## 8.5

isochoric/isometric: $\Delta V = 0$
isothermal: $\Delta T = 0$
isobaric: $\Delta P = 0$

Kassandra Molina 2B
Posts: 43
Joined: Wed Nov 16, 2016 3:03 am

### 8.5

In the question, it says 340 kJ of work is compressed so shouldn't it be negative (because it is being compressed) when we add it to 524 kJ of heat? (so instead of 864 kJ it is 184 kJ?)

Hubert Tang-1H
Posts: 44
Joined: Thu Jul 27, 2017 3:00 am

### Re: 8.5

We are measuring the internal energy of the system. Compressing the gas with 340 kJ of work. Remember that the Work done in a a system is w= - P*delta_V. The sign is negative, and because the system is compressed, delta_V is negative, meaning work done is positive, since compressing the system does work on the system, meaning it puts energy into the system. Therefore, compressing causes the work to be positive.

Nhan Nguyen 2F
Posts: 52
Joined: Thu Jul 13, 2017 3:00 am
Been upvoted: 1 time

### Re: 8.5

Delta U (change in internal energy)= q (heat) + w (work)
It said the piston is doing the work of 340 KJ ONTO the gas in the system.
Since we're focusing on the gas, this means that the gas is receiving positive work done to it (+340 KJ)
so, U= 524 KJ heat + 340 KJ work = 864 KJ

Ishita Monga 1B
Posts: 31
Joined: Thu Jul 27, 2017 3:00 am

### Re: 8.5

The "work" isn't being compressed, the system as a whole (in this case the gas in the piston) is compressed due to a certain amount of work being done to it. When work is done to the system, the value of the work is positive.

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