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?

  • $\begingroup$ @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 '18 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. $\endgroup$ Aug 17 '18 at 19:14

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.

  • $\begingroup$ But why is heat of the reaction equal to change in enthslpy here... I don't understand the logic $\endgroup$
    – Aladdin
    Aug 17 '18 at 23:32
  • $\begingroup$ Are you talking about the overall change, or the enthalpy change for vaporization? $\endgroup$ Aug 17 '18 at 23:35
  • $\begingroup$ You are aware that, for a constant pressure process, Q is equal to Delta H, right? $\endgroup$ Aug 17 '18 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 '18 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. $\endgroup$ Aug 17 '18 at 23:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.