The standard state Gibbs free energies of formation of ) C(graphite and C(diamond) at T = 298 K are
`Delta_(f)G^(@)["C(graphite")]=0kJ mol^(-1)`
`Delta_(f)G^(@)["C(diamond")]=2.9kJ mol^(-1)`
The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [ ) C(graphite ] to diamond [C(diamond)] reduces its volume by `2xx10^(-6)m^(3) mol^(-1).` If ) C(graphite is converted to C(diamond) isothermally at T = 298 K, the pressure at which ) C(graphite is in equilibrium with C(diamond), is
`["Useful information:"1J=1kg m^(2)s^(-2),1Pa=1kgm^(-1)s^(-2),1"bar"=10^(5)Pa]`

Options

A. `58001` bar
B. `1450` bar
C. `14501` bar
D. `29001` bar

Answer

C

Explanation

`DeltaG=VDeltap-SDeltaT,AsDeltaT=0,DeltaG=VDeltap`
`:.G_(gr)-G_(gr)^(@)=V_(gr)DeltaP "and"G_(dim)-G_(dim)^(@)=V_(dim)DeltaP`
`:.(V_(dim)-V_(gr))DeltaP=(G_(dim)-G_(gr))+(G_(gr)^(@)-G_(dim)^(@))`
at eqb. `G_(dim)=G_(gr)`
`:.(V_(dim)-V_(gr))DeltaP=G_(dim)^(@)-G_(gr)^(@)=2.9xx10^(3)`
Calculate `Deltap,i.e.,p-p_(0)`,"then p"i.e.,(p_(0)=1).`