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lecture20.updated

lecture20.updated - Lecture 20 Gases wrap up Lecture 20...

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Lecture 20: Gases: wrap up Combined Gas Law: 2-state problems P 1 V 1 n 1 T 1 = R P 2 V 2 n 2 T 2 = If n remains constant (same system): P 1 V 1 T 1 P 2 V 2 T 2 = combined gas law Always use T in Kelvins

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P 1 V 1 T 1 P 2 V 2 T 2 = Combined Gas Law A gas occupies 401 mL at P = 1.000 atm. What will be its volume if P is decreased to 0.750 atm? T is held constant. T 1 = T 2 (cancel out) P 1 V 1 P 2 V 2 = 1.000 atm (401 mL) = (0.750 atm) V 2 V 2 = 401 mL (1.000 atm/ 0.750 atm) = 535 mL Gas Density and Molar Masses For a gas with mass, m, and molecular mass, M: PV = nRT = RT m M d is proportional to M . M (g/mol) d ( g / L) He 4.003 0.179 air 28.8* 1.28 O 2 31.999 1.428 SF 6 146.06 6.516 STP densities; *Average mass. m V PM RT d = =