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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. Boyle’s Law Shows the relationship between volume and pressure Temperature and amount of gas is held constant

Gas Laws. Boyle’s Law Shows the relationship between volume and pressure Temperature and amount of gas is held constant

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