At `5xx10^(5)` bar pressure density of diamond and graphite are `3 g//c c` and `2g//c c` respectively, at certain temperature `'T'`.Find the value of `DeltaU-DeltaH` for the conversion of 1 mole of graphite to 1 mole of diamond at temperature `'T' :`

Options

A. 100 kJ/mol
B. 50 kJ/mol
C. `-100 kJ//mol`
D. None of these

Answer

A

Explanation

C (graphite) `rarr` C (diamond)
`DeltaH=DeltaU+P.DeltaV`
`V_(m)("diamond")=(12)/(3)mL`
`V_(m)("graphite")=(12)/(2)mL`
`DeltaH-DeltaU=(500xx10^(3)xx10^(5)N//m^(2)) ((12)/(3)-(12)/(2))xx10^(-6)`
`=-100 " kJ" //"mol"`
`DeltaU-DeltaH=+100 " kJ"//"mol"`