The initial state consists of a closed system containing one mole of water (liquid) at 85°C and 1 atmosphere pressure. The immediate surroundings are also at 85°C and 1 atmosphere pressure.
The change in state is described as this one mole of liquid forming vapor at 85°C, as shown by the following reaction at 1 atmosphere pressure.
1 mole, H2O, liquid, 85°C ---> 1 mole, H2O, vapor, 85°C
a) Calculate the Enthalpy change for the system at 85°C.
b) Calculate the Entropy change for this system at 85°C.
c) Assuming that the surroundings behave as a constant temperature reservoir,
calculate the Enthalpy change for the surroundings
d) Assuming that the surroundings behave as a constant temperature reservoir,
calculate the Entropy change for the surroundings
e) Calculate the Entropy change for the universe.
Given the following:
Cp (liquid) = 75.33 J/mole K,
Cp (vapor) = 37.47 J/mole K,
∆H vaporization (100°C) = 40790 J/mole
To calculate the enthalpy change for the system at 85°C (a), we need to consider the enthalpy change of vaporization. The enthalpy change of vaporization (∆Hvap) is given as 40790 J/mol. However, this value is given at 100°C, so we need to account for the change in temperature from 100°C to 85°C.
First, we calculate the energy required to raise the temperature of 1 mole of liquid water from 85°C to 100°C. The heat capacity of liquid water (Cp liquid) is given as 75.33 J/mol K. The temperature difference is 100°C - 85°C = 15°C.
Energy required = Cp (liquid) * ∆T
Energy required = 75.33 J/mol K * 15 K
Energy required = 1129.95 J/mol
Next, we calculate the enthalpy change for vaporization at 85°C by subtracting the energy required to raise the temperature from 85°C to 100°C from the enthalpy change of vaporization at 100°C.
Enthalpy change for vaporization at 85°C = ∆Hvap (100°C) - Energy required
Enthalpy change for vaporization at 85°C = 40790 J/mol - 1129.95 J/mol
Enthalpy change for vaporization at 85°C = 39660.05 J/mol
Therefore, the enthalpy change for the system at 85°C is 39660.05 J/mol.
To calculate the entropy change for the system at 85°C (b), we can use the formula:
∆S = ∆H / T
Where ∆S is the entropy change, ∆H is the enthalpy change, and T is the temperature in Kelvin.
∆S = (39660.05 J/mol) / (85 + 273.15) K
∆S = 39660.05 J/mol / 358.15 K
Therefore, the entropy change for the system at 85°C is 110.71 J/mol K.
Now, let's move on to calculating the enthalpy change for the surroundings when they behave as a constant temperature reservoir (c). Since the temperature of the surroundings remains constant, the enthalpy change for the surroundings (∆Hsurroundings) is equal in magnitude but opposite in sign to the enthalpy change for the system.
∆Hsurroundings = -∆Hsystem
∆Hsurroundings = -39660.05 J/mol
Therefore, the enthalpy change for the surroundings is -39660.05 J/mol.
Similarly, the entropy change for the surroundings when they behave as a constant temperature reservoir (d) is equal in magnitude but opposite in sign to the entropy change for the system.
∆Ssurroundings = -∆Ssystem
∆Ssurroundings = -110.71 J/mol K
Therefore, the entropy change for the surroundings is -110.71 J/mol K.
Finally, to calculate the entropy change for the universe (e), we sum the entropy changes for the system and surroundings.
∆Suniverse = ∆Ssystem + ∆Ssurroundings
∆Suniverse = 110.71 J/mol K + (-110.71 J/mol K)
∆Suniverse = 0 J/mol K
Therefore, the entropy change for the universe is 0 J/mol K.