ThermodynamicsHard
Question
At 500 kbar and T K, the densities of graphite and diamond are 2.0 and 3.0 g/cm3, respectively. The value of (ΔH − ΔU) for the conversion of 1 mole of graphite into diamond at 500 k bar and T K is
Options
A.100 kJ
B.−100 kJ
C.1000 kJ
D.−1000 kJ
Solution
$\Delta H - \Delta U = P.\Delta V = P\left( V_{\text{diamond}} - V_{\text{graphite}} \right)$
$= 500 \times 10^{3} \times 10^{5}\frac{N}{m^{2}}\left( \frac{12}{3.0} - \frac{12}{2.0} \right) \times 10^{- 6}m^{3} = - 100\text{ kJ}$
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