
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
577.5k+ views
Hint: 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.
\[\begin{array}{*{35}{l}}
{{V}_{d}}=12\div 3=4cc \\
{{V}_{g}}=12\div 2=6cc \\
\end{array}\]
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.
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.
\[\begin{array}{*{35}{l}}
{{V}_{d}}=12\div 3=4cc \\
{{V}_{g}}=12\div 2=6cc \\
\end{array}\]
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.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

