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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?. (101.3 kPa) x (500 dm 3 ) = (93.3 kPa) x (V 2 ). 273 K. 273 K. (101.3) x (500) = (93.3) x (V 2 ). P 1 = 101.3 kPa T 1 = 273 K
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The Combined Gas Law When measured at STP, a quantity of gas has a volume of 500 dm3. What volume will it occupy at 0 oC and 93.3 kPa? (101.3 kPa) x (500 dm3) = (93.3 kPa) x (V2) 273 K 273 K (101.3) x (500) = (93.3) x (V2) P1 = 101.3 kPa T1 = 273 K V1 = 500 dm3 P2 = 93.3 kPa T2 = 0 oC + 273 = 273 K V2 = X dm3 V2 = 542.9 dm3
P T Gay-Lussac’s Law The pressure and absolute temperature (K) of a gas are directly related • at constant mass & volume Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
P T Gay-Lussac’s Law The pressure and absolute temperature (K) of a gas are directly related • at constant mass & volume Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
P1V1 T1 P2V2 T2 = Combined Gas Law P T V T PV T PV = k (COMBINED GAS LAW) (Gay-Lussac’s LAW) (CHARLES’ LAW) (BOYLE’S LAW) P1V1T2 =P2V2T1 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Gas Law Calculations Boyle’s Law PV = k Charles’ Law V T = k P and V change n, R, T are constant Ideal Gas Law PV = nRT T and V change P, n, R are constant P, V, and T change n and R are constant Combined Gas Law PV T = k