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An open bulb containing air at 19^(@)C was cooled to a certain temperature at which the number of moles of the gaseous molecules increased by 25%. What is the final temperature ? |
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Answer» Solution :Suppose the VOLUME of the bulb is V containing n moles at `19^(@)C`, i.e, 292 K. Let the temperature be T K when n moles increases to `1.25` n (i.e., by 25%). SINCE 1.25 n moles at T K occupy a volume V `therefore` n moles at T K should occupy `(V)/(1.24)`. Thus for n moles of the GAS, `T_(1) = 292 K ""T_(2) = T K` `V_(1) = V""V_(2) = (V)(1.25)` `p_(1) = p""p_(2) = p(p_(1) = p_(2) " as the bulb is OPEN")` `therefore (pV)/(292) = (p XX V//1.25)/(T)` `T = (292)/(1.25) = 233.6 K` `= -39.4^(@)C` |
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