$\Delta_{\text {vap }} \mathrm{H}^{\ominus}$ for water is $+40.79 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at 1 bar and $100^{\circ} \mathrm{C}$. Change in internal energy for this vapourisation under same condition is ________ $\mathrm{kJ} \mathrm{~mol}^{-1}$. (Integer answer) (Given $\mathrm{R}=8.3 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}$)
Answer (integer)
38
Solution
<p>To find the change in internal energy for the vaporization of water under the given conditions, we'll use the following relationship between enthalpy change ($\Delta_{\text{vap}} H$) and internal energy change ($\Delta_{\text{vap}} U$):</p>
<p>$\Delta_{\text{vap}} H = \Delta_{\text{vap}} U + P \Delta V$</p>
<p>For vaporization, the change in volume ($\Delta V$) can be approximated by considering the volume of the vapor because the volume of liquid water is relatively small compared to the volume of the vapor.</p>
<p>The ideal gas law gives us:</p>
<p>$P V = n R T$</p>
<p>Since we are dealing with 1 mole of water:</p>
<p>$V = \frac{R T}{P}$</p>
<p>Plugging this into the enthalpy change equation, we get:</p>
<p>$\Delta_{\text{vap}} H = \Delta_{\text{vap}} U + P \left( \frac{R T}{P} \right)$</p>
<p>This simplifies to:</p>
<p>$\Delta_{\text{vap}} H = \Delta_{\text{vap}} U + R T$</p>
<p>Rearranging for $\Delta_{\text{vap}} U$:</p>
<p>$\Delta_{\text{vap}} U = \Delta_{\text{vap}} H - R T$</p>
<p>Given:</p>
<p>$\Delta_{\text{vap}} H = 40.79 \, \text{kJ mol}^{-1}$ (or 40790 J/mol)</p>
<p>$R = 8.3 \, \text{JK}^{-1} \text{mol}^{-1}$</p>
<p>$T = 100^\circ \text{C} + 273.15 = 373.15 \, \text{K}$</p>
<p>Now, substitute the values into the equation:</p>
<p>$$\Delta_{\text{vap}} U = 40790 \, \text{J mol}^{-1} - (8.3 \, \text{JK}^{-1} \text{mol}^{-1} \times 373.15 \, \text{K})$$</p>
<p>$$\Delta_{\text{vap}} U = 40790 \, \text{J mol}^{-1} - 3097.145 \, \text{J mol}^{-1}$$</p>
<p>$\Delta_{\text{vap}} U = 37692.855 \, \text{J mol}^{-1}$</p>
<p>Converting back to kJ:</p>
<p>$\Delta_{\text{vap}} U = 37.69 \, \text{kJ mol}^{-1}$</p>
<p>Rounding to the nearest integer, the change in internal energy for the vaporization of water under the given conditions is:</p>
<p><strong>38 kJ mol<sup>-1</sup></strong></p>
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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