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Thermodynamics
Gibbs Energy and Pressure-Volume Work
1 question from 1 paper, each with its answer.
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Concepts tested
- Pressure needed to convert graphite to diamond
All Questions
2003 AIPMT 1 question
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The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm$^{-3}$, respectively. If the standard free energy difference ($\Delta G^\circ$) is equal to 1895 J mol$^{-1}$, the pressure at which graphite will be transformed into diamond at 298 K is -Graphite changes into diamond when the work done by the applied pressure on the decrease in volume supplies the free energy difference: $\Delta G^\circ = P\,|\Delta V|$.
Molar volume of graphite $= \dfrac{12}{2.25} = 5.33$ cm$^3$ mol$^{-1}$
Molar volume of diamond $= \dfrac{12}{3.31} = 3.63$ cm$^3$ mol$^{-1}$
$|\Delta V| = 5.33 - 3.63 = 1.71$ cm$^3$ mol$^{-1}$ $= 1.71 \times 10^{-6}$ m$^3$ mol$^{-1}$
$P = \dfrac{\Delta G^\circ}{|\Delta V|} = \dfrac{1895}{1.71 \times 10^{-6}} \approx 1.1 \times 10^9$ Pa
This is of the order of $10^9$ Pa, i.e. about $10^4$ atmospheres. Among the options, the value of this order is $9.92 \times 10^8$ Pa; the other options are 10, 100 or 1000 times too small.
So graphite will be transformed into diamond at about $9.92 \times 10^8$ Pa.