Page 1 of 1

### 4A.3

Posted: Tue Jan 28, 2020 4:50 pm
On part C of this problem, why would the change in internal energy of the system be different then how much work is done in the compression?

### Re: 4A.3

Posted: Tue Jan 28, 2020 6:29 pm
Sorry this is for 4A.1 but I didn't know where else to ask. How can we tell if a system is open, closed, or isolated? Mainly, what is the difference between closed and isolated? I got those two mixed up

### Re: 4A.3

Posted: Tue Jan 28, 2020 6:52 pm
Micah3J wrote:Sorry this is for 4A.1 but I didn't know where else to ask. How can we tell if a system is open, closed, or isolated? Mainly, what is the difference between closed and isolated? I got those two mixed up

Micah- a closed system by definition according to the textbook "has a fixed amount of matter, but it can exchange energy with the surroundings" like a shake activated icepack. An isolated system by textbook definition "can exchange neither matter nor energy with the surroundings" like an insulated thermos that is sealed with the lid. In an isolated system the matter is constant and is not effected by anything in its surroundings, and a closed system stays constant with matter but can be affected by the energy of its surroundings.

### Re: 4A.3

Posted: Tue Jan 28, 2020 6:59 pm
BeylemZ-1B wrote:On part C of this problem, why would the change in internal energy of the system be different then how much work is done in the compression?

w= -Pex(deltaV)
where Pex= constant external pressure
delta V= Vfinal-Vinitial ( the expansion/compressiom of the system)

when deltaV is positive , the internal energy of the system decreases when the system expands and vice a versa.

they would be the same but you need to express them with the value of w as work done, the deltaU (internal energy) with a + or - in front of the value to express how the internal energy changed.

### Re: 4A.3

Posted: Tue Jan 28, 2020 11:19 pm
KaleenaJezycki_1I wrote:
BeylemZ-1B wrote:On part C of this problem, why would the change in internal energy of the system be different then how much work is done in the compression?

w= -Pex(deltaV)
where Pex= constant external pressure
delta V= Vfinal-Vinitial ( the expansion/compressiom of the system)

when deltaV is positive , the internal energy of the system decreases when the system expands and vice a versa.

they would be the same but you need to express them with the value of w as work done, the deltaU (internal energy) with a + or - in front of the value to express how the internal energy changed.

They aren’t the same in the answer key though.

### Re: 4A.3

Posted: Wed Jan 29, 2020 9:26 am
BeylemZ-1B wrote:They aren’t the same in the answer key though.

That's strange. In my answer key, they are indeed the same (about 28 J).

### Re: 4A.3

Posted: Thu Jan 30, 2020 1:28 pm
How do we know that the pressure is constant for this system?

### Re: 4A.3

Posted: Thu Jan 30, 2020 1:29 pm
Also, how would we know the sign of the internal energy?

### Re: 4A.3

Posted: Thu Mar 05, 2020 1:56 am
Angela Patel 2J wrote:Also, how would we know the sign of the internal energy?

If energy enters the system: +
If energy leaves the system: -
If work is done on the system: +
If work is done by the system: -

### Re: 4A.3

Posted: Sat Mar 14, 2020 9:42 pm
Charlyn Ghoubrial 2I wrote:
Angela Patel 2J wrote:Also, how would we know the sign of the internal energy?

If energy enters the system: +
If energy leaves the system: -
If work is done on the system: +
If work is done by the system: -

Thank you for this quick check for the internal energy