# in the equation Enthalpy = Q + P * delta V...

Discussion in 'MCAT Study Question Q&A' started by masterMood, Jun 20, 2008.

1. ### masterMood Membership Revoked Removed

Joined:
Jun 25, 2004
Messages:
1,758
22
Status:
Post Doc
if Delta H = U + P*delta V

and U being internal energy is = q + w

where w is p*delta v

Why is work taken into account twice? This has been been boggling my mind for some time.

Because if a reaction, say releases a gas molecule into a solution, it does work on the surroundings and can increase the volume of the surroundings. But do we take that into account twice?

also,
examkrackers says 1) that under constant pressure, the enthalpy change of a reaction equals heat.

But hten it says that 2) for condensed phases (i'm assuming liquids and solids) that enthalpy change is approximately = to internal energy change.

Is there an inherent contradiction in statement 1 and 2?

The way I'm trying to understand this is that for 2) under low pressure, a reaction can still do work because the pressure is low enough that the reaaction can actually do work on the surroundings (but if it was too high, the surroundings would resist the reactions attempt to do work and thus doing no work)

however for 1), i don't understand why at constant pressure delta H = q. Is it becasue the work done by the reaction is same in magnitude but opposite of the constant pressure*volume done by the system? so the two get cancelled out and all you're left with is enthalpy = q?

and if this is the case, then why is the work produced by the reaction = in magnitude but opposite in sign from that of the work done by the system?

hopefully i'm not confusing myself lol

Stop hovering to collapse... Click to collapse... Hover to expand... Click to expand...
#1
Last edited: Jun 20, 2008
2. ### engineeredout 10+ Year Member

Joined:
May 11, 2008
Messages:
3,426
590
Status:
Fellow [Any Field]
Okay I'll try to explain it as best as I can.

There are two different kinds of work. Internal energy is not the same as PdeltaV work. Internal energy is a function of temperature, so I can have two gases which have the same internal energy (same temperature), but different enthalpies because they are at different pressures.

If the gas at the lower pressure is expanded, it will do less work than the gas at the higher pressure that is expanding.

Am I making any sense?

Stop hovering to collapse... Click to collapse... Hover to expand... Click to expand...
3. OP

### masterMood Membership Revoked Removed

Joined:
Jun 25, 2004
Messages:
1,758
22
Status:
Post Doc
so they both have the same amount of internal energy. But at different pressure they are able to do different amounts of work (due to the resistance from the external pressure being applied to each respective gas).

So are you saying that the gas has an "innate" ability to do work, but to take into account the different pressures that it may be under, we have to include it again?

I think that makes sense.

Where I get confused, however, is hwo delta H = q when pressure is constant.

I mean the pressures can't be different between the p*v from the internal energy term and the p*v from the "external pressure being applied to gas" term otherwise delta H can't be = to q.

I don't get why they they are exactly equal to each other in magnitude but opposite in signs (because the external pressure like you said can be at different constant pressures, so then why does that make the internal energy's P*V have the same magnitude too?)

in other words:

Delta H at constant pressure = U + P*delta V
"" = q + w + P*delta V
"" = q - P*delta V + P*delta V
"" = q

Thanks again man, you're awesome.

Stop hovering to collapse... Click to collapse... Hover to expand... Click to expand...
#3
Last edited: Jun 20, 2008
4. OP

### masterMood Membership Revoked Removed

Joined:
Jun 25, 2004
Messages:
1,758
22
Status:
Post Doc
okay this is how i'm "reasoning" it out (as best i can).

When there is a constant pressure we assume that enthalpy is equal to heat gained or released, because the work done by the reaction is equal to in magnitude but opposite to the work done by the system on the reaction.

When we have a reaction with constant pressure but its very low, then we say that change in enthalpy is APPROXIMATELY equal to change in internal energy.

Why? Because the work that can be done by the reaction can "override" the very low pressure that is being applied to it, so the reaction can do work on teh system because there isn't as much resistance. That is why change in enthalpy = change in internal energy when pressure is low and constant.

When the pressure is just constant and we don't know enough about what the external pressure's value is then we just assume that change in enthalpy is = q (and assume tha t no work can be done by the reaction because it is opposes in same magnitude the pressure by the system on it).

I guess the only thing left to my stupid littel puzzle is why these two work variable opppose each other exactly in magnitude but opposite in side

(i.e. -20 J of work done by the reaction counters the 20 J of work done by the system [say a surrounding liquid] at a constant pressure.

Would this be an example of Newton's law of "for every force there's an equal and opposite force"?

Is that it?!?!?!?!?!?!

Stop hovering to collapse... Click to collapse... Hover to expand... Click to expand...
5. ### physics junkie 5+ Year Member

Joined:
Nov 20, 2006
Messages:
539
2
edit5

We’ve been on the Internet for over 20 years doing just one thing: providing career information for free or at cost. We do this because we believe that the health education process is too expensive and too competitive.

We believe that all students deserve the same access to high quality information. We believe that providing high quality career advice and information ensures that everyone, regardless of income or privilege, has a chance to achieve their dream of being a doctor.