Question

At $5\times {{10}^{4}}$bar pressure density of diamond and graphite are $3g/cc$ and $2g/cc$ respectively, at a certain temperature ‘T’. Find the value of $\mathbf{\Delta U}\text{ }\text{ }\mathbf{\Delta H}$ for the conversion of 1 mole of graphite to 1 mole of diamond at temperature ‘T’.
A. 100kJ/mol
B. 50kJ/mol
C. -100kJ/mol
D. none of these

Answer 1

Use the first law of thermodynamics which states that the change in internal energy of a particular system is equal to the difference between the heat added and the work done by the system. Use them for graphite and diamond to obtain your answer.

Complete answer:
Let us first look into the values we have been provided with, in the question. We have
$\mathbf{\Delta H}$= which represents the heat added to the system
$\mathbf{\Delta U}$= which represents the change in internal energy of the system
$\mathbf{\Delta }\left( \mathbf{PV} \right)$= which represents work done by the system, as work is product of pressure and volume.
Hence, using first law of thermodynamics, we can say that
$\Delta U-\text{ }~\Delta H=-\text{ }~\Delta \left( PV \right)$
or, $\Delta U-\Delta H=-P\left( {{V}_{d}}-{{V}_{g}} \right)$, where ${{V}_{d}}$ and ${{V}_{g}}$ are volumes of diamond and graphite respectively.
or, $\Delta U-\Delta H=\text{ }-5\times {{10}^{4}}\times {{10}^{5}}(\left( {{V}_{d}}-{{V}_{g}} \right)$………( I ) , we have converted bar to pascal also.
Now we will find the values of ${{V}_{d}}$ and ${{V}_{g}}$for diamond and graphite respectively.
We know,
$Volume=mass\div density$
Now , as one mole is given, so we can assume the mass of one mole of carbon i.e 12g. Therefore we obtain the mass as 12g and with this we will calculate the volumes.

{{V}_{d}}=12\div 3=4cc \\\ {{V}_{g}}=12\div 2=6cc \\\ \end{array}$$ Now using ( I ), we have $$~\Delta U-\text{ }~\Delta H=-5\times {{10}^{4}}\times {{10}^{5}}\times \left( 4-6 \right)\times {{10}^{-6}}^{{}}$$, we have converted cc into $${{m}^{3}}$$ which is the SI unit. On solving the calculation, we get the answer as $$+100kJ/mol.$$ That means $$+100kJ/mol.$$ is required to convert 1 mole of graphite to 1 mole of diamond at temperature ‘T’. **Note:** While noting down the values from the question, try to check for the SI system of units and change them immediately, if they are not in the SI system. This will make the calculation easier. Also recheck the calculation as there might be errors and look for the sign also.