Presentation on theme: "Converting Pressures 760 mm Hg = 760 torr 1 atm = 760 mm Hg 1 atm = 101.3 kPa 760 mm Hg = 101.3 kPa Converting Pressures: Convert 740 torr to kPa 740."— Presentation transcript:
Converting Pressures 760 mm Hg = 760 torr 1 atm = 760 mm Hg 1 atm = 101.3 kPa 760 mm Hg = 101.3 kPa Converting Pressures: Convert 740 torr to kPa 740 torr (current pressure) x 101.3kPa (1 atm in kPa) /760 torr (atm in torr) (740)(101.3kPa /760) 740/760 is the fraction of 1 atm in torrs. Convert 2 atm to mm Hg 2 atm (760 mm Hg/1 atm)
Dalton’s Law of Partial Pressure The pressure of each gas in a mixture is called the partial pressure of that gas. Daltons Law of Partial Pressure states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.
Dalton’s Law of Partial Pressure PT = P1 + P2 + P3 + …….PT = total pressure P# = the partial pressures of the individual gases
Dalton’s Law of Partial Pressure 1e. A mixture of gases has the following partial pressure for the component gases at 20.0 C in a volume of 2.00L: oxygen 180.torr, nitrogen 320.torr, and hydrogen 246torr. Calculate the pressure of the mixture. P tot = P A + P B + P C + … P tot = 180 + 320 + 246 P tot = 746
Dalton’s Law A metal tank contains three gases: oxygen, helium, and nitrogen. If the partial pressures of the three gases in the tank are 35 atm of O2, 5 atm of N2, and 25 atm of He, what is the total pressure inside of the tank? 65 atm (35 + 25 + 5)
Dalton’s Law Blast furnaces give off many unpleasant and unhealthy gases. If the total air pressure is 0.99 atm, the partial pressure of carbon dioxide is 0.05 atm, and the partial pressure of hydrogen sulfide is 0.02 atm, what is the partial pressure of the remaining air? 0.92 atm (.99 -.05 -.02)
Dalton’s Law of Partial Pressure 3e. Oxygen gas from the decomposition reaction of potassium chlorate was collected by water displacement at a pressure of 731torr and a temperature of 20.0 C. What was the partial pressure of the oxygen gas collected? 4e. Solid magnesium and hydrochloric acid react producing hydrogen gas that was collected over water at a pressure of 759mmHg and measured 19.0mL. The temperature of the solution at which the gas was collected was 25.0 C. What would be the pressure of the dry hydrogen gas? What would be the volume of the dry hydrogen gas at STP?
Daltons Law applied to Gases Collected by Water Displacement – Figure 10-15 page 324 P atm or P T = P gas + P H2O P atm or P T = barometric pressure or total pressure P gas = pressure of the gas collected P H2O = vapor pressure of water at specific temperature (Found on page 899 of you textbook.)
Vapour Pressure Defined Vapour pressure is the pressure exerted by a vapour. E.g. the H 2 O(g) in a sealed container. Yet, molecules both leave and join the surface, so vapour pressure also pushes molecules up. Eventually the air above the water is filled with vapour pushing down. As temperature , more molecules fill the air, and vapour pressure . To measure vapour pressure we can heat a sample of liquid on top of a column of Hg and see the pressure it exerts at different °C.
Measuring Vapour Pressure When the vapour pressure is equal to the atmospheric pressure (P atm ), the push out is enough to overcome P atm and boiling occurs. Vapour pressure Temperature Vapour pressure for H 2 O °CkPa°CkPa 101.235012.33 202.347538.54 304.17100 See pg. 464 for more 101.3 Thus, water will boil at a temperature below 100 °C if the atmospheric pressure is reduced.
Collecting gases over water Many times gases are collected over H 2 O Often we want to know the volume of dry gas at STP (useful for stoichiometry). For this we must make 3 corrections: 1. The level of water inside and outside the tube must be level (so pressure inside is equal to the pressure outside). 2. The water vapour pressure must be subtracted from the total pressure (to get the pressure of the dry gas). 3. Finally, values are converted to STP using the combined gas law.
Sample calculation A gas was collected over 21°C H 2 O. After equal- izing water levels, the volume was 325 mL. Give the volume of dry gas at STP (P atm =102.9 kPa). Step 1: Determine vapour pressure (pg. 464) At 21°C vapour pressure is 2.49 kPa Step 2: Calculate the pressure of dry gas P gas = P atm - P H2O = 102.9 - 2.49 = 100.41 kPa Step 3: List all of the data T 1 = 294 K, V 1 = 325 mL, P 1 = 100.41 kPa Step 4: Convert to STP (P 1 )(V 1 )(T 2 ) (P 2 )(T 1 ) V2=V2= (100.4 kPa)(325 mL)(273 K) (101.325 kPa)(294 K) = = 299 mL