1 mole of ice at 0°C and 4.6 mm Hg pressure is converted to water vapour at a constant temperature and pressure. Find Δ U (in kcal) if the latent heat of fusion of ice is 80 Cal / gm and latent heat o

Question

1 mole of ice at 0°C and 4.6 mm Hg pressure is converted to water vapour at a constant temperature and pressure. Find Δ U (in kcal) if the latent heat of fusion of ice is 80 Cal / gm and latent heat of vaporisation of liquid water at 0°C is 596 Cal per gram and the volume of ice in comparison to that of water (vapour) is neglected.

TestHub · Accepted Answer

Total Enthalpy Change ( Δ H ): Δ H = 18 g / m o l × ( 80 + 596 ) C a l / g = 12168 C a l = 12.168 k c a l Work Done ( P Δ V ): Neglecting solid/liquid volume, P Δ V ≈ n RT = 1 × 1.987 × 1 0 − 3 k c a l / ( m o l . k ) × 273.15 K = 0.543 k c a l Internal Energy Change ( Δ U ): Δ U = Δ H − P Δ V = 12.168 − 0.543 = 11.62 k c a l

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