The densities of graphite and diamond at 298 K are 3.31 `g//cm^(3)`, respectively. If standard free entropy difference `(DeltaG^(@))` is equal to 1895 `J"mol"^(-1)`, the pressure at which graphite will be transformed into diamond at 298 K is

Options

A. `9.92 xx 10^(7)` Pa
B. `9.92 xx 10^(6)` Pa
C. `9.92 xx 10^(5)` Pa
D. `11.12 xx 10^(8)` Pa

Answer

D

Explanation

Molar volume of graphite `=12/(2.25) = 5.33 cm^(3)`
Molar volume of diamond `=12/(3.31) = 3.625 cm^(3)`
`therefore` Change in molar volume in the conversion of graphite to diamond
`=3.625 - 5.33 = -1.704 cm^(3) = -1.704 xx 10^(-3)` litre.
Work done during conversion of graphite into diamond `=-PDeltaV`
`=-P xx (-1.704 xx 10^(-3))` lit. atm
`=P xx 1.704 xx 10^(-3) xx 101.3` Joules.
Standard free energy difference `(DeltaG^(@))` is a measure of work done.
`therefore 1895 J = P xx 1.704 xx 10^(-3) xx 101.3` J
`therefore P=10.97` atm
`=10.97 atm xx 1.013 xx 10^(5) Pa = 11.12 xx 10^(8)` Pa
Hence, (D) is the correct answer.