table 14.1 on page 435 has several values used for R. The most commonly used value of R is 0.0821 (L·atm)/(mol·K). This leads to the famous form of the ideal gas law: PV = nRT

  27. PV n = = ( ) RT L·atm 0.0821 mol·K 41 on page 437 If the pressure exerted by a gas at 25oC in a volume of 0.044L is 3.81 atm, how many moles of gas are present? What is given? P = 3.81 atm n = ? V = 0.044 L T = 25oC = 298 K Since only one value of V, P, and T are given, this requires the use of PV = nRT. Solve for “n” (3.81 atm)(0.044 L) (298 K) = 0.006852013832 mol What is the correct answer, and how do the units work out?

  28. Complete any other two practice problems on page 437.

  29. mass mass n = D = Molar mass volume m m n = D = MW V mRT PV = MW m Rearrange the equation until you get on one side by themselves. V m PMW = V RT PV = nRT m = mass, MW = molar mass and volume = V so, Substitute (m/MW) for (n) in the ideal gas law: PV = nRT This equation is for gas density. D =

  30. DRT MW = P This density equation is especially useful for predicting gas density if molar mass is known. PMW D = RT Rearranging for MW gives: This density equation is especially useful for predicting molar mass if gas density is known. Complete practice problems 49 and 50 on page 438.

  31. ( ) 3.0 L O2 Gas Stoichiometry This is much like the stoichiometry problems in chapter 12. CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g) From the equation, 2 moles of oxygen will react exactly with 1 mole of methane. Now we can say 2 L of oxygen will react with 1 L of methane as long as all the gases are under the same temperature and pressure conditions. This idea works because of Avogadro’s principle. Since equal volumes of gas under the same conditions contain equal numbers of particles, they must contain equal number of moles as well. What volume of CO2 will be produce when 3.0 L of oxygen react according to the balanced equation above. ( ) 1 L CO2 = 1.5 L CO2 2 L O2

  32. ( ) O2 (g) + S (s) SO2 (g) 3.5 L SO2 ( ) 1 L O2 1 L SO2 Problem 56 on page 441 What volume of oxygen is needed to react with solid sulfur to form 3.5 L of SO2? First: Write the balanced equation. Second: Start with what was given. = 3.5 L SO2 Third: Use the mole ratio from the balanced equation (stoichiometry) Complete any other two practice problems on page 441.

  33. ( ) 0.10 L N2O ( ) ( ) ( ) 1 mole NH4NO3 80.05 g NH4NO3 1 mole N2O 1 mol N2O 1 mole NH4NO3 22.4 L N2O Gas Stoichiometry problems can be extended into all other varieties of stoichiometry problems. 60 on page 443 Calculate the mass of NH4NO3 solid that must be used to produce 0.10 L of N2O gas at STP based on the balanced equation below. NH4NO3 (s) → N2O (g) + 2 H2O (g) Molar Volume (C14) Mole Ratio (C12) Molar Mass (C11) Complete problems 61-64 (any two) on page 443.

  34. Ideal Versus Real Gases If all gases were ideal, then 1 mole of every gas at STP would occupy 22.4 L. Why do some gases deviate from ideal? The answer lies in the assumptions that were made about gas behavior in chapter 13. If gases are large molecules, the volume of the particle is not insignificant. If gases are polar, they can have significant attractive forces (dipole-dipole). Which of these gases is least ideal? Why?

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