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

The densities of graphite and diamond at 298 K are 3.31 g//cm^(3), res

1.	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
Answer» `9.92 xx 10^(7)` P.a `9.92 xx 10^(6)` Pa `9.92 xx 10^(5)` Pa `11.12 xx 10^(8)` Pa Solution :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.