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Calculate the values of ΔU, ΔH, and ΔS for the following process:
1 mole of liquid water at 25 C and 1 atm ➡️ 1 mole of steam at 100 C and 1 atm
The molar heat of vaporization of water at 373 K is 40.79 kJ mol-1, and molar heat capacity of water is 75.3 J K-1 mol-1. Assume molar heat capacity to be temperature independent and ideal-gas behavior.

Respuesta :

ogorwyne

The values of the changes are

ΔH = 46.44kJ

Δu =  43.34Kj

ΔS = 126.6 J/K

How to find change in H

= 1 mol + 75.3 (100 + 273) - 25 + 273 + 1 * 40.79

= 5.6475 + 40.79

= 46.44kJ

How to find change in S

1 mol + 75.3 x ln 373/298 + 1 mol x 40.79

= 0.1263

ΔS = 126.6 J/K

How to find the change in U

46.44 - 0.00831 * 373

= 43.34Kj

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