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

Answer

A

Explanation

`C("graphite") rarr C("diamond")`
`Delta H = Delta U + PDeltaV implies DeltaU - DeltaH = -PDeltaV implies -5 xx 10^(10) xx (V_("diamond") - V_("graphite")) [1 bar = 10^(5)N//m^2]`
`=-5 xx 10^(10) (N)/(m^2) xx (12/3 - 12/2) xx 10^(-6) m^(3)` [For 1 mole) = 100 kJ/mol.