Thermochemistry and the internal energy equation?

[thestormpetrel]

If we add heat to system and that system does work on its' surrounding then the change in internal energy of the system will be:
deltaU= q-w = q-p(delta volume)
Now, the main thing that confuses me is that constant P in the formula of work. Till now, I thought P was constant because the added heat increases the temp which INCREASES THE PRESSURE but since volume is also increasing, that increase in volume will DECREASE THE PRESSURE. Both effects will cancel out and our P will be constant.

But, then I watched one of the KhanAcademy videos on internal energy and it said that P was constant because it was a quassi-static process and volume was changing slowly so P will be constant. And I don't really understand that. If we ignore temp and if volume is changing slowly then pressure should also change (albeit slowly).

Then, there was this PV work diagram graph like this one:

And in my mind, this diagram basically says that pressure WILL change since the inital and final values of pressure are very different. Even if we break the graph into very small steps so that pressure is nearly constant in every step, the overall work DOES involve the change of pressure so why don't we account for that change of pressure in that work equation?

Long story short: i first thought that pressure was constant because of increase in temp but i later found out that i was wrong and the change in temp has no role in pressure. why?
if temp has no role then why doesn't pressure change? and if it is true that pressure does not change then why does the graph indicate otherwise?

Thanks a ton in advance!

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[ninjiliu]

Pressure IS constant. And you are right in your theory. When the gas heats up the pressure does build but it instantaneously does work on the system by expanding the volume of the container. Its essentially expanding at a rate that keeps the pressure constant in the vessel.

I think the graph you're showing is for a different scenario. In this case Temperature is constant while Pressure and Volume are changing. One of your variables must always be held constant in order to solve for the MCAT level questions involving that formula.

And come to think of it, for your original proposed scenario of increasing heat both Volume and Pressure have the capability change, but you need to figure out whats help constant depending on what information you're given in the question. A rigid container will have constant volume and increasing pressure while a balloon will have a constant pressure and changing volume.

 

[mehc012]

Pressure IS constant. And you are right in your theory. When the gas heats up the pressure does build but it instantaneously does work on the system by expanding the volume of the container. Its essentially expanding at a rate that keeps the pressure constant in the vessel.

I think the graph you're showing is for a different scenario. In this case Temperature is constant while Pressure and Volume are changing. One of your variables must always be held constant in order to solve for the MCAT level questions involving that formula.

And come to think of it, for your original proposed scenario of increasing heat both Volume and Pressure have the capability change, but you need to figure out whats help constant depending on what information you're given in the question. A rigid container will have constant volume and increasing pressure while a balloon will have a constant pressure and changing volume.

Exactly. OP, you can't just ask about PV work...in order to do these problems, you need to fully understand the conditions throughout the process. The equations are different for different scenarios. The main ones to consider are:

Adiabatic conditions: Q = 0
Isothermal conditions: ΔT = 0
Isochoric conditions: ΔV = 0
Isobaric conditions: ΔP = 0

In each of those scenarios, the equations used to calculate ΔU, Q, ΔT, ΔP, W, etc...they'll all by very slightly different from each other.

If you want a more in depth look, try this link, though right now it seems to be down for maintenance and so I've given you the cached page.