Ethyl acetate has a vapour pressure of $94\mathrm{.}7mmHg$ at $25C.$ If ${}_{\text{V}\text{A}\text{P}}H=31\mathrm{.}94k\frac{J}{m}ol,$ at what temperature would the vapour pressure be $165\mathrm{.}1mmHg?$

$ln(\frac{P_2}{P_1})=\frac{-H_{\text{i}\text{v}\text{a}\text{p}}}{R}(\frac{1}{{}_2}-\frac{1}{{}_1})$