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Gas Laws

Gas Laws. Charles’s Law. V 1 x T 2 = V 2 x T 1. As T increases. V increases. At constant Pressure and Amount of gas. 1.54 L x 398 K. V 2 x T 1. =. 3.20 L. V 1.

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Gas Laws

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  1. Gas Laws

  2. Charles’s Law V1 x T2 = V2 x T1 As T increases V increases At constant Pressure and Amount of gas

  3. 1.54 L x 398 K V2 x T1 = 3.20 L V1 A sample of carbon monoxide gas occupies 3.20 L at 125°C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1 x T2 = V2 x T1 *Temperature must always be in Kelvin* V1 = 3.20 L V2 = 1.54 L T1 = 125°C + 273 = 398 K T2 = ? T2 = = 192 K

  4. 6.91 L x 452 K V1 x T2 = 385 K T1 A sample of carbon dioxide gas occupies 6.91 L at 112°C. What volume will the gas occupy when the temperature is raised to 179°C if the pressure remains constant? V1 x T2 = V2 x T1 *Temperature must always be in Kelvin* V1 = 6.91 L V2 = ? T1 = 112°C + 273 = 385 K T2 = 179°C + 273 = 452 K V2 = = 8.11 L

  5. (0.581 atm)(405mL)(273K) V2 = (1.00 atm)(363K) Combined Gas Law P1V1T2 = P2V2T1 If the initial volume of a gas was 405 mL at 90.0°C and 0.581 atm, what would the volume be at STP? T1 = 90.0°C + 273 = 363 K V2 = P1V1T2 P2T1 P1 = 0.581 atm V1 = 405 mL V2 = ? T2 = 0 0C + 273 = 273 K P2 = 1.00 atm V2 = 177 mL

  6. (1.25 atm)(236 L)(298K) P2 = (652 L)(348K) Combined Gas Law P1V1T2 = P2V2T1 If the initial volume of SO2 was 236 L at 75.0°C and 1.25 atm, what would the pressure be when the volume is raised to 652 L while the temperature is decreased to 25.0°C? P2 = P1V1T2 V2T1 P1 = 1.25 atm V1 = 236 L T1 = 75.0°C + 273 = 348 K T2 = 25.00C + 273 = 298 K V2 = 652 L P2 = 0.387 atm

  7. Ideal Gas Law • ideal gas - follows all assumptions of kinetic theory (particles have no volume, no attraction between particles) • used to find one factor that effects gas when the other 3 factors are known • about the gas at one moment in time (no changing conditions) • only law that allows you to determine the amount of gas • PV = nRT • R is ideal gas law constant

  8. 1 mol HCl V = n = 49.8 g x = 1.37 mol 36.45 g HCl 1.37 mol x 0.08206 x 273 K V = 1 atm nRT L•atm P mol•K Ideal Gas Law Remember: n = moles! PV = nRT What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0°C + 273 = 273 K PV = nRT P = 1 atm R = 0.08206 L•atm/mol•K V = 30.6 L

  9. 1 mol O2 P = n = 28.0 g x = 0.875 mol 32.0 g O2 0.875 mol x 0.08206 x 288 K P = 2.50 L nRT L•atm V mol•K Ideal Gas Law Remember: n = moles! PV = nRT What is the pressure occupied by 28.0 g of O2 at 15.0°C and 2.50 L? T = 15°C + 273 = 288 K PV = nRT V = 2.50 L R = 0.08206 L•atm/mol•K P = 8.27 atm

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