$\Delta U=q+w$

Posts: 53
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 1 time

For question 8.7, how does one differentiate work being done on a system vs. work being done by a system? Also, how does one calculate work once this distinction has been made?

Jaewoo Jo 2L
Posts: 31
Joined: Fri Sep 29, 2017 7:06 am

### Re: Question about Question 8.7

Internal energy increased more than the amount of heat added, so the extra energy is from the work done to the system.
Since the equation to calculate work is "w= deltaU-q", plug in the values of change in internal energy and heat absorbed given in question to get "982 J - 492 J = +4.90*10^2 J

Christy Zhao 1H
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

### Re: Question about Question 8.7

On the other hand, if work was done by the system, the internal energy of the system would have decreased.

Nickolas Manipud 1C
Posts: 60
Joined: Fri Sep 29, 2017 7:07 am

### Re: Question about Question 8.7

Work is done ON a system when the system's internal energy increases. This makes sense if you think about a system being compressed (which is work being done ON the system). Since it's being compressed, more pressure is created, leading to more internal energy in that system.