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. `9.92xx10^(5)Pa`
B. `11.094xx10^(8)Pa`
C. `10.952xx10^(7)Pa`
D. `9.92xx10^(6)Pa`

Answer

B

Explanation

`DeltaG=-PDeltaV=` Work done
`DeltaV` is the change in molar volume in the conversion of graphite to diamond.
`DeltaV=(12/3.31-12/2.25)xx10^(-3)L=-1.708xx10^(-3)L`
Work done `=-(1.708xx10^(-3))xxPxx101.3J`
`1895J"mol"^(-1)=-(-1.708xx10^(-3))xxPxx101.3J`
`:.P=(1895J"mol"^(-1))/(1.708xx10^(-3)xx101.3)=10.952xx10^(3)` atm
1 atm `=10^(5)xx1.013` Pa
`impliesP=10.952xx10^(3)xx1.013xx10^(5)Pa`
`=11.094xx10^(8)Pa`