Work in a Closed System

[collegelife101]

Kaplan says that when work is done on a closed system, the internal energy does not change and the work done results in the production of heat. Is this always the case?

I'm confused because I know that there are instances where work is done on a system, resulting in an increase in internal energy rather than in heat.

Can someone explain?

Thanks!

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[el_duderino]

Work done ON a closed system, or IN a closed system? It wouldn't make sense to me that you could do work ON a closed system. If you can do work on it, it's not closed.

 

[collegelife101]

It is necessary to have a good understanding of the process described in paragraph 2 and to grasp the relationship
of this process to the First Law of Thermodynamics. When work is done on a closed system (the vapor bubbles
are small, closed systems), the internal energy does not change, and the work done results in the production of
heat.

It was specifically from their AAMC 9 practice test solution #37.

 

[SuperSneaky24]

Work done ON a closed system, or IN a closed system? It wouldn't make sense to me that you could do work ON a closed system. If you can do work on it, it's not closed.

I think you're confusing closed systems with isolated systems. In closed systems the molecules inside the system don't interact with the surroundings but the energy is still exchanged with the surroundings.

And hm... I'm not exactly sure if Kaplan's right. I know when work is done by the closed system, the internal energy decreases, for obvious reasons. I'm not exactly sure why the opposite is false though? Are there any other factors involved? Maybe if temperature was held constant in the system, this might be true.

 

[el_duderino]

Ah, right.

 

[collegelife101]

I think you're confusing closed systems with isolated systems. In closed systems the molecules inside the system don't interact with the surroundings but the energy is still exchanged with the surroundings.

And hm... I'm not exactly sure if Kaplan's right. I know when work is done by the closed system, the internal energy decreases, for obvious reasons. I'm not exactly sure why the opposite is false though? Are there any other factors involved? Maybe if temperature was held constant in the system, this might be true.

That's also why I was confused because in the passage it states that the temperature inside of the system increases as it gets compressed. But the AAMC solution also states that the work done on the system causes it to gain heat. So I wasn't sure how you would know that there is a transfer of heat?

 

[Czarcasm]

Kaplan says that when work is done on a closed system, the internal energy does not change and the work done results in the production of heat. Is this always the case?

I'm confused because I know that there are instances where work is done on a system, resulting in an increase in internal energy rather than in heat.

Can someone explain?

Thanks!

The only way the internal energy would remain constant is if the energy gained by doing work on the system is lost as heat. Internal Energy = change in heat + work. If you consider a gas in a piston, if we were to compress the piston, we're doing work on the gas. More specifically, the surroundings are doing work on the system (gas). If the container itself was adiabatic, meaning no heat is gained or lost (indicating q=0), then the internal energy would equal the energy gained or lost due to work (in this example, it would be energy gained; if you were to increase the volume, it would be energy lost). If they instead are referring to a non-adiabatic system, then change in internal energy would be zero because energy gained is lost as heat. This is generally the case for an isothermal process, where temperature is constant (indicating change in internal energy is zero).

 

[MCAT Tachycardia]

Do you guys know whether or not a reaction could be adiabatic and isothermal at the same time?