in this lab, you will determine the vapour pressure (p) of water at various temperatures and use this data to find its enthalpy of vaporization ($\delta_{vap}h$).
more generally, for an unknown liquid: if a plot of $\ln p$ vs. $1/t$ gives a slope of -3105 k, what is its $\delta_{vap}h$ in kj/mol? do not worry about how realistic the value is. assume $r = 8.314 j mol^{-1}k^{-1}$.

Question

Accepted Answer

# Explanation:
## Step1: Recall the Clausius - Clapeyron equation
The Clausius - Clapeyron equation in the form of a linear equation $y = mx + c$ for $\ln P$ vs. $\frac{1}{T}$ is $\ln P=-\frac{\Delta_{vap}H}{R}\times\frac{1}{T}+C$. Here, the slope $m =-\frac{\Delta_{vap}H}{R}$.
## Step2: Solve for $\Delta_{vap}H$
Given $m=- 3105\space K$ and $R = 8.314\space J\space mol^{-1}\space K^{-1}$. From $m =-\frac{\Delta_{vap}H}{R}$, we can solve for $\Delta_{vap}H$. Rearranging the formula gives $\Delta_{vap}H=-m\times R$.
Substitute the values: $\Delta_{vap}H=-(-3105\space K)\times8.314\space J\space mol^{-1}\space K^{-1}$
$$
\begin{align*}
\Delta_{vap}H&=3105\times8.314\space J\space mol^{-1}\
&=3105\times8.314\times10^{- 3}\space kJ\space mol^{-1}\
\end{align*}
$$
$$
\begin{align*}
\Delta_{vap}H&=(3105\times8.314)\times10^{-3}\space kJ\space mol^{-1}\
&=(3105\times8 + 3105\times0.314)\times10^{-3}\space kJ\space mol^{-1}\
&=(24840+975.97)\times10^{-3}\space kJ\space mol^{-1}\
&=25815.97\times10^{-3}\space kJ\space mol^{-1}\
&\approx25.8\space kJ\space mol^{-1}
\end{align*}
$$

# Answer:
$25.8\space kJ/mol$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the Clausius - Clapeyron equation

Step2: Solve for \(\Delta_{vap}H\)

Answer: