What is the value of change in internal energy at 1 atm in the process? $\mathrm{H}_{2} \mathrm{O}(1,323…

What is the value of change in internal energy at 1 atm in the process?
$\mathrm{H}_{2} \mathrm{O}(1,323 \mathrm{~K}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}, 423 \mathrm{~K})$
Given : $\mathrm{C}_{\mathrm{v}, \mathrm{m}}\left(\mathrm{H}_{2} \mathrm{O}, 1ight)=75.0 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
$\mathrm{C}_{\mathrm{p}, \mathrm{m}}\left(\mathrm{H}_{2} \mathrm{O}, \mathrm{g}ight)=33.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
$\Delta \mathrm{H}_{\text {vap }}$ at $373 \mathrm{~K}=40.7 \mathrm{~kJ} / \mathrm{mol}$
  1. $42.91 \mathrm{~kJ} / \mathrm{mol}$
  2. $43086 \mathrm{~kJ} / \mathrm{mol}$
  3. $42.6 \mathrm{~kJ} / \mathrm{mol}$
  4. $49.6 \mathrm{~kJ} / \mathrm{mol}$

Solution

Given,
\(\begin{aligned} & \mathrm{H}_2 \mathrm{O}(\ell, 323 \mathrm{~K}) ightarrow \mathrm{H}_2 \mathrm{O}\left(9, \mathrm{H} 2 3 \mathrm{~K}ight) \\ & \mathrm{C} \mathrm{v}, \mathrm{m}\left(\mathrm{H}_2 \mathrm{O}, \mathrm{\ell}ight)=75.0 \mathrm{J~k}^{-1} \mathrm{~mol}^{-1} \\ & \mathrm{C} \mathrm{p}, \mathrm{m}\left(\mathrm{H}_2 \mathrm{O}, \mathrm{g}ight)=33.31 \mathrm{H}~ \mathrm{K}~ \mathrm{K}^{-1} \mathrm{~mol}^{-1} \\ & \Delta \mathrm{H} \text { vap at } 373 \mathrm{~K}=40.7 \mathrm{~kJ} / \mathrm{~mol}^{-1}\end{aligned}\)
The steps involved in the above conversion is
\(\mathrm{H}_2 \mathrm{O}(\ell, 323 \mathrm{~K}) \underset{\Delta \mathrm{U}_1}{\longrightarrow} \mathrm{H}_2 \mathrm{O}(\ell, 373 \mathrm{~K}) \underset{\Delta \mathrm{U}_2}{\longrightarrow} \mathrm{H}_2 \mathrm{O}(9,373 \mathrm{~K}) \underset{\Delta \mathrm{U}_3}{\longrightarrow} \mathrm{H}_2 \mathrm{O}(9, \mathrm{H} 23 \mathrm{k})\)
\(\begin{aligned} & \Delta U_{\text {total }}=\Delta U_1+\Delta U_2+\Delta U_3 \\ & \Delta U_1=\mathrm{C} \mathrm{v}, \mathrm{m} \left(H_2 O, \ellight) \times \Delta T=75 \times 50 \times 10^{-3} \mathrm{~kJ} / \mathrm{mol} \\ & \Delta U_2=\Delta H-\Delta D R T=40.7-1 \times 8.314 \times 373 \times 10^{-3} \\ & =37.598 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta U_3=\left(v, m\left(H_2 \mathrm{O}, gight) \Delta T=\left(C p, m\left(\mathrm{H}_2 \mathrm{O}, gight)-\mathrm{R}ight) \Delta Tight. \\ & =(33.314-8.314)(423-373) \times 10^{-3} \\ & =25 \times 50 \times 10^{-3} \mathrm{~kJ} / \mathrm{mol} \\ & \Delta U=75 \times 50 \times 10^{-3}+37.598+25 \times 50 \times 10^{-3} \mathrm{~kJ} \mid \mathrm{mol} \\ & \Delta U=42.598 \mathrm{~kJ} / \mathrm{mol}=42.6 \mathrm{~kJ} / \mathrm{mol} \\ & \end{aligned}\)

Asked in: JEE-TOPICTESTS-CHEMISTRY

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