Internal energy, work, and heat

[plzNOCarribbean]

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So I looked up previous threads and they cleared almost everything up. Thanks a lot MDodysey and many others. What I gathered was that the way you set the equation up depends on what's happening, and you need to be careful of the sign that you use for work. You must specify whether is being done BY the system or ON the system.

if work is being done BY the system then
U= Q - W ; the system is doing work & therefore the systems internal energy decreases

U= Q + Q ; there is work being done on the system and therefore the systems internal energy is increasing.

dU refers do the change in the internal energy of our system.


So, I just want to make sure I understand this correctly by applying it to a few scenarios that I am confused about. The system to consider is the conventional piston with gas in it.

1) if you apply heat to the system, and the gases move and push the piston upward, then is it correct to say that the Initially, before the piston move, the internal energy of the system increased?

2) Then, by as the piston was pushed upward by the gas, work was done BY our system (the gas). So basically, the heat that was put into our system was equal to the amount of work done BY our system, so overall, the internal energy change, dU=0? ie, whatever energy we gained in heat was lost due to the work done by our system. is this correct?

3) Lastly, say we had the same piston, but instead now the piston was completely insulated, so that heat could not be lost to the surrounding and the contained WALLS COULD NOT EXPAND; Also, there is a pin on the piston to prevent it from moving; say, theoretically that we added heat to the system with a flame (yes, even though its insulated, or say we put the insulation right after it was heated and assume no heat loss); Then, would it be correct in saying that the internal energy increase based on U=Q + W and that the internal energy increase is proportional to the heat added , since no energy could be lost due to work that could be done by the system ( since the container walls cannot expand and the piston cannot move )?

I have a few more questions, but il wait until these are adressed and Il keep them in the same thread.

 

[MD Odyssey]

1) if you apply heat to the system, and the gases move and push the piston upward, then is it correct to say that the Initially, before the piston move, the internal energy of the system increased?

Not really. As you heat the gas, it is going to expand and do work on the surroundings. Of course, as you heat the gas, the internal energy of the system will absolutely change because you are heating it and it is doing work on its surroundings.

2) Then, by as the piston was pushed upward by the gas, work was done BY our system (the gas). So basically, the heat that was put into our system was equal to the amount of work done BY our system, so overall, the internal energy change, dU=0? ie, whatever energy we gained in heat was lost due to the work done by our system. is this correct?

It seems that you are separating the two events. You heat the gas and as you do, it does work on the surroundings. They aren't separate events
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3) Lastly, say we had the same piston, but instead now the piston was completely insulated, so that heat could not be lost to the surrounding and the contained WALLS COULD NOT EXPAND; Also, there is a pin on the piston to prevent it from moving; say, theoretically that we added heat to the system with a flame (yes, even though its insulated, or say we put the insulation right after it was heated and assume no heat loss); Then, would it be correct in saying that the internal energy increase based on U=Q + W and that the internal energy increase is proportional to the heat added , since no energy could be lost due to work that could be done by the system ( since the container walls cannot expand and the piston cannot move )?

If I understand you correctly, you're asking about what happens if I have a gas in an insulted box that is unable to increase its volume or exchange particles with the environment around it. If I heat the gas in this configuration, then you are correct - the change in internal energy is just the heat transferred into the system. This internal energy would show up as in increase in temperature and pressure, in accordance with the ideal gas law.

 

[plzNOCarribbean]

It seems that you are separating the two events. You heat the gas and as you do, it does work on the surroundings. They aren't separate events


^So yeah, whatever energy you put into the system in terms of heat was lost to the surrounding when the piston moved. Therefore, it is correct to say that dU=0 when this happens.

I understand that the processes weren't separate, but I was asking if I was correct with my assumptions if the two were assumed to be separate.

1)add head = internal energy increases
2) piston moves = work is done by the gas (our system) = internal energy decreases

therefore, dU= 0, right?

 

[indianjatt]

I thought it would depend..... dU isn't always 0. The heat that enters isn't always equal to the work out. I believe that happens when we consider an isothermal process.

For example, when we undergo an isobaric process, when a gas expands (or does work), heat is added into the system, but doesn't necessarily mean dU = 0. In an adiabatic process, dU = W, so again not zero. Isochoric, dU = Q, again not zero.

 

[MD Odyssey]

It seems that you are separating the two events. You heat the gas and as you do, it does work on the surroundings. They aren't separate events


^So yeah, whatever energy you put into the system in terms of heat was lost to the surrounding when the piston moved. Therefore, it is correct to say that dU=0 when this happens.

I understand that the processes weren't separate, but I was asking if I was correct with my assumptions if the two were assumed to be separate.

1)add head = internal energy increases
2) piston moves = work is done by the gas (our system) = internal energy decreases

therefore, dU= 0, right?

Not necessarily. It depends upon how you get there. For cyclical processes, the work done is always less than the heat transfer. That's pretty much what the second law of thermodynamics is saying.

 

[costales]

Another way: Kinetic theory says internal energy varies only as temperature, so dU = 0 only if dT = 0 i.e. in isothermal process.

 

[MD Odyssey]

Another way: Kinetic theory says internal energy varies only as temperature, so dU = 0 only if dT = 0 i.e. in isothermal process.

That's a really insightful perspective. I hadn't thought to express it that way. Thanks.

 

[Majik]

So I looked up previous threads and they cleared almost everything up. Thanks a lot MDodysey and many others. What I gathered was that the way you set the equation up depends on what's happening, and you need to be careful of the sign that you use for work. You must specify whether is being done BY the system or ON the system.

if work is being done BY the system then
U= Q - W ; the system is doing work & therefore the systems internal energy decreases

U= Q + Q ; there is work being done on the system and therefore the systems internal energy is increasing.

dU refers do the change in the internal energy of our system.


So, I just want to make sure I understand this correctly by applying it to a few scenarios that I am confused about. The system to consider is the conventional piston with gas in it.

1) if you apply heat to the system, and the gases move and push the piston upward, then is it correct to say that the Initially, before the piston move, the internal energy of the system increased?

2) Then, by as the piston was pushed upward by the gas, work was done BY our system (the gas). So basically, the heat that was put into our system was equal to the amount of work done BY our system, so overall, the internal energy change, dU=0? ie, whatever energy we gained in heat was lost due to the work done by our system. is this correct?

3) Lastly, say we had the same piston, but instead now the piston was completely insulated, so that heat could not be lost to the surrounding and the contained WALLS COULD NOT EXPAND; Also, there is a pin on the piston to prevent it from moving; say, theoretically that we added heat to the system with a flame (yes, even though its insulated, or say we put the insulation right after it was heated and assume no heat loss); Then, would it be correct in saying that the internal energy increase based on U=Q + W and that the internal energy increase is proportional to the heat added , since no energy could be lost due to work that could be done by the system ( since the container walls cannot expand and the piston cannot move )?

I have a few more questions, but il wait until these are adressed and Il keep them in the same thread.

There's a few key concepts you want to understand for pistons:

Expanding Gases COOL (Lower Temperature ~ Internal Energy)
Compressed Gases HEAT UP (Higher Temperature ~ Internal Energy)

This concept will help you answer questions much more quickly than the equation. However, the equation is conceptually useful in rare occasions. The problem however is a discrepancy with the sign (Chem / Physics conventions).

Whether:
E = Q + W vs. E = Q - W is true depends solely on what work is defined as. From a chemistry perspective, we take work as: -PdeltaV, where P is the external pressure and V is the volume of the container. Regardless, both equations ultimately result in the same change in internal energy (work however, since its defined differently - will be different). For simplicity, recognize that this equation will ALWAYS be true: E = Q - PdeltaV.

Now, backtracking a bit. Why do expanding gases cool? Well, assuming that we're in an adiabatic container (that is, there's no heat loss or gain; q=0), then E = -PdeltaV. Therefore, if we expand the volume of the container (that is, a positive deltaV), then the overall -PdeltaV term will be a negative value. This would equate to a loss of internal energy (E). Being that internal energy is essentially the same thing as temperature, a loss of internal energy would imply a decrease in temperature (ie. it cools down).

The scenario you are describing in situation 1 (and 2) is isothermal conditions (E=0). When E=0, heat input = work output (or vice versa). Therefore the heat you put into the system is used up to expand the container (it does useful work). The energy of the system however, remains constant. (-q = W).

 

[plzNOCarribbean]

The scenario you are describing in situation 1 (and 2) is isothermal conditions (E=0). When E=0, heat input = work output (or vice versa). Therefore the heat you put into the system is used up to expand the container (it does useful work). The energy of the system however, remains constant. (-q = W).

Right, that is what my question was. You would assume, for the mcat (without getting so theoretical) that the amount of heat you put in to the system results in the gas pushing on the system (our system is doing work) and therefore the heat in + work done by our system = the change in the internal energy of our system, which in this case would be dU=0.

Without getting so technical and theoretical, i don't understand why "the work done is less than the heat transfer". I know it depends on what kind of cyclical process is occurring, and I appreciate everyones input. Can someone check if this is right:

dU= Q + W

1) Isobaric (constant P) ; dU= heat added + work done by system; the pressure is kept constant, but the heat added can translate into work being done by the system, so dU= 0? (if the magnitude of the heat added = the amount of work done by the system)

2) Isochoric (constant V) ; dU = heat added (or lost); if heat is added, the gas pressure increases and T increases, right?

3) Isothermal (constant temp); sorta stumped with this one; could use some help. dU= ?

4) adiabatic - (from what I can remember, this is a system that has a constant pressure, and cannot have an exchange in heat with the environment) dU= ?

The last thing is the whole expanding or compressing idea. I think this is what may be confusing me; I read it in TPR and majik stated it

compressed gases heat up
and expanding gases cool

^so by this, does it mean that someone or something (say the surrounding) pushed down on our piston. Since the gas is being compressed, it is gaining energy (dU increases) and this seen due to the heating of the gas. Basically, the increasing in the temp of the gas AND increase in pressure is proportional to increase in internal energy??

and the reverse process would be our system pushing the piston up, causing our gas to expand; THis is the cause of the decrease in internal energy. Therefore, the decrease in internal energy is proportional to the decrease in temperature of our gas (the gas is cooling) AND the decrease in pressure (since volume is increasing)

 

[indianjatt]

1) Isobaric (constant P) ; dU= heat added + work done by system; the pressure is kept constant, but the heat added can translate into work being done by the system, so dU= 0? (if the magnitude of the heat added = the amount of work done by the system)

In an isobaric process, when a gas expands and does work to the surroundings we know that the internal energy decreased due to negative work and the volume increased, BUT since the volume increased and pressure is constant, then by PV=nRT the only variable that can change is temperature. Since temperature and volume are proportional, the temperature of the system must increase. Also noting the fact that the kinetic molecular theory of gases states that KE = 3/2KbT, we know that an increase in temperature will cause the internal energy of our system to increase. Therefore, when we undergo an isobaric process, when the system does work, heat must go into the system in order to increase the internal energy.
2) Isochoric (constant V) ; dU = heat added (or lost); if heat is added, the gas pressure increases and T increases, right?

3) Isothermal (constant temp); sorta stumped with this one; could use some help. dU= ?

Isothermal is simply when temperature is constant. Again, internal energy only change dues to temperature. Since we have no temperature change, the internal energy change should be 0. Therefore, W=-Q because U = Q+W and U = 0 in this case.

4) adiabatic - (from what I can remember, this is a system that has a constant pressure, and cannot have an exchange in heat with the environment) dU= ?

Adiabatic is when heat is constant. So U = W (heat doesn't affect the change in internal energy because it stays constant).

The last thing is the whole expanding or compressing idea. I think this is what may be confusing me; I read it in TPR and majik stated it

compressed gases heat up
and expanding gases cool

^so by this, does it mean that someone or something (say the surrounding) pushed down on our piston. Since the gas is being compressed, it is gaining energy (dU increases) and this seen due to the heating of the gas. Basically, the increasing in the temp of the gas AND increase in pressure is proportional to increase in internal energy??


Well, one actually results form the other. When we increase the pressure of our system, more and more collisions between particles are occurring. Therefore, as pressure increases we have more kinetic energy (more frequent collisions). As a result, our temperature increases (KE = 3/2KbT), so the gases are hotter. Vice versa for the other case.

 

[plzNOCarribbean]

1)
In an isobaric process, when a gas expands and does work to the surroundings we know that the internal energy decreased due to negative work and the volume increased, BUT since the volume increased and pressure is constant, then by PV=nRT the only variable that can change is temperature. Since temperature and volume are proportional, the temperature of the system must increase. Also noting the fact that the kinetic molecular theory of gases states that KE = 3/2KbT, we know that an increase in temperature will cause the internal energy of our system to increase. Therefore, when we undergo an isobaric process, when the system does work, heat must go into the system in order to increase the internal energy.

Are you sure about that; I see what your saying from the equation but i'm a little skeptical.....are you sure that the temperature increase is for the gas in our system? I thought EXPANDING gases COOLED? since our system is doing work, wouldn't the increase in temperature be the increase in temp for the surrounding??

 

[indianjatt]

Are you sure about that; I see what your saying from the equation but i'm a little skeptical.....are you sure that the temperature increase is for the gas in our system? I thought EXPANDING gases COOLED? since our system is doing work, wouldn't the increase in temperature be the increase in temp for the surrounding??

No, its right. It is actually referred to as isobaric compression and expansion. Wikipediaexplains it well, but they just derive everything using the law of equipartition of gases and stating that q, heat is molar heat capacity or thermal energy.