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Combined Gas Law

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The Combined Gas Law When measured at STP, a quantity of gas has a volume of 500 dm 3. What volume will it occupy at 0 o C and 93.3 kPa? P 1 = 101.3 kPa T 1 = 273 K V 1 = 500 dm 3 P 2 = 93.3 kPa T 2 = 0 o C + 273 = 273 K V 2 = X dm 3 (101.3 kPa) x (500 dm 3 ) = (93.3 kPa) x (V 2 ) 273 K V 2 = 542.9 dm 3 (101.3) x (500) = (93.3) x (V 2 )

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The Combined Gas Law When measured at STP, a quantity of gas has a volume of 500 dm 3. What volume will it occupy at 0 o C and 93.3 kPa? P 1 = 101.3 kPa T 1 = 273 K V 1 = 500 dm 3 P 2 = 93.3 kPa T 2 = 0 o C + 273 = 273 K V 2 = X dm 3 = P 2 x V 2 T 2 P 1 x V 1 T 1 (101.3 kPa) x (500 dm 3 ) = (93.3 kPa) x (V 2 ) 273 K V 2 = 542.9 dm 3

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The pressure and absolute temperature (K) of a gas are directly related –at constant mass & volume P T Gay-Lussacs Law Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem Temperature (K) Pressure (torr) P/T (torr/K) 248691.62.79 273760.02.78 298828.42.78 3731,041.22.79

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Gay-Lussacs Law P T Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem The pressure and absolute temperature (K) of a gas are directly related –at constant mass & volume

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= kPV PTPT VTVT T Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2 P 1 V 1 T 2 = P 2 V 2 T 1 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem (BOYLES LAW)(CHARLES LAW)(Gay-Lussacs LAW)(COMBINED GAS LAW)

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The Combined Gas Law Keys The Combined Gas Law http://www.unit5.org/chemistry/GasLaws.html

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Gas Law Calculations Boyles Law PV = k Boyles Law PV = k Charles Law V T Charles Law V T Combined Gas Law PV T Combined Gas Law PV T Ideal Gas Law PV = nRT Ideal Gas Law PV = nRT = k T and V change P, n, R are constant P, V, and T change n and R are constant P and V change n, R, T are constant

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