# Enthalpy vs internal energy

1. Mar 12, 2012

### HWGXX7

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.
$$U=E_{K}+E_{p}$$
The related first law of thermodynamics (for closed systems): $$dU=dQ+dW$$
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: $$H=U+p.V$$

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

Do I miss something important in my derivation?

grtz

2. Mar 12, 2012

### kmarinas86

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.

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

Last edited: Mar 12, 2012
3. Mar 12, 2012

### HWGXX7

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, wich 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. Mar 12, 2012

### Ted Burgess

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

Last edited: Mar 12, 2012
5. Mar 13, 2012

### HWGXX7

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. Mar 13, 2012

### Ted Burgess

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. Mar 13, 2012

### Q_Goest

I think what you mean here regards the microscopic energy. From Hyperphysics:
Ref: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/inteng.html

Regarding enthalpy, the MIT.edu web site has a nice explanation:
Ref: http://web.mit.edu/16.unified/www/FALL/thermodynamics/chapter_6.htm

Another good page on the MIT web site that might help:
http://web.mit.edu/16.unified/www/FALL/thermodynamics/chapter_4.htm

Last edited by a moderator: Apr 18, 2017
8. Mar 14, 2012

### HWGXX7

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