Unit 6: Gases and Atmospheric Chemistry Pressure, Volume, and Temperature Boyle’s Law

Pressure, temperature and volume Pop can demonstration - Watch this! A bigger can … - Watch this!

Pressure, temperature and volume Pressure is a measure of force per area. As the particles strike the walls of their container, they exert a force. The force per area is the pressure of the gas. Units : The metric unit for pressure is the Pascal:

Atmospheric pressure The pressure of the atmosphere can be measured with a barometer: 760 mm Hg = 760 Torr = 1.00 atm = 101.3 kPa Since all of these units are equivalent, we can use them as conversion factors: e.g. Convert a pressure of 700 mm Hg into kPa:

Temperature Temperature is a measure of the average kinetic energy possessed by the particles of a substance. Celcius (°C) Kelvin (K) 100 373 To convert from °C to K:   TK = TC + 273 25 298 273 To convert from K to °C:   TC = TK - 273 -273 Absolute zero (where all particle motion stops) = 0 K = -273.15 °C.

STANDARD TEMPERATURE AND PRESSURE (STP) Volume Units: 1 mL = 1 cm3 1 L = 1000 mL = 1000 cm3 25 cm3 mL 182 cm3 L 3.680 L mL 𝑥 1 𝑚𝐿 1 𝑐𝑚 3 = 25 𝑥 1 𝐿 1000 𝑐𝑚 3 = 0.182 𝑥 1000 𝑚𝐿 1 𝐿 = 3680 STANDARD TEMPERATURE AND PRESSURE (STP)   Temperature = 0oC = 273 K Pressure = 101.3 kPa

Boyle’s Law If the pressure on a specific amount of gas is increased, the volume will decrease.  V  P  V  P

Boyle’s law “The volume of a fixed mass of gas is INVERSELY PROPORTIONAL to pressure, provided temperature remains constant.” 𝑉= 𝑘 𝑃 𝑤ℎ𝑒𝑟𝑒 𝑘=𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑉 1 𝑃

Boyle’s Law Since 𝑉= 𝑘 𝑃 , then 𝑃𝑉=𝑘  𝑃 1 𝑉 1 = 𝑃 2 𝑉 2 e.g. A 100 L volume of gas is at a pressure of 32 kPa. If the pressure increases to 44 kPa, what is the final volume?

e. g. A 100 L volume of gas is at a pressure of 32 kPa e.g. A 100 L volume of gas is at a pressure of 32 kPa. If the pressure increases to 44 kPa, what is the final volume? Given: P1 = P2 = V1 = V2 = 32 kpa 44 kpa Always list given information. 100 L ? 𝑃 1 𝑉 1 = 𝑃 2 𝑉 2 𝑉 2 = 𝑃 1 𝑉 1 𝑃 2 = 32 𝑘𝑃𝑎 (100 𝐿) (44 𝑘𝑃𝑎) =72.727…𝐿 = 73 L Rounded to 2 sig. digits.

Unit 6: Gases and Atmospheric Chemistry Charles’ Law and Gay-Lussac’s Law

Charles’ Law “The volume of a fixed mass of gas is DIRECTLY PROPORTIONAL to temperature, provided pressure remains constant.” 𝑉=𝑘𝑇 𝑤ℎ𝑒𝑟𝑒 𝑘=𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑉𝑇

Charles’ Law

Charles’ Law Since 𝑉=𝑘𝑇 , then 𝑉 𝑇 =𝑘  𝑉 1 𝑇 1 = 𝑉 2 𝑇 2 e.g. If the volume of a balloon is 10.0L at 0.0°C, what is its volume at room temperature (25°C)?

e. g. If the volume of a balloon is 10. 0L at 0 e.g. If the volume of a balloon is 10.0L at 0.0°C, what is its volume at room temperature (25°C)? Given: T1 = T2 = V1 = V2 = 0.0°C = 273.0 K 25°C = 298 K Always list given information. Always convert from Celsius to Kelvin 10.0 L ? 𝑉 1 𝑇 1 = 𝑉 2 𝑇 2 𝑉 2 = 𝑇 2 𝑉 1 𝑇 1 = 298 K (10.0 𝐿) (273.0 𝐾) =10.9157…𝐿 = 10.9 L Rounded to 3 sig. digits.

Gay-Lussac’s law “The pressure of a fixed mass of gas is DIRECTLY PROPORTIONAL to temperature, provided volume remains constant.” P =𝑘𝑇 𝑤ℎ𝑒𝑟𝑒 𝑘=𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 P 𝑇

Gay-Lussac’s Law

Gay-Lussac’s Law Since 𝑃=𝑘𝑇 , then 𝑃 𝑇 =𝑘  𝑃 1 𝑇 1 = 𝑃 2 𝑇 2 e.g. A can of hairspray contains a gas which exerts a pressure of 900 kPa at 25°C. If the can can only withstand a maximum of 1800 kPa, at what temperature will it explode?

e.g. A can of hairspray contains a gas which exerts a pressure of 900 kPa at 25°C. If the can can only withstand a maximum of 1800 kPa, at what temperature will it explode? Given: T1 = T2 = P1 = P2 = 25°C = 298K ? Always list given information. Always convert from Celsius to Kelvin. 900 kPa 1800 kPa 𝑃 1 𝑇 1 = 𝑃 2 𝑇 2 𝑇 2 = 𝑃 2 𝑇 1 𝑃 1 = 1800 kPa (298K ) (900 𝑘𝑃𝑎) =596 K =323 °C

The Combined Gas Law

Combining the gas laws P1V1 P2V2 T1 T2 = V1 T1 = V2 T2 P1 T1 = P2 T2 So far we have seen three gas laws: Robert Boyle Jacques Charles Joseph Louis Gay-Lussac V1 T1 = V2 T2 P1 T1 = P2 T2 P1V1 = P2V2 These are all subsets of a more encompassing law, the combined gas law: P1V1 P2V2 T1 T2 =

Combined Gas Law Equations: P1 = P2T1V2 T2V1 P2 = P1T2V1 T1V2 T1 = P1T2V1 P2V2 T2 = P2T1V2 P1V1 V1 = P2T1V2 T2P1 V2 = P1T2V1 P2T1

Ex. 1) A 10. 0L volume of gas is at 27°C and 101. 3 kPa Ex.1) A 10.0L volume of gas is at 27°C and 101.3 kPa. What would be its volume if the gas was at 0°C and 50 kPa? Given: T1 = T2 = P1 = P2 = V1 = V2 = 27°C = 300K 0°C = 273K Always list given information. Always convert from Celsius to Kelvin. 101.3 kPa 50 kPa 10.0 L ? 𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2 𝑉 2 = 𝑃 1 𝑉 1 𝑇 2 𝑃 2 𝑇 1 = (101.3 𝑘𝑃𝑎)(10.0 𝐿)(273𝐾) (50 𝑘𝑃𝑎)(300𝐾) =18.4366 𝐿 ≐18 L  

(87.5 𝑘𝑃𝑎)(820 𝑚𝐿)(270𝐾) (1.05 𝑎𝑡𝑚)(1010 𝑚𝐿) Ex.2) 1010 mL of a gas is at 1.05 atm and -3°C. What would the temperature be if the pressure is changed to 87.5 kPa and the volume become 820 mL? Given: T1 = T2 = P1 = P2 = V1 = V2 = -3°C = 270K ? 1.05 atm 87.5 kPa 1010 mL 820 mL 𝑃 1 =1.05 𝑎𝑡𝑚 × 101.3 𝑘𝑃𝑎 1.00 𝑎𝑡𝑚 ≐106 𝑘𝑃𝑎 𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2 𝑇 2 = 𝑃 2 𝑉 2 𝑇 1 𝑃 1 𝑉 1 = (87.5 𝑘𝑃𝑎)(820 𝑚𝐿)(270𝐾) (1.05 𝑎𝑡𝑚)(1010 𝑚𝐿) =180.949935…𝐾 ≐181𝐾 If the original question uses °C you must answer in the same units. ≐−92°C  

(1.0 𝑎𝑡𝑚)(820 𝑚𝐿)(270𝐾) (1.05 𝑎𝑡𝑚)(1010 𝑚𝐿) Ex.3) A solid steel cylinder is full of gas at a temperature of 25°C and pressure of 1.6 atm. When the gas was released into a balloon under STP conditions the volume was 865 L. What was the original volume of the cylinder? Given: T1 = T2 = P1 = P2 = V1 = V2 = 25°C = 298K 273K 1.6 atm 101.3 kPa = 1.0 atm ? 865 L 𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2 𝑉 1 = 𝑃 2 𝑉 2 𝑇 1 𝑃 1 𝑇 2 = (1.0 𝑎𝑡𝑚)(820 𝑚𝐿)(270𝐾) (1.05 𝑎𝑡𝑚)(1010 𝑚𝐿) =590.13278…𝐿 ≐590 𝐿  

Ex. 4) A 2000 L volume of gas at STP was compressed into 0 Ex.4) A 2000 L volume of gas at STP was compressed into 0.500 L at 35°C. What would the final pressure be in the cylinder? Given: T1 = T2 = P1 = P2 = V1 = V2 = 273K 35°C = 308K 101.3 kPa ? 2000 L 0.500 L 𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2 𝑃 2 = 𝑃 1 𝑉 1 𝑇 2 𝑉 2 𝑇 1 = (101.3 𝑘𝑃𝑎)(2000 𝐿)(308𝐾) (0.500 𝐿)(273𝐾) =457 158.7179…𝑘𝑃𝑎 ≐457 000 𝑘𝑃𝑎  

Ex. 5) A balloon filled with 0 Ex.5) A balloon filled with 0.0303 mol of helium at 30°C and a pressure of 1.0 atm occupies a volume of 0.75 L and has a density of 0.161 g/L. What would the density of the helium gas be if the balloon was placed in the freezer at -10°C and a pressure of 2.0 atm? Given: T1 = T2 = P1 = P2 = V1 = V2 = nHe = D1= D2 = mHe= 30°C = 303K -10°C = 263K 1.0 atm 2.0 atm 0.75 L ? 0.0303 mol Not important 0.161 g/L ?   𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2 𝑉 2 = 𝑃 1 𝑉 1 𝑇 2 𝑃 2 𝑉 2 = (1.0 𝑎𝑡𝑚)(0.75 𝐿)(263𝐾) (2.0 𝑎𝑡𝑚)(303𝐾) =0.325495…L ≐0.33 𝐿 𝐷 2 = 𝑚 𝐻𝑒 𝑉 2 = 0.12 𝑔 0.33 𝐿 ≐0.37 𝑔/𝐿