What is the difference between enthalpy and internal energy?

In summary: If you heat the gas until its particles are moving, the gas will release its internal energy again as its pressure rises and the volume decreases (again, the pressure will drop to 0 when all of the particles have moved). This process is also called exothermic.In summary, internal energy is composed of two main parts, kinetic energy and potential energy. The first law of thermodynamics states that the change in internal energy is equal to the change in heat energy, the change in work energy, and the change in pressure energy.
  • #1
HWGXX7
46
0
Hello,

I'am new here and having some troubles with understanding the difference between 2 related subjects.

I learned that internal energy contains 2 main components: kinetic energy and potential energy.
[tex]U=E_{K}+E_{p}[/tex]
The related first law of thermodynamics (for closed systems): [tex]dU=dQ+dW[/tex]
wich means practically: within the systems heat can be exchanged for work and vice versa. Suppose an ideal system with no loss. Ok so far no problems.

But now the definition of ethalpy (closed system) as I learned it: [tex]H=U+p.V[/tex]

I don't understand why the component [tex]p.V[/tex] is added.
This component is already a component to the internal energy under the form of [tex]dW[/tex]

Do I miss something important in my derivation?

grtz
 
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  • #2
HWGXX7 said:
Hello,

I'am new here and having some troubles with understanding the difference between 2 related subjects.

I learned that internal energy contains 2 main components: kinetic energy and potential energy.
[tex]U=E_{K}+E_{p}[/tex]
The related first law of thermodynamics (for closed systems): [tex]dU=dQ+dW[/tex]
wich means practically: within the systems heat can be exchanged for work and vice versa. Suppose an ideal system with no loss. Ok so far no problems.

But now the definition of ethalpy (closed system) as I learned it: [tex]H=U+p.V[/tex]

I don't understand why the component [tex]p.V[/tex] is added.
This component is already a component to the internal energy under the form of [tex]dW[/tex]

Actually it's not. In conditions of constant pressure, [itex]p.V[/tex] is the work done by that the system against the environment to establish its volume. The dW term, if it is positive, refers to the amount of work that was done onto that system.

Keep in mind what a "closed system" actually means in thermodynamics.

Wikipedia said:
Interactions of thermodynamic systems
Code:
Type of system 	Mass flow	Work 	Heat
Open 			Y		Y		Y
Closed 			N		Y		Y
Isolated	 		N		N		N

See also:
http://en.wikipedia.org/wiki/Internal_energy#Internal_energy_of_a_closed_thermodynamic_system

Wikipedia said:
==Internal energy of a closed thermodynamic system==
This above summation of all components of change in internal energy assume that a positive energy denotes heat added to the system or work done on the system, while a negative energy denotes work of the system on the environment.

Typically this relationship is expressed in infinitesimal terms using the differentials of each term. Only the internal energy is an exact differential. For a system undergoing only thermodynamics processes, i.e. a closed system that can exchange only heat and work, the change in the internal energy is
:[itex] d U = \delta Q + \delta W \, [/itex]
which constitutes the first law of thermodynamics.<ref name=signconvention group=note/> It may be expressed in terms of other thermodynamic parameters. Each term is composed of an intensive variable (a generalized force) and its Conjugate variables (thermodynamics)|conjugate infinitesimal extensive variable (a generalized displacement).

For example, for a non-viscous fluid, the mechanical work done on the system may be related to the pressure ''p'' and Volume (thermodynamics)|volume ''V''. The pressure is the intensive generalized force, while the volume is the extensive generalized displacement:
:[itex]\delta W = - p \mathrm{d}V\,[/itex].
This defines the direction of work, ''W'', to be energy flow from the working system to the surroundings, indicated by a negative term.<ref name=signconvention group=note/> Taking the direction of heat transfer ''Q'' to be into the working fluid and assuming a Reversible process (thermodynamics)|reversible process, the heat is
:[itex]\delta Q = T \mathrm{d}S\,[/tex].
::[itex]T[/itex] is temperature
::[itex]S[/itex] is entropy

and the change in internal energy becomes
:[itex]\mathrm{d}U = T\mathrm{d}S-p\mathrm{d}V \![/itex]
 
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  • #3
I still have trouble to understand the concept. Another approach: wath remains of the system (any kind of system) if
a) I (surrounding) get it's internal energy
b) I (surrounding) get it's enthalpy

I correlate the work done by the system to maintain its volume as potential energy. If the systems collapses, the work I can use for driving a turbines for example...The systems volume will be exchanged for energy to it's surrounding and what remains is than only (microscopic) kinetic energy?

But I from the equation of the enthalpy I conclude that only internal energy wil be left, which has still a potential energy component and therefore also ability to delever work...

So I hope that de difference between situation a) en b) will clear my confusion.

grtz
 
  • #4
From the little I know,

The idea of the enthalpy is that you can consider a system as having some energy by virtue of it existing in a volume V at a pressure p. The extra amount of energy that it is considered to have is equal to pV.

Take, for example, a perfect gas. Say a perfect gas occupies a volume V at a pressure p. One can extract the perfect gas's internal energy by cooling it until all of its particles are stationary while keeping the volume constant at V (its pressure will of course drop to 0). The particles are stationary and thus have no kinetic energy nor potential energy (particles in a perfect gas don't have any potential energy). This cooling process therefore extracts all of the systems internal energy.

Having extracted the perfect gas's internal energy, one can extract no more energy from the gas unless the surroundings are allowed to do work on it. If the volume of the gas is allowed to drop to 0, the surroundings do pV of work to the gas (which increases the gas's internal energy from 0 to pV). This extra energy can be again extracted by someone by cooling until the particles are stationary. By extracting this extra energy which your system has received from its environment, you have extracted the systems enthalpy.

Ted
 
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  • #5
If the volume of the gas is allowed to drop to 0, the surroundings do pV of work to the gas (which increases the gas's internal energy from 0 to pV). This extra energy can be again extracted by someone by cooling until the particles are stationary. By extracting this extra energy which your system has received from its environment, you have extracted the systems enthalpy

After you couled down the ideal gas the internal energy is zero. Understand, but also the pressure is 0. How can the ideal gas than maintain its volume? I was thought that pressure is the reason why the gas has its volume, big pressure mean great volume (if system is allowed to expand).


Also suppose volume is the same, why have the surrounding work tot do to decrease the volume, pressure is 0 already (because you couling down the ideal gas..) and therefore p.V=0...

Isn't that a contradiction in the 2 situations..?

thanks for help!
 
  • #6
Again, as far as i know,

In the first situation where the gas is cooled, it can't maintain its volume but the container maintains the same volume (because we assume it's rigid). By allowing the walls of the container to move, the wall(s) impart kinetic energy to the particles (because they are accelerated by the moving wall(s)).

When the surroundings compress the gas of zero pressure to zero volume, the work is equal to pV rather than zero due to the conservation of energy. Consider the cyclic process where the gas is compressed by the surroundings but rather than cooling it, we use its kinetic energy to do work against the walls by expanding to its original volume. The change in internal energy in this process is equal to zero because the gas returns to its initial state. No heat is exchanged in the process so the total amount of work done on the system must be zero (the first law). The work done by the system against the surroundings is equal to pV when it expands against the constant pressure p so the work done by the surroundings to compress it originally must be equal to pV too.

Ted
 
  • #7
HWGXX7 said:
I learned that internal energy contains 2 main components: kinetic energy and potential energy.
[tex]U=E_{K}+E_{p}[/tex]
I think what you mean here regards the microscopic energy. From Hyperphysics:
Internal energy is defined as the energy associated with the random, disordered motion of molecules. It is separated in scale from the macroscopic ordered energy associated with moving objects; it refers to the invisible microscopic energy on the atomic and molecular scale. For example, a room temperature glass of water sitting on a table has no apparent energy, either potential or kinetic . But on the microscopic scale it is a seething mass of high speed molecules traveling at hundreds of meters per second. If the water were tossed across the room, this microscopic energy would not necessarily be changed when we superimpose an ordered large scale motion on the water as a whole.
inteng1.gif


...

MICROSCOPIC ENERGY
Internal energy involves energy on the microscopic scale. For an ideal monoatomic gas, this is just the translational kinetic energy of the linear motion of the "hard sphere" type atoms , and the behavior of the system is well described by kinetic theory. However, for polyatomic gases there is rotational and vibrational kinetic energy as well. Then in liquids and solids there is potential energy associated with the intermolecular attractive forces. A simplified visualization of the contributions to internal energy can be helpful in understanding phase transitions and other phenomena which involve internal energy.
Ref: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/inteng.html

Regarding enthalpy, the MIT.edu website has a nice explanation:
It is also more convenient to divide the work into two terms: 1) the flow work done by the system which is p2v2-p1v1, and 2) any additional work which we will term external work or shaft work, ws. Then we have
or
We will call this the steady flow energy equation.

...

Enthalpy is most useful for separating flow work from external work (as might be produced by a shaft crossing the control volume boundary for instance). In the figure shown below. Heat is added, a compressor is doing work on the system, the flow entering the system does work on the system (work = -p1V1), and work is done by the system through pushing out the flow (work = +p2V2). The first law relates the change in energy between states 1 and 2 to the difference between the heat added and the work done by the system. Frequently, however, we are interested only in the work that crosses the system boundary, not the volumetric or flow work. In this case it is most convenient to work with enthalpy.

This also leads to a direct physical interpretation for enthalpy. In an open flow system, enthalpy is the amount of energy that is transferred across a system boundary by a moving flow. This energy is composed of two parts: the internal energy of the fluid (u) and the flow work (pv) associated with pushing the mass of fluid across the system boundary.
Ref: http://web.mit.edu/16.unified/www/FALL/thermodynamics/chapter_6.htm

Another good page on the MIT website that might help:
http://web.mit.edu/16.unified/www/FALL/thermodynamics/chapter_4.htm
 
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  • #8
I think what you mean here regards the microscopic energy.
Correct, but the notation for internal energy :U appears also in the expression for the first law of thermodynamics.
So to speak: adding heat and adding work increases the internal energy U, but for the eqaution of enthalpy h=u+p.v , only u changes...? I do understand the physical meaning of enthalpy now.

Like u states with the example of the glas of water, to give rise to it's volume, work has to be done to force the volume expanding against atmospheric pressure. Ok, but in therms of enthalpy, U changes because of the first law of thermodyanmics but also p.V changes because de rise of volume...

The 2 formulas: first law of thermodynamics and enthalpy get me rather confused here.

grtz
 

1. What is the difference between enthalpy and internal energy?

Enthalpy (H) is the sum of a system's internal energy (U) and the product of its pressure (P) and volume (V), while internal energy is the total energy of a system's particles, including the kinetic and potential energy. Essentially, enthalpy takes into account the pressure-volume work done by the system, while internal energy does not.

2. How are enthalpy and internal energy related?

Enthalpy and internal energy are related through the equation H = U + PV, where H is enthalpy, U is internal energy, P is pressure, and V is volume. This equation shows that enthalpy is dependent on internal energy, but also takes into account the work done by the system.

3. When is enthalpy more useful than internal energy?

Enthalpy is more useful than internal energy when studying systems that undergo changes in pressure and volume, such as chemical reactions or phase changes. This is because enthalpy takes into account the work done by the system, which can affect the overall energy of the system.

4. Can enthalpy and internal energy be measured directly?

No, enthalpy and internal energy cannot be measured directly. They are both state functions, meaning they only depend on the initial and final states of a system, not the process in between. This means that changes in enthalpy and internal energy can be measured, but not the values themselves.

5. How is enthalpy represented on a thermodynamic diagram?

On a thermodynamic diagram, enthalpy is represented as a vertical axis, while internal energy is represented as a horizontal axis. The slope of the enthalpy line is equal to the pressure of the system, while the slope of the internal energy line is equal to the temperature of the system.

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