2. State Boyle’s law, and use it to solve problems involving pressure and volume. State Charles’s law, and use it to solve problems involving volume and temperature. State Gay-Lussac’s law, and use it to solve problems involving pressure and temperature. State Avogadro’s law, and explain its importance in determining the formulas of chemical compounds. Objectives
3. The Gas Laws Simple mathematical relationships between Volume Temperature Pressure Amount of gas Gas Law Program: Shows the relationship of all four of the above on gases http://intro.chem.okstate.edu/1314F00/Laboratory/GLP.htm Constant : Volume and amount of gas Shows: Change in pressure and temperature
4. Boyle’s Law States – volume of a fixed mass of gas varies inversely with the pressure at constant temperature. Constant temperature and amount of gas If you double volume, pressure is cut in half If you cut volume in half, pressure doubles
5. Boyle’s Law Pressure caused by Moving molecules hitting container walls Speed of particles (force) and number of collisions Both increase pressure Mathematically: Volume-Pressure Data for Gas Sample 1 Volume Pressure P x V (mL) (atm) V k = PV or = k P k is constant 1200 0.5 600 P is pressure 600 1.0 600 V is volume 300 2.0 600 200 3.0 600 150 4.0 600 120 5.0 600 100 6.0 600 Interactive graph
6. Boyle’s Law Boyle’s Law Equation: P1V1 = k P2V2 = k So: P1V1 = P2V2 Sample Problem 1 A sample of oxygen gas has a volume of 150. mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant. P1 = 0.947 atm P1V1 = P2V2 V1 = 150. mL (0.947 atm) (150. mL) P2 P2 V2 = P2 = 0.987 atm 0.987 atm P1V1 V2 = ? = V2 = 144 mL P2
7. Charles’s Law Volume-temperature Relationship When using temperature, Must use absolute zero, Kelvin Temperature scale Temperature -273.15oC is absolute zero All molecular movement would stop. K = oC + 273 212 100 373 So : How many Kelvin is 10 oC? K = oC + 273 = 10oC + 273 32 0 273 = 283 K Fahrenheit Celsius Kelvin Temperature Scales
9. Charles’s Law States that the volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature. Constant Pressure and amount of gas If you double temperature, the volume will double If you cut the temperature in half, the volume will be half as much. V V k = or kT = T k is constant T is temperature V is volume Charles’s Law Equation: V1 V2 = k = k T1 T2 So: V2 V1 = T2 T1
10. Charles’s Law Sample Problem 2 A sample of neon gas occupies a volume of 752 mL at 25oC. What volume will the gas occupy at 50oC if the pressure remains constant? V2 V1 V1 = 752 mL T2 = T2 x x T2 T1 T1 = 25oC + 273 = 298 K V2 = ? V1 T2 (752 mL) (323 K) T2 = 50oC + 273 = 323 K V2 = = T1 298 K = 815 mL **Always convert to Kelvin!!
11. Gay-Lussac’s Law Pressure-temperature Relationship Increasing temperature, increases the speed of the gas particles Thus, more collisions with the container walls Causing an increase in pressure
12. Gay-Lussac’s Law States that the pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature. Constant volume and amount of gas If you double temperature, pressure doubles If you cut temperature in half, pressure is also cut in half P P k = or kT = T k is constant T is temperature P is pressure P1 P2 = k = k T1 T2 So: P2 P1 = T1 T2
13. Gay-Lussac’s Law Sample Problem 3 The gas in an aerosol can is at a pressure of 3.00 atm at 25oC. Directions on the can warn the user not to keep the can in a place where the temperature exceed 52oC. What would the gas pressure in the can be at 52oC? P1 P2 P1 = 3.00 atm T2 = T2 x x T2 T1 T1 = 25oC + 273 = 298 K P2 = ? P1 T2 (3.00 atm) (325 K) T2 = 52oC + 273 = 325 K P2 = = T1 298 K **Always convert to Kelvin!! = 3.27 atm
14. Avogadro’s Law Equal Volumes of gases at the same temperature and pressure contain equal numbers of molecules. How does it work! Hydrogen Gas + Chlorine gas Hydrochloric Acid H2Cl2 2 HCl 1 Volume 1 Volume 2 Volumes 1 Molecule 1 Molecule 2 Molecules 1 Mol 1 Mol 2 Mol V = kn Direct Relationship n = amount of gas
15. Molar Volume of Gas Recall 1 mol = 6.022 x 10 23 molecules 1 mol O2 = 6.022 x 10 23 molecules O2 = 32.00 g O2 1 mol H2 = 6.022 x 10 23 molecules H2 = 2.02 g H2 Standard Molar Volume of gases Volume occupied by 1 mol of a gas at STP At STP, 1 mol of any gas = 22. 4 L
