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 ?

Explanation

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`