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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)
`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"`
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