Answer :
The change in entropy in the system when 39.3 grams of steam at 1 atm condenses to a liquid at the normal boiling point is 237.4 J/K.
The normal boiling point of a substance is the temperature at which its vapor pressure equals the pressure of the surroundings. In the case of water, the normal boiling point is 100.0 °C at a pressure of 1 atm.
The molar enthalpy of vaporization is the amount of energy required to convert one mole of a liquid into a gas at a constant temperature and pressure. For water, this value is 40.67 kJ/mol.
To determine the change in entropy when 39.3 grams of steam at 1 atm condenses to a liquid at the normal boiling point, we can use the equation ΔS = q/T, where ΔS is the change in entropy, q is the heat transferred, and T is the temperature.
In this case, the heat transferred is equal to the molar enthalpy of vaporization multiplied by the number of moles of water condensed, which is equal to the mass of steam divided by the molar mass of water.
First, we need to convert the mass of steam to moles. The molar mass of water is 18.015 g/mol, so 39.3 g of steam is equal to 39.3/18.015 = 2.183 mol of water.
Next, we can calculate the heat transferred using the molar enthalpy of vaporization:
q = ΔHvap × n = 40.67 kJ/mol × 2.183 mol = 88.76 kJ
Finally, we can calculate the change in entropy:
ΔS = q/T = 88.76 kJ / (373.15 K) = 237.4 J/K
Therefore, the change in entropy in the system when 39.3 grams of steam at 1 atm condenses to a liquid at the normal boiling point is 237.4 J/K.
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