標題:
F.6 Partial Pressure of ideal gas calculations
A gas cylinder of 55.0dm3 volume contains N2(g) at a pressure of 20.0atm and 23。C. How many grams of Ne(g) must be introduced into this same cylinder to raise the total pressure to 75.0 atm? (Relative atomic mass: Ne=20.2)
最佳解答:
Rewriting the units in standard units: P = 20 1.01 105 = 2.02 105 Pa T = 273 + 23 = 296 K V = 55 10-3 = 0.055 m3 By n = PV/RT, we have: n = 4.515 moles By Avogadro's and Dalton's law, no. of moles of neon that should be introduced into the cylinder = (75 - 55)/55 4.515 = 1.641 moles which gives mass = 33.16 g 2008-09-21 23:27:49 補充： Correction to the last sentence: By Avogadro's and Dalton's law, no. of moles of neon that should be introduced into the cylinder = (75 - 20)/20 x 4.515 = 12.42 moles which gives mass = 250.9 g Thanks again to Uncle Michael.
其他解答:
F.6 Partial Pressure of ideal gas calculations
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發問:A gas cylinder of 55.0dm3 volume contains N2(g) at a pressure of 20.0atm and 23。C. How many grams of Ne(g) must be introduced into this same cylinder to raise the total pressure to 75.0 atm? (Relative atomic mass: Ne=20.2)
最佳解答:
Rewriting the units in standard units: P = 20 1.01 105 = 2.02 105 Pa T = 273 + 23 = 296 K V = 55 10-3 = 0.055 m3 By n = PV/RT, we have: n = 4.515 moles By Avogadro's and Dalton's law, no. of moles of neon that should be introduced into the cylinder = (75 - 55)/55 4.515 = 1.641 moles which gives mass = 33.16 g 2008-09-21 23:27:49 補充： Correction to the last sentence: By Avogadro's and Dalton's law, no. of moles of neon that should be introduced into the cylinder = (75 - 20)/20 x 4.515 = 12.42 moles which gives mass = 250.9 g Thanks again to Uncle Michael.
其他解答:
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