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At 500 kbar pressure density of diamond...

At 500 kbar pressure density of diamond and graphite are `3 "g"//"cc" and 2 "g"//"cc"` respectively, at certain temperature T. Find the value of |DeltaH-DeltaU|`("in kJ"//"mole")`for the conversion of 1 mole of graphite 1 mole of diamond at 500 kbar pressure.
(Given : `1 "bar" = 10^(5) "N"//"m"^(2))`

Text Solution

Verified by Experts

[100]
C(graphite) ` to ` C (diamond)
`DeltaH=DeltaU+P_(2)V_(2)-P_(1)V_(1)`
`DeltaH-DeltaU=(500xx10^(3)xx10^(5)N//m^(2)((12)/(2)-(12)/(3))xx10^(-6)`
`=500xx2xx10^(3)xx10^(5)xx10^(-6)=10^(5) J mol`
=100 kJ/mol
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