[21385]

I really don't understand the differences between these two. Can someone explain this to me?

also, can someone explain this specifically to abiabatic processes?

[Rabn]

Look at the definition of each, each one has a specific thermodynamic definition.  The only way we can help you is if you tell us what is causing the confusion, tell us your thoughts. That is the only way to get an answer. 

[Yggdrasil]

Although the two quantities are often confused, ΔH and q are completely different values.  First, definitions:

q is the amount of heat transfered to the system.  It is one of the two means of energy transfer during most processes studied in thermodynamics.  The other means of transferring energy is through work.  Since these are the only means by which we can transfer energy between the system and the surroundings, we can write the change of internal energy of our thermodynamic system as:

ΔE= q + w
(note: some books denote internal energy as U)

Enthalpy (H), on the other hand, is a thermodynamic potential, much like internal energy (E).  It is essentially a measure of the potential energy of the system.  Enthalpy is defined by the following equation:

H = E + PV

or

ΔH = ΔE + Δ(PV)

By adding in the definition of ΔE, we can see that

ΔH = q + w + Δ(PV)

Why do ΔH and q often get confused.  Because, lets consider what happens in a system at constant pressure.  Because pressure is constant Δ(PV) = PΔV.  Also, for a process occurring at constant pressure, w = -PΔV.  So, our equation for the change in enthalpy simplifies to:

ΔH = q  (valid for constant pressure ONLY!)

Since most chemical reactions occur at constant pressure (i.e. in open flasks exposed to atmospheric pressure), ΔH is very useful to chemists because it readily relates an easily measured quantity (heat) to a thermodynamic potential (enthalpy).  Of course, once you start looking at transformations that occur under variable pressure, calculating ΔH is no longer very simple.


Now, there is an important fundamental difference between enthalpy and heat.  Lets say you are studying a process that takes your system from P₁, V₁ and T₁ to P₂, V₂ and T₂.  Now, there are many different paths one can take between the two thermodynamic states.  Along each of these paths, the values of q and w will differ.  However, no matter what path you take, ΔH for the transformation will always be the same because ΔH depends only on the initial and final states of the transformation.

[21385]

Thank you very much, Yggdrasil. I have one more question. In an isothermal expansion, why is deltaH=0? and if deltaH is 0 and deltaE is also 0, how does deltaE=deltaH - delta(PV) work?

Thanks

[Yggdrasil]

In an isothermal expansion, ΔT = 0.  The internal energy (E) of an ideal gas depends only on the temperature of the gas:

ΔE = nC_vΔT

So, you can see that for an isothermal expansion, ΔE = 0.

Now, recall our earlier definition of enthalpy:

ΔH = ΔE + Δ(PV)

To find ΔH, we need to look at Δ(PV).  For an ideal gas:

PV = nRT

which implies that

Δ(PV) = nRΔT

(as long as n is fixed).  Therefore, for an isothermal process, Δ(PV) is also equal to zero and therefore, ΔH = 0 as well.