The equation below will be used to

determine the vapor pressure of isopropanol

in this experiment:

$lnP=-(\frac{H_{\text{v}\text{a}\text{p}}}{RT})+C$

Let's assume that you run the experiment at

$21\mathrm{.}1C,27\mathrm{.}2C,33\mathrm{.}9C$ and $13\mathrm{.}1C$ with

the measured vapor pressure $56\mathrm{.}4mmHg,$

$80\mathrm{.}0mmHg,95\mathrm{.}1mmHg,$ and $48\mathrm{.}1mmHg,$

respectively. What is the $\frac{1}{T}$ value for the

temperature at $27\mathrm{.}2($keep $5$ SIGs$)$

$3\mathrm{.}6765{10}^{-2}{T}^{-1}$

$3\mathrm{.}6765{10}^{-3}{T}^{-1}$

$3\mathrm{.}3295{10}^{-3}{T}^{-1}$

$3\mathrm{.}3295{10}^3{T}^{-1}$

$D$ View hint for Question $3$