# relationship between enthalpy and internal energy

Hi

Enthalpy is defined as
H = U + PV where U is internal energy of system and PV are pressure and volume of system

You can define the difference between H and U like so according to a book i'm reading:

H(initial) = U(initial) + PV(initial)
H(initial) = U(initial) + n(initial)RT <==replace PV with nRT

H(final) = U(final) + PV(final)
H(final) = U(final) + n(final) RT<==replace PV with nRT

D(H) = H(final) - H(initial)
D(H) = D(U) + (n(final) - n(initial))RT

When working out PV or nRT in the initial conditions I understand that you can use the initial temperature

But when you calculate PV for the final conditions, how can you use the initial temperature? The result for PV, at constant pressure, will depend on how many moles of gas have been produced or consumed and also the temperature of the system which will have changed during the reaction. In other words, the temperature of the system will be changing as the gas is expanding due to energy released/absorbed by the reaction. Is the use of the initial temperature is a necessary simplification or am I missing a key point?

Thanks
###### Who is Participating?
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

Commented:
Assumptions: nuclear energy and chemical energy are not considered. No nuclear or chemical changes will occur. A perfect gas is assumed.
OK... now

Short answer: use of initial temperature is only necessary if you need to calculate PV from the mass of the gas and its temperature as part of initial conditions. Calculation of final PV depends on how much energy is put into or taken from the system thru chemical reactions or work,  how much mass is added or taken out.

Long answer (Oh... forget it... it's too long to read any way)
In physics, very often one of the easiest ways of keeping track of what is going on in a series of formulas is to keep track of the units.  Your enthalpy H represents joules of energy. U  or your internal energy also represents joules of energy It can also be considerd here to be the thermal energy of the system. .  p and V represent pressure (pascals) and volume (cubic meters) and heat energy in a perfect gas (in joules) equals pV.  The conversion of nHT with pV is merely a different way of representing the same thing.  nHT = pV... is a relationship. Increase pressure without increasing volume, temperature goes up... etc.  Ok.. it's all a matter of energy. How?  Enthalpy H is equal to Internal Energy plus the energy represented by the "springiness" or compressibility of the gas  [PV or nRT]. You compress a gas you put energy into the "spring" .  If no energy is added or subtracted, then an equilibrium exists.  Temperature increase only occurs with a volume decrease.

Now to your question. Calculating a PV for final conditions.  You start with an initial condition. Initial temperature, initial pressure, initial volume, initial moles of gas.  You change the system by changing pressure, volume, moles of gas, or a combination. You can also add energy (heat the system) or remove energy (cool the system).  That is where your consuming or creating moles of gas comes in and the heat in the system. I assume no chemical reactions here.  Now. Given an H and a U  How do you calculate the end PV?  You can not. Sorry....

In the end, if you have defined the total internal energy (U) and enthalpy (H) than to calculate P you must know V and to calculate V you must know P.  You can not solve for both simultaneously. Temperature is immaterial. In fact, once n is known and the internal energy and entropy are known, you can calculate the final temperature without calculating P or V In fact to solve for any variable in the initial equation for enthalpy, you only need to know the other variables.

That is how they are related.

,
Author Commented:
Hi

D(H) = D(U) + (n(final) - n(initial))RT

In the book i am reading you can convert between DU and DH by the final term (n(final) - n(initial))RT. The temperature used is the initial temperature. I don't understand how you can use the initial temperature unless it is a simplification
Commented:
The formula assumes that the process is isothermal if T is the initial temperature as well as the final temperature. With an ideal gas, PV will increase without a change in temperature only if mass is added. The change in mass ( (n(final) - n(initial) ) times (R*T) equals the increase in enthalpy. note:If only volume increases and P is constant then work is done on the environment.

Can the end temperature be different from the beginning temperature? You bet.. but then not only is U changed (more thermal energy added) but the change in PV will no longer be

Delta(PV) =  (n(final) - n(initial) )RT      but (Delta(PV) =  R(n(final)T(final)-n(initial)T(initial))

It would seem your book is simplifying for an example

It should be clear that in the isothermal reaction,  as extra gas is added from the chemical reaction, PV will increase and internal energy will decrease (loss of chemical energy).

Experts Exchange Solution brought to you by

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Author Commented:
Thanks for your answer. This is really what I was trying to ask in the other question you answered but I got muddled as i was writing it. I assumed it was a simplification. I initially thought it was saying the process was isothermal but it uses this formula in relation to enthalpy of combustion reactions, and heat is surely released too quickly in that situation to be isothermal? Here is an example:

D(U) for the combustion of 1 mol of glucose is -2559 kj at 298 K
What is the change in enthalpy?

D(H) = D(U) + (n(final) - n(initial))RT
n final - n initial = 12 - 6 = 6 moles of gas
D(H) = -2559 + (6 * 8.314 * 298)
D(H) = -2544 kj/mol
Author Commented:
I presume another way to derive the final term is as follows:

D(H) = D(U) + PD(V) at constant pressure

calculate PD(V) at constant pressure

V(1) = n(1) RT / P
V(2) = n(2) RT/P

D(V) = V2 - V1
= (n2 - n1) RT/P

PD(V) = P * (n2 - n1) RT/P
P(DV) =  (n2 - n1) RT

This is assuming constant temperature like you said.

But the book definitely doesn't mention constant temperature. Is this something i should know and it is taken as read? For example, are enthalpies generally determined in such a way that the system stays at constant temperature?
Author Commented:
Karl, will you be around for the next couple of days? I have loads of questions i could ask you. If you are, what time zone are you in
Commented:
I am east coast... I can keep up some correspondence today... and evenings during the week after I get back from work. If my time gets short. I'll let you know.
Commented:
Answer to question of constant temperature, pressure, etc.

In any problem starting and ending conditions have to be known to calculate results. In a textbook problem you are normally safe to assume that unless specifically stated a starting condition will be an ending condition. It simplifies things.
Author Commented:
How could you assume combustion of glucose was isothermal? The heat would be released too fast wouldn't it?
Commented:
Since you specify the reaction, you no longer have to make assumptions. Combustion of glucose in an oxidizing atmosphere will be exothermal.  Now you can get a better idea of what the final end states will be.    However.... while the glucose reaction may release energy, the system may perform work by expansion and at the end of the expansion the overall temperature may not have changed... ergo the overall action could still be isothermal.   Start and End states must be known (except for at most one variable) for that one variable to be calculated.

One can keep going in circles here.  You throw in more variables, you have to make more assumptions or tie down more degrees of freedom to be able to calculate answers. In a textbook question, such things as speed of reaction, internal friction of gases, heat transfer to container, etc... are not considered.
Author Commented:
Hi

I think its saying the temperature is constant as any heat generated/lost is immediately removed/replaced and the final term calculates the energy needed to expand the gas at that temperature. I dont think its any more complicated than that. Like you say its a book
Author Commented:
many thanks
###### It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Math / Science

From novice to tech pro — start learning today.