Question about change in internal energy

In summary: ENDED:In summary, the system undergoes vaporization, releasing energy to the environment. The resultant liquid has less internal energy than the original vapor even though the temperature remains the same. q decreases as the molecules crowd together and ΔH-vaporization of H2O @ 100 = 40.7 kJ/K*mol.
  • #1
modx07
2
0
Fair warning: This is more chemistry than physics but I know that thermodynamics and heat and work are all covered in General Physics but I'll post in the Other Science forum too.

1. Assume that one mole of H2O(g) condenses to H2O(l) at 1.00atm and 95 Celcius. Calculate q, w, ΔH, ΔS of the system, ΔS of surroundings. BUT I AM NOT ASKING HOW TO CALCULATE THESE VALUES, SEE LAST SENTENCE OF POST.

Homework Equations


q = nCΔT
ΔH = n(Cp)ΔT
W = -PΔV
ΔS = [q_reversible] / T

Cp of H2O(l) = 75.3 J / K*mol
Cp of H2O(g) = 36.4 J / K*mol
ΔH-vaporization of H2O @ 100 = 40.7 kJ / K*mol

The Attempt at a Solution



Basically, I have the solution but I don't understand why it's the case. When I asked my TA, he said that the condensation is actually occurring at 95 Celcius, which confused me right off the bat since I knew that at 1atm, the condensation point should be 100 degrees.

Regardless, I thought that the step that was occurring should be simply:

H2O (g) -> H2O (l) @ 95 degrees.

Thus, I thought that since the system stays at the same temperature, then ΔE = 0 and that q = -w. However, this assumption is wrong. I am simply asking why this is wrong (and not to calculate the values).
 
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  • #2
When water vapor condenses the vapor releases energy to the environment. The resultant liquid has less internal energy than the original vapor even though the temperature remains the same. As the molecules crowd together when forming the liquid state, the potential energy of interaction between the molecules decreases.
 
  • #3
modx07 said:
Regardless, I thought that the step that was occurring should be simply:

H2O (g) -> H2O (l) @ 95 degrees.

Thus, I thought that since the system stays at the same temperature, then ΔE = 0 and that q = -w. However, this assumption is wrong. I am simply asking why this is wrong (and not to calculate the values).
You are assuming that ΔE (ΔU) is a function of temperature only. ie. ΔE = nCvΔT. But this is true only for an ideal gas. Water vapour is NOT an ideal gas. Its internal energy is very much a function of temperature and volume/pressure. It takes energy to increase the separation between water molecules.

AM
 

Related to Question about change in internal energy

1. How is internal energy defined?

Internal energy is the sum of the kinetic and potential energies of all the particles within a system. It is a measure of the total energy that is present in a system.

2. What factors can cause a change in internal energy?

A change in internal energy can be caused by several factors, such as changes in temperature, pressure, volume, or the addition or removal of heat or work to or from the system.

3. What is the relationship between internal energy and heat?

Heat is the transfer of energy from one system to another due to a temperature difference. When heat is added to a system, the internal energy of the system increases. Similarly, when heat is removed from a system, the internal energy decreases.

4. How is the change in internal energy calculated?

The change in internal energy (ΔU) of a system can be calculated using the first law of thermodynamics, which states that the change in internal energy is equal to the heat added to the system (Q) minus the work done by the system (W): ΔU = Q - W.

5. Can internal energy be negative?

Yes, internal energy can be negative. This indicates that the system has lost energy, either through the transfer of heat or the performance of work. However, the total change in internal energy (ΔU) must always be positive, as energy cannot be created or destroyed, only transferred or converted.

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