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^(5)` Pa
B. `1.1 xx 10^(9)` Pa
C. `9.92 xx 10^(7)` Pa
D. `9.92 xx 10^(6)` Pa

Answer

B

Explanation

Gibb.s free energy is the measure of useful work done at constant temperature and pressure. `Delta G^(@)= - P Delta V`
Volume difference `(Delta V)` for 12g of carbon (1 mole)
`=(-12g)/(2.25 g//cm^(3))+ (12g)/(3.31g//cm^(3))`
`Delta V= -1.7 cm^(3)= -1.7 xx 10^(-6) m^(3)`/mole
`P= (-Delta G)/(Delta V)= (1895 "J/mole")/(1.7 xx 10^(-6) m^(3)//"mole")= 1.1 xx 10^(9)` Pa