MCAT CBT8 Item 28

[rls303]

Hi,
I was wondering for this question...why would work being by the gas in the balloon expanding lower it's internal energy? For some reason I just don't see it?

Thanks in advance!

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[wrigley09]

I just did this exam yesterday and thought it was brutal! Anyway...

U = Q - W (U is internal energy, Q is heat exchange, W is work)

Work= PdeltaV If work is being done BY the gas, then work is positive. The question doesn't say anything about heat exchange, so assume it's constant. So U = -W Since work is positive, U is negative (it decreased)

 

[crazyasian]

The internal energy is being converted to work. W=FD. Since the balloon is expanding work is being done. Just remember conservation of energy. The energy to do work has to come from somewhere. Most often times when a gas expands, if there's no external energy flowing back into the system, then the gas loses energy in the form of work being done (which is why the temp goes down).

 

[rls303]

Man! Thanks guys! I keep on getting screwed with minor things like these! eek!

 

[rls303]

I just did this exam yesterday and thought it was brutal! Anyway...

U = Q - W (U is internal energy, Q is heat exchange, W is work)

Work= PdeltaV If work is being done BY the gas, then work is positive. The question doesn't say anything about heat exchange, so assume it's constant. So U = -W Since work is positive, U is negative (it decreased)

Hi,
Just wanted to clarify, if we use the equation:

dE=Q+W

Work done by the gas is considered as negative and work done on the gas is positive, correct?

Thanks!

 

[wrigley09]

Well hang on. I wrote something and now I'm not sure...

 

[wrigley09]

Okay, after flipping through a bunch of my old lecture notes and googling, I see that apparently there are two different sign conventions. I was taught this way both in classes at my university and in the Kaplan course:


Work done by the gas is positive, always. Work done on the gas is always negative. I think you're confused because I said U=-W. This equation only holds if Q is zero (Q is heat exchange. If no heat is exchanged with the environment, Q = 0). U = Q-W is the equation that always holds. If Q = 0, the equation becomes U = 0-W or U = -W. Since the work was done by the gas, work is positive. U must be the opposite sign, so it is negative.

So if, for example, volume was constant, that makes W = 0 (because W = PdeltaV. DeltaV = 0, then W = 0). So U = Q in that case.

BUT, if you're using the formula that you wrote, dE=Q+W (plus, not minus like in my formula), then you switch the sign convention and work done by the gas is negative and work done on the gas is positive.

I know that doing it the way I was taught always gets me the right answer. I don't know about that other formula. Someone correct me if I'm wrong...

 

[rls303]

Okay, after flipping through a bunch of my old lecture notes and googling, I see that apparently there are two different sign conventions. I was taught this way both in classes at my university and in the Kaplan course:


Work done by the gas is positive, always. Work done on the gas is always negative. I think you're confused because I said U=-W. This equation only holds if Q is zero (Q is heat exchange. If no heat is exchanged with the environment, Q = 0). U = Q-W is the equation that always holds. If Q = 0, the equation becomes U = 0-W or U = -W. Since the work was done by the gas, work is positive. U must be the opposite sign, so it is negative.

So if, for example, volume was constant, that makes W = 0 (because W = PdeltaV. DeltaV = 0, then W = 0). So U = Q in that case.

BUT, if you're using the formula that you wrote, dE=Q+W (plus, not minus like in my formula), then you switch the sign convention and work done by the gas is negative and work done on the gas is positive.

I know that doing it the way I was taught always gets me the right answer. I don't know about that other formula. Someone correct me if I'm wrong...

Ahh! That makes perfect sense now. Thanks so much for explaining it. All these book companies have so many variations that it gets hard to keep things straight at times.

Thanks again!