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F.4 一條Phy題 明測驗用10分 謝

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Question ; http://i.na.cx/e13Cj.jpg http://i.na.cx/350OP.jpg thx ar!!! 更新: Question b ans : 500ms^-1 c i ans; 2.09x10^5 Pa c ii ans : 2 x 10^-3 mol I don't understand how to calculate c part only

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22. (a) P = 12 x 10? Pa V = 100 x 10?? m3 R = 8.31 J mol?1 K?1 T = 273 K PV = nRT (12 x 10?) x (100 x 10??) = n x 8.31 x 273 Number of moles of gas in container A, n = 0.0529mol (b) R = 8.31 J mol?1 K?1 T = 273 K M = (4.52 x 10?2?) x(6.02 x 1023) kg mol?1 Root-mean-square speed of the gas in B = √{3 x 8.31 x 273 / [(4.52 x 10?2?) x (6.02 x 1023)]} = 500 m s?1 (c)(i) Let nA mol and nB mol be the numbers of moles in containersA and B respectively. PV = nRT In A : P x (100 x 10??) =nA x 8.31 x 373 nA = P x (100 x 10??) / (8.31 x 373) In B : P x (500 x 10??) =nA x 8.31 x 273 nB = P x (500 x 10??) / (8.31 x 273) nA + nB = 0.0529 [P x (100 x 10??) /(8.31 x 373)] + [P x (500 x 10??) / (8.31 x 273)] = 0.0529 P = 0.0529 ÷ {[(100 x 10??) / (8.31 x 373)] + [(500 x 10??) / (8.31 x 273)]} P = 2.09 x 10? Pa c(ii) Before the heating process : nA = 0.0529 x [100/(100 + 500)] = 8.82 x 10?3 mol After the heating process nA = (2.09 x 10?) x (100 x 10??) / (8.31 x 373) = 6.72 x 10?3 mol The net amount of gas that passed through the connecting tube during theheating process = (8.82 x 10?3) -(6.72 x 10?3) = 2.1 x 10?3 mol ≈ 2 x 10?3 mol

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