Question

Calculate w and $\mathbf{∆}\mathit{E}$when 1 mole of a liquid is vaporized at its boiling point $\left({80}^{0}C\right)$and 1.00 atm pressure. for the liquid is 30.7 kJ mol 21 at ${80}^{0}C$.

Short answer

Work (w) = 2940 J and Internal energy $\left(∆E\right)=27.8kJ$

Step-by-step solution

First we have to find out the number of moles that change when a liquid change into gas/vapor

Given:

$∆H_{vap}=30.7Kj$; temperature$={80}^{0}C$ or $\left(80+273k\right)=353K;pressure=1.00atm$

Reaction: $X\left(l\right)\to X\left(g\right)$

Where change in number of moles $\left(∆n\right)=n\left(product\right)--n\left(reactant\right)$

$∆n=1--0=1$

Now, by using the formula of work;$w=-P∆V$

Since, from the ideal gas equation:$PV=nRT;P=\frac{nRT}{V}$

Sincepressure is constant:

$$\begin{gathered} w=-nRT \\ w=-\left(1mol\right)\left(8.3145J{K}^{-1}mo{l}^{-1}\right)\left(353K\right)=-2940J \end{gathered}$$

Now we can calculate the change in internal energy by using formula:

$$\begin{gathered} ∆E=q+w=w+nH \\ ∆E=\left(-2.94kJ\right)+30.7kJ \\ ∆E=27.8kJ \end{gathered}$$

Hence, the $w=-2940Jand∆E=27.8kJ$