temp changing in PV work?

[fluoropHore]

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ok so i'm a big confused here: when a gas expands during isotherm expansion, V changes and it does work.. how can you use w=PdeltaV if P IS changing during this time? (P is changing if T is contant and V is increasing... pv=nrt) youd have to integrate and cant use it simply straight. or am i missing something?

 

[MedPR]

ok so i'm a big confused here: when a gas expands during isotherm expansion, V changes and it does work.. how can you use w=PdeltaV if P IS changing during this time? (P is changing if T is contant and V is increasing... pv=nrt) youd have to integrate and cant use it simply straight. or am i missing something?

I was having this same problem until someone explained to me that isothermal just means no heat in/out. It doesn't mean the heat within the system is constant. Isothermal expansion with constant pressure leads to decreased temperature because the heat within the system is being converted to PV work (to expand it by increasing the volume) and no additional heat is being put into the system (hence, isothermal) to compensate for the heat lost (converted) to work.

 

[fluoropHore]

wait are you sure isotherm means deltaq=0? i just found this ani: http://science.sbcc.edu/physics/flash/heatengines/Carnot cycle.html

i thought if you add heat, it has to "dissipate" (i know wront word) it by expanding in a gas..therefore: same T.

 

[fluoropHore]

(in the animation isothermal part of the cycle, a heat source is added.)

 

[fluoropHore]

this doesnt sound right: adiabatic means no change in q, isothermal means no change in T,

and i am assuming they mean the system (the gas itself due to compression.expansion) and not an additional heat source.

huhhh

 

[fluoropHore]

oh jeez i think yr right. so i had it backwards??

 

[kasho11]

The pressure is NOT changing because it is the pressure against which the gas pushes - meaning the EXTERNAL pressure. Pressure inside the chamber is indeed changing and can be related by Boyle's Law in the isothermal case, P1V1 = P2V2.

But for the purposes of W = PdeltaV, P is the external pressure on the system and does not change


edit - and yes adiabatic means q = 0, So Energy = -W

double edit - and to be clear about 1st law of thermo, because this confused me, U = Q + W, when work is done ON the system, which gives a + value for W. U = Q - W when work is done BY the system.

 

[fluoropHore]

Can you explain that animation then? because it sounds like what you are describing is the flipped of those processes on the graph (ie isothermal should really say adiabatic? or the animation of the piston is wrong? or i am wrong..)

 

[kasho11]

In the first part of the Carnot cycle, heat is added isothermally to the system from a heat reservoir. In this situation the heat is added slowly enough to prevent a temperature change by allowing the gas to dissipate the heat continuously through isothermal expansion.

The Red and Blue things at the bottom are Heat and Cold reservoirs respectively. They will equilibrate with the system they are put in contact with, so the heat reservoir will add heat energy, the cold reservoir will take away heat energy.

At the second part (adiabatic expansion) no more heat energy (q) is added to the system, q = 0 because here the piston and system are assumed to be insulated (separated from the reservoirs) but there is still extra thermal energy within the piston, so it continues to do work until it reaches a cooler temperature.

At the third part work is done ON the system while in contact with the cool reservoir, it is isothermal so heat leaves to the cool reservoir while the volume decreases.

In the fourth part, work is done ON the system while insulated (not in contact with a reservoir) so q = 0 but W = positive.

Sorry if this was a little scattered, but basically the reservoirs on the bottom (blue and red bars) when in contact with the system result in an isothermal situation, because the change is occurring slowly enough to allow the temperature to stay constant. But when the reservoirs are separated from the system, they are thought to be insulated, and thus no heat exchange can occur.

 

[fluoropHore]

no thats what i thought! thanks! buuutt what about the original q: if its isothermal, how can it be constant P if V is increasing? I know intuitively since the heat used to increase the KE of the gas turns into Pv work (hence P=const) but EK says that since T=cons and Vincreases, P cannot be constant.

 

[kasho11]

Can you quote the statement? It seems like they are referring to the equation P1V1 = P2V2, an isothermal situation. Again for W = PdeltaV the pressure is the external pressure and is constant.

 

[MedPR]

Oh I still have it backwards. Wonderful.

 

[MedPR]

So in isothermal, pressure can't remain constant if volume is changing. So you can't find the area under the curve (PV work done) without calculus or extreme approximation.

Adiabatic is U=W since U =Q+W and Q=0 for adiabatic. Sorry for the confusion!

 

[TXKnight]

ok so i'm a big confused here: when a gas expands during isotherm expansion, V changes and it does work.. how can you use w=PdeltaV if P IS changing during this time? (P is changing if T is contant and V is increasing... pv=nrt) youd have to integrate and cant use it simply straight. or am i missing something?

This was the same question i was going to post a few hrs ago. I learned a lot here:
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

 

[MedPR]

Also, even if P and V are both changing, the work is still defined as PV.

 

[TXKnight]

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/imgheat/eng4.gif

 

[fluoropHore]

thanks kasho! thats what i originally thought

 

[fluoropHore]

ok stil having issues:
heres a simple q:

which of the following changes during the free adiabatic expansion of an ideal gas?

internal E
Temp
P

answer is P only.. some people have been saying that when it expands adiabatically since q=0 work=0 the temp changes as it expands ..but EK keeps saying it doesnt since there is not exchange of q/work and therefore the moleecules dont change in energy= therefore no T change.

WHUT.

 

[MedPR]

ok stil having issues:
heres a simple q:

which of the following changes during the free adiabatic expansion of an ideal gas?

internal E
Temp
P

answer is P only.. some people have been saying that when it expands adiabatically since q=0 work=0 the temp changes as it expands ..but EK keeps saying it doesnt since there is not exchange of q/work and therefore the moleecules dont change in energy= therefore no T change.

WHUT.

Free adiabatic expansion is different than normal adiabatic expansion.

Ok I think I figured out where all the confusion is coming from in regard to this.

Free adiabatic expansion of an ideal gas means the ideal gas is allowed to expand into a preset volume container, such as the one shown here:

When I initially read this thread, I thought the container was expanding (as in a piston), but that is not the case. The volume of the CONTAINER remains constant, while the volume of the GAS (real or ideal) changes. Sorry if I'm the only one just now realizing this, but it makes things very clear for me so I imagine some of you are encountering the same confusion I was.

The reason it is irreversible is because of entropy. You are not going to get 10 particles to squeeze into a 10mL space if they all have access to a 100mL space unless you put in some energy. Adiabatic = no energy input, so your reaction is not going to reverse to the original state barring being in some universe where entropy is disfavored.

No work is done by the gas because it is simply increasing entropy spontaneously and under no force or energy absorption.

So to answer the OP. Free adiabatic expansion of ideal gas does involve constant temperature. Free adiabatic expansion of real gases will involve a temperature decrease.

There's another thread where it is explained very well. http://forums.studentdoctor.net/showthread.php?t=898865

 

[fluoropHore]

ah that makes sense thanks! 🙂