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The densities of graphite and diamond at \[298K\]are 2.25 and $3.31gc{{m}^{-3}}$ respectively. If the standard free energy difference is$1895Jmo{{l}^{-1}}$, the pressure at which graphite will be transformed into diamond is:
A. $9.92\times {{10}^{8}}Pa$
B. $9.92\times {{10}^{7}}Pa$
C. $9.92\times {{10}^{6}}Pa$
D. None of these

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Last updated date: 13th Jun 2024
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Answer
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Hint: As we know that both graphite and diamond are all allotropes of carbon. When the graphite is converted into the diamond and thermal reaction takes place. When the graphite is converted into diamond then high temperature and high pressure is required.

Formula used:
$\Delta G=-P\Delta V$
Where $\Delta G$- Standard free energy
P - Pressure
$\Delta V$ -Volume

Complete step by step answer:
Given that density of diamond =$3.31gc{{m}^{-3}}$
Density of graphite = $2.25gc{{m}^{-3}}$
Standard free energy difference =$1895Jmo{{l}^{-1}}$
Energy = $\Delta G$
P =?
As we know that Density = $\dfrac{Mass}{Volume}$
Mass of graphite & diamond = 12
(Because both are allotropes of carbon)
Volume of diamond = $\dfrac{Mass\,of\,carbon}{Density\,of\,diamond}$
${{V} {d}} $ $=\dfrac{12}{3.31}\times {{10} ^ {-3}} L$
$=3.6253\times {{10} ^ {-3}} L$
Volume of graphite = $\dfrac{Mass\,of\,carbon}{Density\,of\,graphite}$
${{V} {g}} $ $=\dfrac{12}{2.25}\times {{10} ^ {-3}} L$
$=5.33\times {{10} ^ {-3}} L$
$\Delta V= {{V} {d}}-{{V} {g}} $
$=(3.625-5.33)\times {{10}^{-3}}L$
$=-1.71\times {{10} ^ {-3}} L$
Now $\Delta G=-P\Delta V$
$1895=-P (-1.71\times {{10} ^ {-3}})$ $(1atm=101.3Pa)$
$ 1895=P\times 101.3\times 1.7\times {{10} ^ {-3}} \\
P=\dfrac{1895}{101.3\times 1.7\times {{10} ^ {-3}}} = 10.93\times {{10} ^ {2}}atm \\
\,\,\,\,\,=11.07\times {{10} ^ {8}}Pa \\
$

So, the correct answer is “Option D”.

Note:
Convert volume into litres from centimetres to use it in the density formula. Standard free energy is the energy change during the formation of a substance from its elements in their most stable state.
The expression for standard free energy must be known.
The knowledge of how the unit Pascal is converted to atm & atm is converted to Pascal must be remembered.