The densities of graphite and diamond at `298K` are `2.25` and `3.31gcm^(-3)` , respectively. If the standard free energy difference `(DeltaG^(0))` is equal to `1895Jmol^(-1)` , the pressure at which graphite will be transformed into diamond at `298K` is

Options

A. `9.92 xx 10^(6) `pa
B. `9.02 xx 10^(5) ` pa
C. `9.92 xx 10^(8)` pa
D. `9.92 xx 10^(7)` pa

Answer

C

Explanation

Volume of graphite =` ("Mass")/("Density")=2.4 K`
Volume of diamond `=(12)/(3.31)`
Change in volume,
`DeltaV=((12)/(331)-(12)/(225))xx10^(-3)L =-1.91 xx 10^(-3) L`
`DeltaG^(@)`= work done =-`pDeltaV`
`p=-(DeltaG)/(DeltaV)=(1895 J "mol"^(-1))/(1.19 xx 10^(-3) xx 101.3)`
`=9734` atm [ `because ` 1 atm =`10^(5)xx1.013 ` Pa]
`9.92xx 10^(8)` Pa