The molar volumes of ice and water are 0.0196 and 0.0180 litres per mole at 273K. If ΔH for the transition of ice to water is 1440 calories per mole at 1 atm pressure, find ΔU.

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Vikas TU · Answer

From eqn. ΔU + Δ(PV) ΔH = = ΔU + ΔPV + PΔV Given, 1440 = ΔU + 0.0196*273 + 0.0180*273 ΔU = 1440 – 10.2648 = 1429.73 calories.

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The molar volumes of ice and water are 0.0196 and 0.0180 litres per mole at 273K. If ΔH for the transition of ice to water is 1440 calories per mole at 1 atm pressure, find ΔU.

The molar volumes of ice and water are 0.0196 and 0.0180 litres per mole at 273K. If ΔH for the transition of ice to water is 1440 calories per mole at 1 atm pressure, find ΔU.

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The molar volumes of ice and water are 0.0196 and 0.0180 litres per mole at 273K. If ΔH for the transition of ice to water is 1440 calories per mole at 1 atm pressure, find ΔU.

The molar volumes of ice and water are 0.0196 and 0.0180 litres per mole at 273K. If ΔH for the transition of ice to water is 1440 calories per mole at 1 atm pressure, find ΔU.

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