At 5 × 105 bar pressure, density of diamond and graphite are 3 g/cc and 2 g/cc respectively, at certain temperature T. Find the value of \left(\frac{ΔU - ΔH}{20}\right) for the conversion of 1 mole of graphite to 1 mole of diamond (in KJ) at temperature T. (1L.atm = 100 J)

Engineering

Chemistry

Enthalpy and Internal Energy Relation

Pressure Volume and Temperature

Ideal Gas Equation

Question

At 5 × 10⁵ bar pressure, density of diamond and graphite are 3 g/cc and 2 g/cc respectively, at certain temperature T. Find the value of $(\frac{ΔU - ΔH}{20})$ for the conversion of 1 mole of graphite to 1 mole of diamond (in KJ) at temperature T. (1L.atm = 100 J)

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C_(graphite) → C_(diamond)

1 mole 1 mole

ρ = 2 g/cm³ ρ = 3 g/cm³

$V = \frac{12}{2} ml V = \frac{12}{3} ml$

ΔH = ΔU + PΔV

ΔU – ΔH = – PΔV

$= - 5 \times 10^{5} (\frac{12}{3} - \frac{12}{2}) \times 10^{- 3} L . atm = - 5 \times 10^{5} \times \frac{12}{6} \times 10^{- 1} J = 100 KJ / mole = 10 \times 10^{4} J = 100 KJ / mole$

$\frac{ΔU - ΔH}{20} = \frac{100}{20} = 5$