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What is the value of change in internal energy at $\pu{1atm}$ in the process $$\ce{H2O(l,$323\ \mathrm K$)->H2O(g,$423\ \mathrm K$)}$$ given $C_{\mathrm m,V}(\ce{H2O,l})=\pu{75.0 J K^{-1} mol^{-1}}$ and $C_{\mathrm m,p}(\ce{H2O,g})=\pu{33.314 J K^{-1} mol^{-1}}$

a. $\pu{42.91 kJ/mol}$
b. $\pu{43086 kJ/mol}$
c. $\pu{42.6 kJ/mol}$
d. $\pu{49.6 kJ/mol}$

I tried to use the First Law here. For chemical reaction I found $\Delta n_\mathrm g$ to be $1$ and the heat of the reaction $\Delta H$ is given as $\pu{40.7 kJ mol-1}$. However, I am not able to use the specific heat data given in the question. What am I doing wrong here?

I am getting stuck in phase change reactions, particularly in using $C_p$ and $C_V$ here. I know the First Law will work, but how do we use $C_p$ and $C_V$ in such reactions?

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  • $\begingroup$ @GENESECT I got you are started by replacing the image. That should help to format the other parts of the post. Also, option d was cut off in your image. $\endgroup$
    – Tyberius ♦
    Aug 17, 2018 at 17:12
  • $\begingroup$ It seems like you might have to use the density of water and the ideal gas law to approximate the change in volume as water goes from the liquid to vapor state. $\endgroup$
    – a-cyclohexane-molecule
    Aug 17, 2018 at 19:14

1 Answer 1

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The easiest way to do this problem is to first determine the change in enthalpy for this constant pressure process. Do you think you can do that? Then determine the change in PV between the initial and final states.

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  • $\begingroup$ But why is heat of the reaction equal to change in enthslpy here... I don't understand the logic $\endgroup$
    – Aladdin
    Aug 17, 2018 at 23:32
  • $\begingroup$ Are you talking about the overall change, or the enthalpy change for vaporization? $\endgroup$
    – Chet Miller
    Aug 17, 2018 at 23:35
  • $\begingroup$ You are aware that, for a constant pressure process, Q is equal to Delta H, right? $\endgroup$
    – Chet Miller
    Aug 17, 2018 at 23:44
  • $\begingroup$ I got that. It's a constant pressure process so we can use ncp∆T.Will this be overall heat $\endgroup$
    – Aladdin
    Aug 17, 2018 at 23:45
  • $\begingroup$ No. Heat is being added during vaporization at constant temperature and pressure, and this portion of the heat is equal to the enthalpy change for that step in the process. $\endgroup$
    – Chet Miller
    Aug 17, 2018 at 23:52

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