Charles’s Law – Gas Volume and Temperature

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. Gasvolume as a function of temperature V at V = kT General Chemistry: Chapter 6

4. Charles’s Law • We can take two points from the last graph. • V1 = kT1 and V2 = kT2 • Combined these two give • = • This is obviously an equation used for “initial state → final state” problems (similar to Boyle’s Law). We will eventually consider these empirical laws in the context of kinetic theory.

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 oC or -491.67 oF (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(oC) + 273.15

6. Class Example • At a pressure of 0.986 bar and -22.0 oC a sample of ethane gas , C2H6(g) has a volume of 2.75 L. What volume would the gas occupy if the temperature were raised to + 22.0 oC without changing the pressure? • Class demonstration: Effect of temperature on gas volume at constant pressure.

7. Standard Temperature and Pressure • Gas properties depend on conditions. • IUPAC defines standard conditions of temperature and pressure (STP). P = 1 Bar = 105 Pa T = 0°C = 273.15 K General Chemistry: Chapter 6

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

9. At fixed T and P V  n or V = c n At STP 1 mol gas = 22.711 L gas Molar volume of a gas visualized Figure 6-9 General Chemistry: Chapter 6

10. Formation of Water – actual observation and Avogadro’s hypothesis Figure 6-8 General Chemistry: Chapter 6

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 H2(g) + O2(g) → 2 H2O(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 law V  1/P • Charles’s law V  T • Avogadro’s law V  n } V  General Chemistry: Chapter 6

13. PV R= nT The Ideal Gas Equation PV=nRT General Chemistry: Chapter 6

14. Applying the ideal gas equation General Chemistry: Chapter 6

15. = = P1V1 P1 P2V2 P2 n1T1 T1 n2T2 T2 If we hold the amount and volume constant: The General Gas Equation R = General Chemistry: Chapter 6

16. Using the Gas Laws General Chemistry: Chapter 6

17. Class Example – Ideal Gas Law Eq. • Find (a) the density of CO2(g) at 55.0oC and a pressure of 64.3 kPa and (b) the number of gas molecules per cm3 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 (O2(g)) has a volume of 8.00 L. At the same T and P 10.0 L of a gas having the molecular formula XO2 has a mass of 20.0 g. Identify element X.