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Q.

At 5×105 bar pressure, density of diamond and graphite are 3 glee and 2 glee respectively, at certain temperature T. Find the value of ΔUΔH for the conversion of 1 mole of graphite to 1 mole of diamond at temperature T:

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a

- 100 kJ/mol

b

100 kJ/mol

c

50 kJ/mol

d

None of these

answer is A.

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Detailed Solution

Graphite  Diamond

1 mole =12gV=6cc V=4ccΔH=ΔU+Δ(PV)ΔUΔH=Δ(PV)=5×105×100×1031000×[46]×106=+100kJ

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