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The Gas Laws

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Directly proportionalas one variable goes up/down the other goes up/down. Both variable do the same thing. Indirectly proportional– as one variable goes up the other goes down. The two variable do the opposite thing.

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Boyles Law P 1 V 1 = P 2 V 2 Temperature is constant Pressure (P) and volume (V) are indirectly proportional

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Example 1 A helium balloon was compressed from 4.0L to 2.5L at a constant temperature. If the pressure of the gas in the 4.0L balloon is 210 kPA, what will the pressure be at 2.5L? Given: V 1 = 4.0LV 2 = 2.5L P 1 = 210 kPaP 2 = ? P 1 V 1 = P 2 V 2 (210 kPa) × (4.0L) = P 2 × (2.5L) P 2 = 336 kPa 340 kPa

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Example 2 A sample of neon gas occupies 0.200L at 0.860 atm. What will be its volume at 29.2 kPa pressure? Given: V 1 = 0.200L V 2 = ? P 1 = 0.860 atm P 2 = 29.2 kPa P 1 V 1 = P 2 V 2 (87.1 kPa) × (0.200L) = (29.2 kPa) × V 2 V 2 = 0.597L **Units must match for each variable (doesnt matter which one is converted) 0.860 atm 101.3 kPa 1 atm = 87.1 kPa

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Charles Law V 1 V 2 T 1 T 2 Pressure is constant Temperature must be in Kelvin Volume (V) and temperature (T) are directly proportional =

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Example 1 A gas ample at 40.0°C occupies a volume of 2.32L. If the temperature is raised to 75.0°C, what will the volume be, assuming the pressure remains constant? Given: T 1 = 40.0°C = 313KT 2 = 75.0°C = 348K V 1 = 2.32LV 2 = ? V 1 V 2 T 1 T 2 = 2.32L V 2 313 348 = V 2 = 2.58L

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Example 2 A gas ample at 55.0°C occupies a volume of 3.50L. At what new temperature in kelvin will the volume increase to 8.00L? Given: T 1 = 55.0°C = 328KT 2 = ? V 1 = 3.50LV 2 = 8.00L V 1 V 2 T 1 T 2 = 3.50L 8.00 328 T 2 = T 2 = 750K

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Gay-Lussacs Law P 1 P 2 T 1 T 2 Volume is constant Temperature must be in Kelvin Pressure (P) and temperature (T) are directly proportional =

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Example 1 The pressure of a gas in a tank is 3.20 atm at 22.0°C. If the temperature rises to 60.0°C, what will be the gas pressure in the tank? Given: P 1 = 3.20 atmP 2 = ? T 1 = 22.0°C = 295KT 2 = 60.0°C =333K P 1 P 2 T 1 T 2 = 3.20 atm P 2 295K 333K = P 2 = 3.61 atm

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Example 2 A rigid container has a gas at constant volume at 665 torr pressure when the temperature is 22.0C. What will the pressure be if the temperature is raised to 44.6C? Given: P 1 = 665 torrP 2 = ? T 1 = 22.0°C = 295KT 2 = 44.6°C =317.6K P 1 P 2 T 1 T 2 = 665 torr P 2 295K 317.6K = P 2 = 716 atm

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Gas Laws 10-2 and 10-3. Ideal Gas Law PV = nRT PV = nRT P = Pressure, in atm V = volume, in L n = number of moles T =Temperature, in Kelvins (K = C +

Gas Laws 10-2 and 10-3. Ideal Gas Law PV = nRT PV = nRT P = Pressure, in atm V = volume, in L n = number of moles T =Temperature, in Kelvins (K = C +

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