Compressed Adiabatic System

[DrKendrickLamar]

A cylinder with a sample of gas is compressed adiabatically by a piston. Which of the following statements are true?

1. The work done on the system is negative.
2. The work done by the surroundings is positive.
3. The temperature of the system increases.

A. I and II only
B. I and III only
C. II and III only
D. I, II and III

Answer: C

My reasoning:
2.
U=q-w
Since the volume is compressed, the work is negative on the system. So, U=q+w? and therefore, work done by the system is positive? and how would work done by the surroundings be positive?

3.
Compressing a volume will decrease space causing more collisions, which increases the temperature.

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[sleepy425]

um, yeah, well, i think you're on the right track.

first, looking at the choices, you know that you can eliminate A and D because I and II are opposite things, so both can't happen at the same time. In other words, the work is done by the surroundings ON the system, so both the work done by surroundings and the work done on the system should have the same sign. that's just how you can reason through eliminating some answers, but...

to the real thermodynamics! ok, so I don't know about you, but I find that the best way to think of work done and work lost is to think of the system as exerting a force upwards on the piston, while the surroundings exert a downward force on the system. now, if the gas is then compressed, that means that the work done by the surroundings is positive, because the piston went down (which is displacement in the same direction as the force exerted by the surroundings, so you get positive work). another way of saying this is that the work done ON the system is positive. at the same time, the work done BY (not ON) the system should be negative because the system exerts an upward force on the piston but it is still moving downwards (so displacement in the opposite direction of the force), so you get negative work.

so, we now know that II is correct, while I is incorrect.

What about 3?

ok then, if you have deltaU=q-w, well, remember that w is the work done by the system. so w is negative, and in an adiabatic process q=0. so deltaU will equal a positive value (negative of the negative work). Since deltaU=int(CvdT), if deltaU is positive and Cv is a constant, T must increase to give a positive deltaU.

sooo, temperature increases.

hope that helps.

 

[RSAgator]

A cylinder with a sample of gas is compressed adiabatically by a piston. Which of the following statements are true?

1. The work done on the system is negative.
2. The work done by the surroundings is positive.
3. The temperature of the system increases.

A. I and II only
B. I and III only
C. II and III only
D. I, II and III

Answer: C

My reasoning:
2.
U=q-w
Since the volume is compressed, the work is negative on the system. So, U=q+w? and therefore, work done by the system is positive? and how would work done by the surroundings be positive?

3.
Compressing a volume will decrease space causing more collisions, which increases the temperature.

I think questions like this one that assign a sign to the work are a little bit poorly worded and definitely confusing. By convention, work done on a system results in a positive change in energy in the system and a negative change in energy to the surroundings. Work done by a system results in a negative change in energy of the system and a positive change in energy of the surroundings. I think it's much better to understand that than to have the U = q - w equation memorized.

Temperature definitely increases because it's adiabatic. All the heat that is added stays in the system.

Work done on a system, as mentioned before, is positive. This is what's happening in this question. If "negative" work is done "on" a system, then this is the equivalent to the system doing work, so this answer is wrong.

If "positive" work is done "by" the surroundings, this is the same thing as saying positive work is done on the system. It's also the same as simply saying "work is done by the surroundings." If work is done by the surroundings it is done on the system. This answer is correct because work is being done on the system.