Q and W in isobaric expansion

[MrNeuro]

so for the isothermal expansion seen in the path 1 (the horizontal line) the internal energy is increasing meaning that there must have been more heat flow however path1 still fits answer a as there is more work performed by the system on the surroundings during the isovolumetric cooling (vertical line) second part of path 1 exhibits a lower temperature meaning the internal energy must have decreased (which means there is more work done on the surroundings) due to heat flow out not in and it wasn't due to work

tl;dr people is this right?
so for the horizontal line isobaric expansion
Q in > W
and
vertical line isovolumetric expansion
Q out > Q in

hence W on surroundings > Q in

Right??

is there an easier way of going about this problem besides POE???

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[chiddler]

yeah i asked this question some week ago or so. no responses :<

 

[MrNeuro]

yeah i asked this question some week ago or so. no responses :<

I think it's right....

 

[Dasypus]

Edit: this is wrong ==> [A must be wrong, because it violates the second law of thermodynamics. Not only can the amount of work done not exceed the amount of heat put in, it can't even equal it (the theoretical yield of the best possible heat engine, a Carnot engine, is only 1-[t(cold)/t(hot)]).]

Also, in any non-adiabatic expansion, the amount of work done is path-dependent, so we know that D is wrong.

Internal energy definitely isn't constant, because all three cool down up.

Edit: this must be wrong too ==> [So it must be C, and they're talking about net temperature change. If you take &#916;T to be T(final)-T(initial), then definitely all three systems undergo the same &#916;T. This is because the temperature of an ideal gas is fully fixed by its pressure and volume, and all three start and finish at the same P and V.]

 

[chiddler]

A must be wrong, because it violates the second law of thermodynamics. Not only can the amount of work done not exceed the amount of heat put in, it can't even equal it (the theoretical yield of the best possible heat engine, a Carnot engine, is only 1-[t(cold)/t(hot)]).

this question is not about carnot engines. it is completely possible for them to be equal in a frictionless ideal system.

 

[andrewsmack05]

Work = Pressure x Volume. Each case is at adiabatic or isothermal expansion. Thus, no heat is flowing in or out, which is adiabatic and the other two are isothermal, so no heat is being exchanged in any of the cases. So Q = 0. Since volume and pressure are both changing work is not equal to zero.

 

[Dasypus]

this question is not about carnot engines. it is completely possible for them to be equal in a frictionless ideal system.

I know it's not about Carnot engines, but it's still about thermodynamics. And it's worth knowing that you can't get out more than you put in, when you're only talking about heat/mechanical changes (I know you can with chemical ones, but you're still not going to get something for nothing, which is what A requires).

 

[chiddler]

I know it's not about Carnot engines, but it's still about thermodynamics. And it's worth knowing that you can't get out more than you put in, when you're only talking about heat/mechanical changes (I know you can with chemical ones, but you're still not going to get something for nothing, which is what A requires).

that's a good point. looking at it again, i guess C does make the most sense.

 

[MrNeuro]

that's a good point. looking at it again, i guess C does make the most sense.

.....temperatures decreasing.

 

[MrNeuro]

I know it's not about Carnot engines, but it's still about thermodynamics. And it's worth knowing that you can't get out more than you put in, when you're only talking about heat/mechanical changes (I know you can with chemical ones, but you're still not going to get something for nothing, which is what A requires).

I think the energy is being drawn from the internal energy. For example path 2 is an isothermal expansion where work = q in but the second stage is adiabatic expansion so q=0 and the work done by the system is drawn from internal energy. So work > q in

 

[chiddler]

.....temperatures decreasing.

can't temperature increase be negative? just as work in can be negative to indicate work out.

regardless, i'm going to stop talking because i have no idea how to answer this question.

 

[MrNeuro]

Yeah but i think the question's "frame of reference" is the system. So it has to be a decrease in temp of the system.

 

[Dasypus]

I think the energy is being drawn from the internal energy. For example path 2 is an isothermal expansion where work = q in but the second stage is adiabatic expansion so q=0 and the work done by the system is drawn from internal energy. So work > q in

Dammit, I was so sure I was right! 🙁

You're totally right that the work is drawn from internal energy, since otherwise adiabatic expansion would do no work, which isn't right (it's adiabatic free expansion which does no work, which I think is what had me confused).

Bugger.


Also, I'm guessing the reason 3 is wrong is because (1) is heated way up before cooling down to the same temperature as the rest.

 

[Dasypus]

n/m