The initial state consists of a closed system containing one mole of super-heated liquid water at 115°C and 1 atmosphere pressure. The immediate surroundings are also at 115°C and 1 atmosphere pressure.
The change in state is described as this one mole of super-heated liquid water vaporizing to water vapor at 115°C, as shown by the following reaction at 1 atmosphere pressure.
1 mole, H2O, liquid, 115°C ---> 1 mole, H2O, vapor, 115°C
a) Calculate the Enthalpy change for the system at 115°C.
b) Calculate the Entropy change for this system at 115°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
I really have no idea what to do or where to start. I know I need to break the problem down into steps but I'm not sure exactly how to split it up.
To solve this problem, we need to apply the laws and principles of thermodynamics. Let's break it down into steps:
Step 1: Determine the enthalpy change (∆H) for the system at 115°C.
To calculate the enthalpy change, we need to consider the energy required for the phase change from liquid to vapor. The enthalpy change for vaporization (∆Hvaporization) at a given temperature is given as 40790 J/mole.
Step 2: Calculate the entropy change (∆S) for the system at 115°C.
Entropy change is related to the heat transfer in a system. At constant temperature and pressure, the entropy change (∆S) can be calculated using the equation: ∆S = ∆H/T, where ∆H is the enthalpy change and T is the temperature in Kelvin (K).
Step 3: Calculate the enthalpy change for the surroundings.
Since the surroundings are assumed to behave as a constant temperature reservoir, the enthalpy change (∆Hsurroundings) is equal to the heat gained or lost by the surroundings. In this case, the enthalpy change of the surroundings will be equal in magnitude but opposite in sign to the enthalpy change of the system (∆H).
Step 4: Calculate the entropy change for the surroundings.
Using the same equation as in Step 2 (∆S = ∆Hsurroundings/T), we can calculate the entropy change for the surroundings (∆Ssurroundings).
Step 5: Calculate the entropy change for the universe.
To calculate the total entropy change for the universe, we add the absolute values of ∆S for the system and the surroundings (∆Suniverse = |∆Ssystem| + |∆Ssurroundings|).
Now let's apply these steps and calculate the values:
a) Enthalpy change for the system (∆Hsystem):
∆Hsystem = ∆Hvaporization = 40790 J/mole
b) Entropy change for the system (∆Ssystem):
∆Ssystem = ∆Hsystem / T = 40790 J/mole / 115°C + 273.15 = 40790 J/mole / 388.15 K
c) Enthalpy change for the surroundings (∆Hsurroundings):
∆Hsurroundings = -∆Hsystem = -40790 J/mole
d) Entropy change for the surroundings (∆Ssurroundings):
∆Ssurroundings = ∆Hsurroundings / T = -40790 J/mole / 388.15 K
e) Entropy change for the universe (∆Suniverse):
∆Suniverse = |∆Ssystem| + |∆Ssurroundings|
Finally, plug in the values to calculate ∆Ssystem, ∆Hsurroundings, ∆Ssurroundings, and ∆Suniverse.