Thermodynamics and Thermochemistry - Result Question 19
####19. Compute the heat of formation of liquid methyl alcohol in $kJ mol^{-1}$, using the following data. Heat of vaporisation of liquid methyl alcohol $=38 kJ / mol$. Heat of formation of gaseous atoms from the elements in their standard states :
$H=218 kJ / mol, C=715 kJ / mol, O=249 kJ / mol$.
Average bond energies:
$(1997,5 M)$
$C-H=415 kJ / mol, C-O=356 kJ / mol$,
$O-H=463 kJ / mol$
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Solution:
- For isothermal reversible expansion,
$$ |W|=n R T \ln \frac{V _f}{V _i}=n R T \ln \frac{V}{V _i} $$
where, $V=$ final volume, $V _i=$ initial final.
or $|W|=n R T \ln V-n R T \ln V _i$
On comparing with equation of straight line, $y=m x+c$, we get
$$ \begin{aligned} \text { slope } & =m=+n R T \ \text { intercept } & =-n R T \ln V _i \end{aligned} $$
Thus, plot of $|W|$ with $\ln V$ will give straight line in which slope of $2\left(T _2\right)$ is greater than slope of $1\left(T _1\right)$ which is given in all options.
Now, if $V _i<1$ then $y$ intercept $\left(-n R T V _i\right)$ becomes positive and if it is positive for one case then it is positive for other case also. Thus, it is not possible that one $y$-intercept goes above and other $y$-intercept goes below. Thus, option (b) and (d) are incorrect.
If we extent plot given in option (a) it seems to be merging which is not possible because if they are merging they give same + ve $y$-intercept. But they cannot give same $y$-intercept because value of $T$ is different.
Now, if we extent the line of $T _1$ and $T _2$ given in option (c) it seems to be touching the origin. If they touch the origin then $y$-intercept becomes zero which is not possible. Thus, it is not the exactly correct answer but among the given options it is the most appropriate one.
