## 4F.1 7th Edition

$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

Dakota_Campbell_1C
Posts: 51
Joined: Fri Sep 28, 2018 12:15 am

### 4F.1 7th Edition

For part A, when solving for change in enthalpy why do we place a negative in front of energy transferred as heat (q)?

Annalyn Diaz 1J
Posts: 61
Joined: Fri Sep 28, 2018 12:15 am

### Re: 4F.1 7th Edition

I'm not entirely sure, but it may be negative because the question is asking how body heat affects the entropy in your surroundings?

Jessica Castro 2H
Posts: 60
Joined: Fri Sep 28, 2018 12:29 am

### Re: 4F.1 7th Edition

The amount of heat given is the heat generated by your body (which is the system). The question is asking for the entropy of the surroundings. Therefore, when plugging in the heat of the surroundings, remember q system = - q surroundings. The heat of a system is equal and opposite to the heat of its surroundings.

Posts: 60
Joined: Fri Sep 28, 2018 12:19 am

### Re: 4F.1 7th Edition

Because qsystem=-qsurroundings, the amount of heat given off by your body (the system) is the same amount of heat absorbed by the surroundings.

LilyL1C
Posts: 24
Joined: Fri Sep 28, 2018 12:20 am

### Re: 4F.1 7th Edition

The heat of the system is equal and opposite to the heat of the surroundings. qsystem = -qsurroundings. The heat given off by the system is the same amount of heat that is absorbed by the surroundings because the heat of the universe has to stay constant.