1. Charles’s Law – Gas Volume and Temperature A fixed amount of a gas at a fixed pressure will expand if the temperature increases. Under these conditions the volume of gas is proportional to the Celsius temperature (or, any other temperature!). We have, eg. V(t) = mt + b where t is the Celsius temperature. This graph does not pass through the origin of a V vs t plot.

2. Charles’s Law – Kelvin Temperatures The graph of gas volume vs temperature can be made to pass through the origin if we employ an absolute temperature scale, such as the Kelvin scale, where zero degrees is the lowest temperature achievable. With the Kelvin scale (for example) at fixed P the volume of gas becomes directly proportional to temperature. Leads to simpler calculations. V(T) = kT

3. Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 6Slide 3 of 19 Gas volume as a function of temperature V = kT V  t

4. Charles’s Law

5. Kelvin Temperature Scale Celsius and Kelvin degrees have the same size. A temperature of zero degrees Kelvin, or 0 K, corresponds to -273.15 o C or -491.67 o F (sounds colder although it’s not!). We won’t use Fahrenheit temperatures further. To convert from Celsius to Kelvin temperatures we simply “add” 273.15 degrees. T(K) =t( o C) + 273.15

6. Class Example At a pressure of 0.986 bar and -22.0 o C a sample of ethane gas, C 2 H 6 (g) has a volume of 2.75 L. What volume would the gas occupy if the temperature were raised to + 22.0 o C without changing the pressure? Class demonstration: Effect of temperature on gas volume at constant pressure.

7. Standard Temperature and Pressure Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 6Slide 7 of 19 Gas properties depend on conditions. IUPAC defines standard conditions of temperature and pressure (STP). P = 1 Bar = 10 5 Pa T = 0°C = 273.15 K

8. Avogadro’s Law Gay-Lussac 1808 – Small volumes of gases react in the ratio of small whole numbers. Avogadro 1811 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 6Slide 8 of 19 At a fixed temperature and pressure, the volume of a gas is directly proportional to the amount of gas.

9. Molar volume of a gas visualized Figure 6-9 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 6 Slide 9 of 19 At STP 1 mol gas = 22.711 L gas At fixed T and P V  n or V = c n

10. Formation of Water – actual observation and Avogadro’s hypothesis Figure 6-8 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 6Slide 10 of 19

11. Avogadro’s Hypothesis On the previous slide it is assumed that all reactants and products are gases. We would write the chemical reaction describing the change as 2 H 2 (g) + O 2 (g) → 2 H 2 O(g) At “low” temperatures we might expect to see liquid water formed. This reaction produces both heat and light and is featured in many Hollywood movies – eg. The Hindenburgh.

12. 6-3 Combining the Gas Laws: The Ideal Gas Equation and the General Gas Equation Boyle’s lawV  1/P Charles’s lawV  T Avogadro’s law V  n Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 6Slide 12 of 19 }

13. The Ideal Gas Equation Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 6Slide 13 of 19 R =R = PV nT PV = nRT

14. Applying the ideal gas equation Copyright © 2011 Pearson Canada Inc. Slide 14 of 19General Chemistry: Chapter 6

15. The General Gas Equation Copyright © 2011 Pearson Canada Inc. Slide 15 of 19General Chemistry: Chapter 6 R = = P2V2P2V2 n2T2n2T2 P1V1P1V1 n1T1n1T1 = P2P2 T2T2 P1P1 T1T1 If we hold the amount and volume constant:

16. Copyright © 2011 Pearson Canada Inc. Slide 16 of 19General Chemistry: Chapter 6 Using the Gas Laws

17. Class Example – Ideal Gas Law Eq. Find (a) the density of CO 2 (g) at 55.0 o C and a pressure of 64.3 kPa and (b) the number of gas molecules per cm 3 at this T and P. Solution (partial): The problem could be tackled using the Combined Gas Law eqtn. but is better approached using the Ideal Gas Law. Why? “Trick”. No amount of gas is specified. Any amount of gas works since (at a given T and P) density is an intensive quantity.

18. Class Example – Avogadro’s Hypothesis: At a given T and P, 8.00 g of oxygen gas (O 2 (g)) has a volume of 8.00 L. At the same T and P 10.0 L of a gas having the molecular formula XO 2 has a mass of 20.0 g. Identify element X.