The density of ice at 0°C is 0.915 g/cc and that of liquid water at 0°C is 0.99987 g/cc. The work done for melting 1 mole of ice at 1 .0 0 bar (assuming work is done only due to expansion) is approximately

Question

The density of ice at 0°C is 0.915 g/cc and that of liquid water at 0°C is 0.99987 g/cc. The work done for melting 1 mole of ice at 1 .0 0 bar (assuming work is done only due to expansion) is approximately (A) 0.17 J (B) 1.7 × 103 J (C) 1.7 × 10-6 J (D) can’t be determined

Tardigrade Admin · Accepted Answer

W=-PΔ V=105(N/m2)[(18×10-6/0.99987)m3-(18×10-6/0.915)m3] =+105×1.67×10-6J m=0.167J∼0.17J

Q. The density of ice at $0^{\circ} C$ is $0.915 g / cc$ and that of liquid water at $0^{\circ} C$ is $0.99987 g / cc$ . The work done for melting $1$ mole of ice at $1.00$ bar (assuming work is done only due to expansion) is approximately

$0.17 J$

$1.7 \times 1 0^{3} J$

$1.7 \times 1 0^{- 6} J$

can’t be determined

$W = - P Δ V = 1 0^{5} \frac{N}{m ^{2}} [\frac{18 \times 1 0 ^{- 6}}{0.99987} m^{3} - \frac{18 \times 1 0 ^{- 6}}{0.915} m^{3}]$
$= + 1 0^{5} \times 1.67 \times 1 0^{- 6} J m = 0.167 J \sim 0.17 J$

Q. The density of ice at 0∘C is 0.915g/cc and that of liquid water at 0∘C is 0.99987g/cc. The work done for melting 1 mole of ice at 1.00 bar (assuming work is done only due to expansion) is approximately

0.17J

1.7×103J

1.7×10−6J