## 7.15 clarifications

$\Delta U=q+w$

Denise 3L
Posts: 13
Joined: Fri Sep 26, 2014 2:02 pm

### 7.15 clarifications

I'm a little confused on the wording of the problem.
It gives two systems, one showing a solid to liquid process and another showing a vapor to liquid process.
the question asks to state whether work is being done on or by the system, what exactly does that mean? It's hard to visualize.

Sanmeet Atwal 1D
Posts: 17
Joined: Fri Sep 26, 2014 2:02 pm

### Re: 7.15 clarifications

When a solid turns into a liquid you know that heat is absorbed and q is positive and since the change is occurring at a constant temperature you know that ΔU = 0. From here you can use the formula: ΔU = q+w = 0 to solve for w. If w is positive you know that work was done on the system and if w is negative you know that the system did work.

For the solid-->liquid = +q (endothermic) because heat is absorbed by the solid.
so +q+w=0
q= -w
since w is negative, the system did work.

For the vapor---> liquid = -q(exothermic) because heat is released by the vapor.
-q+w=0
-q= -w
q= +w
since w is positive, work was done on the system.

Hope this helps!

martha-1I
Posts: 76
Joined: Fri Sep 26, 2014 2:02 pm

### Re: 7.15 clarifications

I think of it in terms of whether the system lost or gained energy. In a, the system is melting from a solid to a liquid. This mean that it is gaining energy as heat (therefore q is positive).To visualize it you can imagine the particles going from being very rigid in a tight space (solid) to flowing easily with little free space (liquid). To figure out work, you can use the equation W=P$\bigtriangleup$V. Since the pressure is constant, we only need to worry about the change in volume. Here the volume decreases making $\bigtriangleup$V negative since Tf is smaller than TI, so w is negative which means work is done by the system.
In b, the system is cooling from vapor to liquid. This means that it is losing energy (therefore q is negative). To visualize it you can again imagine the particle characteristics of a vapor and a liquid. In order to figure out work, pay attention to the change in volume. The volume increases in this example since the total amount of gas particles is less than the total amount of liquid particles making $\bigtriangleup$V positive. w is also positive which means work is done on the system.
Let me know if I need me to clarify it more.

Neil DSilva 1L
Posts: 70
Joined: Fri Sep 26, 2014 2:02 pm

### Re: 7.15 clarifications

Hey Martha,

Could you clarify how you got your $\Delta V$ values? In part a, you used $T_{f}$ < $T_{i}$ making $\Delta V$ negative (but temperature is constant so wouldn't $T_{f}=T_{i}$?). Then in part b, you said $\Delta V$ is positive, since there are more liquid particles than gas particles (how did you come to this conclusion?).

Justin Le 2I
Posts: 142
Joined: Fri Sep 26, 2014 2:02 pm

### Re: 7.15 clarifications

The problem doesn't specify that there is a constant external pressure so I don't think we can use w=-P$\Delta$V.

Neil DSilva 1L
Posts: 70
Joined: Fri Sep 26, 2014 2:02 pm

### Re: 7.15 clarifications

Justin,
Wouldn't open beakers (in the diagrams) indicate open systems that are exposed to the atmosphere and an implied 1 atm of external pressure?

Even then, I think Skatwal's explanation is the best way to approach this problem. Using $w=-P\Delta V$ just seems too complicated.