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27 de Jan de 2023•0 recomendaciones•3 vistas

27 de Jan de 2023•0 recomendaciones•3 vistas

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Helium gas with a volume of 2.90 L under a pressure of 1.80 atm and at a temperature of 43.0 ?C is warmed until both the pressure and volume of the gas are doubled. What is the final temperature? How many grams of helium are there? The molar mass of helium is 4.00 g/mol Solution a) Using the gas equation, P1V1/T1 = P2V2/T2 Here let P1 = 1.8 atm so P2 = 2P1 atm V1 = 2.90 L so V2 = 2V1 L T1 = 273 + 43 = 316 K , T2 = ? T2 = P2V2/P1V1*T1 = 2P1*2V1/(P1V1)*316 = 1264 K b) By using, PV = nRT where n is the No. of moles of He n = PV/RT = 1.8*2.90/(0.0821*316) = 0.20 Thus the moles of He = 0.20 And the mass of He = 0.20 * 4 = 0.80 g .

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- 1. Helium gas with a volume of 2.90 L under a pressure of 1.80 atm and at a temperature of 43.0 ?C is warmed until both the pressure and volume of the gas are doubled. What is the final temperature? How many grams of helium are there? The molar mass of helium is 4.00 g/mol Solution a) Using the gas equation, P1V1/T1 = P2V2/T2 Here let P1 = 1.8 atm so P2 = 2P1 atm V1 = 2.90 L so V2 = 2V1 L T1 = 273 + 43 = 316 K , T2 = ? T2 = P2V2/P1V1*T1 = 2P1*2V1/(P1V1)*316 = 1264 K b) By using, PV = nRT where n is the No. of moles of He n = PV/RT = 1.8*2.90/(0.0821*316) = 0.20 Thus the moles of He = 0.20
- 2. And the mass of He = 0.20 * 4 = 0.80 g