A) 14501 bar
B) 58001 bar
C) 1450 bar
D) 29001 bar
Correct Answer: A
Solution :
\[{{\operatorname{C}}_{\left( graphite \right)}}\to {{\operatorname{C}}_{\left( diamond \right)}}\left( Isothermally \right)\] |
\[{{\Delta }_{r}}G{}^\circ =\Delta G{{{}^\circ }_{\left( diamond \right)}}-\Delta G{{{}^\circ }_{\left( graphite \right)}}\] |
\[=2.9-0=2.9kJ\,mo{{l}^{-1}}\] |
Gibbs free energy is the maximum useful work, then |
\[-\Delta G={{w}_{ma\,x}}=-P\Delta V\] |
\[-2.9\times {{10}^{3}}=-P\times 2\times {{10}^{-6}}\] |
\[P=\frac{2.9\times {{10}^{3}}}{2\times {{10}^{-6}}}=1.45\times {{10}^{9}}\] |
\[Pa=1.45\times {{10}^{9}}\times {{10}^{-5}}bar\] |
\[=1.45\times {{10}^{4}}bar=14500bar\] |
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