Chemistry - Thermodynamics - II Question with Solution | TestHub

Differential equation of free energy dG = VdP - SdT at constant temp. with change in allotropic modification there is a change in molar volume$\therefore$$\text{d}\left(\Delta \text{G}\right)=\left(\Delta \text{V}\right)\text{dP}$.

$\Delta \text{V}=\frac{{\text{M}}_{\text{Diamond}}}{{\rho }_{\text{Diamond}}}-\frac{{\text{M}}_{\text{graphite}}}{{\rho }_{\text{graphite}}}=\frac{12}{3}-\frac{12}{2}$

$\Delta \text{V}=-2{\text{cm}}^3/\text{mole}=-2\times 10^{-6}{\text{m}}^3{\text{mole}}^{-1}$

$\underset{\Delta {\text{G}}_{\text{f}}^{\text{o}}\text{at }298\text{K, 1 atm}}{\overset{\Delta {\text{G}}_{\text{f}}^{\text{o}}\text{at }298\text{K, p atm}=0}{\int \text{d}\left(\Delta \text{G}\right)}}=-2\times 10^{-6}{\text{m}}^3{\text{mole}}^{-1}\underset{\text{P}=\text{1 atm}}{\overset{\text{P }=\text{ p atm}}{\int \text{dP}}}$

Because at equilibrium when diamonds are formed both the phases are in equilibrium$\Delta {\text{G}}_{\text{f}\left(\text{Diamonds}\right)}^{\text{o}}\text{at }298\text{K ,}\text{ P atm}=0$

$\underset{3000{\text{J mole}}^{-1}}{\stackrel{0}{\int }\text{d}\left(\Delta \text{G}\right)}=-2\times 10^{-6}{\text{m}}^3{\text{mole}}^{-1}\times 10^5{\text{Nm}}^{-2}\underset{\text{P }=\text{ 1 atm}}{\overset{\text{P }=\text{ p atm}}{\int \text{dP}}}$

( 1 atm = 10⁵Nm^{- 2})

0 - 3000 Joules mole^{- 1}= - 2 x 10^{- 1}J mole^{- 1}(P - 1)

$\frac{3000}{0\text{.}2}=15000=\left(\text{P}-1\right)$

$\therefore$P = 15001 atm.