## Lecture question [ENDORSED]

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Parsia Vazirnia 2L
Posts: 44
Joined: Fri Jul 22, 2016 3:00 am

### Lecture question

Can someone further explain the diagram on page 30 of the course reader in which the mass is decreasing slowly? I was having a hard time understanding this during lecture.

Thanks

Vagarshak 2C
Posts: 6
Joined: Wed Sep 21, 2016 2:55 pm

### Re: Lecture question

Basically, the system Professor Lavelle sets up is almost similar to the system in the text book. The mass can be seen as the external pressure, Pex, and the heat reservoir is the heat entering the system to replace the energy lost by doing work. The point he was trying to make was that different systems can reach the same result; it is just a matter of the set up. For the system he sets up, as the weight of the mass decreases, the gas can expand which a type of work in itself
(w=-PdeltaV) which would cause the energy of the system to decrease but because there is a heat reservoir, there is no heat lost in the system and the total energy is 0 and q = -w. Also, the system is used to set up for the equation of work in an isothermal (constant T) system with a changing pressure.

AnkitaNair1E
Posts: 17
Joined: Wed Sep 21, 2016 2:55 pm
Been upvoted: 1 time

### Re: Lecture question  [ENDORSED]

Hi!
So essentially you have a container of an ideal gas with a volume V and a vacuum. (For the most part, you can just ignore the vacuum because its just part of the setup and only thing that happens to the vacuum is that it becomes smaller as the ideal gas expands.) This volume is being held by a external pressure which in this problem is the pulley with the mass. So, when we reduce the amount of mass on the pulley, we are effectively reducing the pressure. When pressure is reduced, the volume is allowed to expand and the gas does work on the surroundings. It's important to understand that when a system (the gas) does work, it loses energy to its surroundings.

Since we know that the gas is an ideal gas and that temperature is constant (isothermal) we can then use the equation (delta) U=(3/2)nR(delta)T to figure out the change in internal energy of the system. This equation is another way you can find the change in internal energy but it only works if you have an ideal gas in three dimensions (which we do). If you use this formula, you get (delta) U=0 since the temperature is constant and delta T =0
Since we know internal energy U=w+q we now know that w=-q. This tells us that the work done by the system is equal to the heat being put back into the system which is why our overall change in internal energy is zero.

Why is this important? Well, you'll see in the diagram directly below there's another set up where a gas is pushing against against a piston while heat is being put into the system. Even though these are two different experiments, they have the same result because both experiments show that the change in internal energy is zero. In both experiments, the amount of heat entering the system is equal to the amount of work the system does to expand. Ultimately, like Vagarshak said, I think the point of this example is to show you that two different systems can have the same result if you set it up the right way. Hope this helps!