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At 5xx10^(4) bar pressure density of dia...

At `5xx10^(4)` 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' :`

A

`100 kJ//mol`

B

`50 kJ//mol`

C

`-100 kJ//mol`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
`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.
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