At `5 xx 10^(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. `100kJ // mol`
B. `50kJ//mol`
C. `-100kJ//mol`
D. none of these

Answer

A

Explanation

`C("graphite") rarr C("diamond")`
`DeltaH=DeltaU+P_(2).DeltaV`
`V_(m)("diamond")=(12)/(3)mL`
`v_(m)("graphite")=(12)/(2)mL`
`DeltaH-DeltaU=(500xx10^(3)N//m^(2))((12)/(3)-(12)/(2))xx10^(-6)`
`=100kJ//mol^(-1)`
`DeltaU-DeltaH=+100kJ//mol`