Difference between internal E and enthelpy

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Homework Help Overview

The discussion revolves around the concepts of internal energy and enthalpy, particularly their distinctions and relationships in the context of thermodynamics. Participants are exploring how internal energy differs from kinetic and potential energy, as well as the implications of these differences in various systems, including ideal and real gases.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to clarify the definitions and distinctions between internal energy, kinetic energy, and potential energy, particularly in the context of ideal versus real gases. Questions are raised about the implications of exothermic reactions on internal energy and the relationship between internal energy and enthalpy.

Discussion Status

Some participants have provided insights into the nature of internal energy in ideal and real gases, while others have expressed concerns about the accuracy of explanations regarding enthalpy. There is ongoing exploration of the concepts, with no explicit consensus reached yet.

Contextual Notes

Participants have noted the complexity of the topic, particularly in relation to the assumptions made in different scenarios, such as constant temperature conditions. There is also a mention of varying comfort levels with calculus-based approaches among participants.

chiddler
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Homework Statement


I'm trying to understand two things:

1. How internal energy is distinct from kinetic and potential energy

2. The difference between internal energy and enthalpy.

Homework Equations


enthalpy ΔH = ΔU + PΔV
internal energy ΔU = Q + W = Q - PΔV

The Attempt at a Solution


I have searched old posts. I found this. And this.

My book asked a question: if an exothermic reaction occurs in a closed system, does internal energy change? The answer is no. So it seems that internal energy includes potential energy. Both pages that I linked say that E = KE + PE + U. This suggests that internal energy and potential energy are completely distinct, yet the book question suggests that they are the same! Also, how is internal energy different from enthalpy?

thank you very much!
 
Last edited:
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chiddler said:
1. How internal energy is distinct from kinetic and potential energy
For ideal gases, internal energy is the total kinetic energy of the molecules. For a real gas, it includes potential and kinetic energies.

In an ideal gas, potential energy (which is energy due to the separation of molecules relative to others) does not exist. So it is only kinetic energy that you have to take into account. Translational kinetic energy is a function of temperature. But diatomic and polyatomic molecules can have kinetic energy other than translational kinetic energy (vibrational, rotational). These other kinetic energies are a function of temperature as well.

For real gases, molecules exert forces on each other. This forces can be relatively strong (eg. water vapour) or weak (vanderwaals forces). It takes energy to overcome those forces to increase the separation of the molecules - ie. potential energy. So when energy is added and volume increases, some of the energy added goes into increasing potential energy, which means it does not go into increasing the kinetic energy of the molecules.
My book asked a question: if an exothermic reaction occurs in a closed system, does internal energy change? The answer is no. So it seems that internal energy includes potential energy. Both pages that I linked say that E = KE + PE + U. This suggests that internal energy and potential energy are completely distinct, yet the book question suggests that they are the same! Also, how is internal energy different from enthalpy?
In any process the change in internal energy is equal to the energy transfer that occurs. This energy transfer can result from Work (W) being done on or by the system or heat flow (Q) into or out of the system.

Since H = U + PV,

dH = dU + PdV + VdP

For an ideal gas PdV = - VdP so dH = dU. But for real gases, PdV ≠ VdP so enthalpy becomes useful.

AM
 
thanks very much for the excellent response.
 
chiddler said:
thanks very much for the excellent response.
The only problem is that the explanation for dH is not quite right!
Andrew Mason said:
For an ideal gas PdV = - VdP so dH = dU. But for real gases, PdV ≠ VdP so enthalpy becomes useful.
This is only true if T is constant, in which case dH = dU = 0.

I'll have a look at this again later.

AM
 
Andrew Mason said:
The only problem is that the explanation for dH is not quite right!
This is only true if T is constant, in which case dH = dU = 0.

I'll have a look at this again later.

AM

appreciate it. I'm not very comfortable in a calculus based approach as my physics was only algebra based. So could you please accommodate me in your response? (ie, more conceptual, please)

I didn't ask immediately because I didn't think this question gathered enough interest for follow up questions!
 

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