Work (sign convention)

[Richard Baddock]

I'm studying from a Kaplan book and it states "Work is done by a gas on its surroundings (positive work) increases the volume of a gas. Work done on a gas (negative work) decreases the volume of the gas". They also give the equation for pressure volume work as:
W = P deltaV

Heres my question. Isn't all that I stated above old convention? When a system does work on its surroundings, shouldn't the sign of the work be negative (the same way it would be if we were talking about heat)? In physical chemistry we always said:
W = -PdeltaV
This would seem to make more sense when considering:
deltaE(sys) = q + w

If anyone could give me a clear explanation of this, that would be much appreciated. Also, if anyone knows which is the convention used on the MCAT, that would be helpful information as well.

---

Members don't see this ad.

 

[koopa_troopa]

I had the same problem with them too. I just used what I was taught in chem class, which is work done on the gas is positive.

 

[cwfergus]

I had the same problem with them too. I just used what I was taught in chem class, which is work done on the gas is positive.

This is what i was learnt in Physics/chem

 

[physics junkie]

Doesn't matter as long as you are consistent. There is no new or old convention...different disciplines(engineering, chemistry, physics, etc.) set different sign conventions.

 

[Richard Baddock]

Doesn't matter as long as you are consistent. There is no new or old convention...different disciplines(engineering, chemistry, physics, etc.) set different sign conventions.

This may be true but it would seem to matter if a problem was asking for a numerical answers and for example one answer was 1000J and another answer was -1000J. There are probably other examples along those lines that would make answering a problem like this very tricky if the convention isn't stated.

 

[cwfergus]

My kaplan book says, "Work done by a gas on its surroundings is positive becasue the volume of the gas increases while work done on a gas is negative because the volume of the gas decreases"

 

[Richard Baddock]

My kaplan book says, "Work done by a gas on its surroundings is positive becasue the volume of the gas increases while work done on a gas is negative because the volume of the gas decreases"

Alright, but Work is a form of Energy (joules). If the gas (system) does work on the surroundings the volume may increase, but the energy of the gas decreases (because it did work on the system). The system losing energy (for example heat in an exothermic reaction) is usually written with a negative sign. I'm not sure how volume increasing or decreasing would make one convention more prevalent than the other. Correct me if any of my thinking is wrong please.

 

[cwfergus]

Alright, but Work is a form of Energy (joules). If the gas (system) does work on the surroundings the volume may increase, but the energy of the gas decreases (because it did work on the system). The system losing energy (for example heat in an exothermic reaction) is usually written with a negative sign. I'm not sure how volume increasing or decreasing would make one convention more prevalent than the other. Correct me if any of my thinking is wrong please.

I know, its confusing because i was ALWAYS taught in physics that W= p(delta)V and that if POSITIVE, work is being done ON the system.

But. what they are trying to say i think is that W= p (delta)V. So also,
W= P (Vf-Vi)
So if the final volume is less than the initial, (volume decreases) it will be negative and thus be done ON the gas. I dont know what we should expect though question wise

 

[Richard Baddock]

Maybe someone can jump in here and tell us which way we should learn this for the actual MCAT?

 

[tncekm]

The only way I was able to make sense of this is to understand how PV work related to changes in internal energy and the energy of the surroundings.

I.e. if we have an adiabatic expansion of a gas we know that the gas is expanding and there is not heat transfer. So, what is happening is that the gas is transferring work energy to the environment meaning that it lost internal energy to its surroundings. Whether its + or - depends on the formula you're given. For the MCAT its much more important just to know where the energy is going.

 

[Quantum Chem]

I think that you should also see how Kaplan defines change in energy. For example, the EK chemistry book (3.4-3.5) that I have says delta(E) = q + w, which means that PV work must be w=P*delta(V).

I know that in my Thermo P. Chem class we always used w=-P*delta(V), but also used delta(E)= q -w. Like one poster has already stated, it doesn't matter as long as you are consistent, or you follow what the passage states.

As for a MCAT type question, imagine that they give you a passage describing some sort of gas cycle sort of like how your car operates, and the passage states "when the fuel is combusted and expands energy leaves the gas through heat and work done on the piston, decreasing the gas's internal energy". Then later a question asks "A car is driving alongside the beach, how much work is done on a piston if the gas inside the piston expands from X liters to 2X liters?" Then you would need to answer the question with the final answer being positive since in the question the system is the piston.

You could also hope that there isn't any answers that are the same, just different signs

 

[physics junkie]

Maybe someone can jump in here and tell us which way we should learn this for the actual MCAT?

The MCAT would not give you the same answer with two different signs for an energy problem. The AAMC is not trying to trick you. Setting up a question like that is not actually testing any knowledge or critical thinking ability so they would not ask it. If they did ask you such a question they would state the sign convention in the passage.

 

[tncekm]

The MCAT would not give you the same answer with two different signs for an energy problem. The AAMC is not trying to trick you. Setting up a question like that is not actually testing any knowledge or critical thinking ability so they would not ask it. If they did ask you such a question they would state the sign convention in the passage.

Yeah, exactly. They'll ask you what happened to the energy or how the energy of the system and the surroundings changed.

 

[Richard Baddock]

Sounds good, thanks to everyone for responding.

 

[unsung]

I'm studying from a Kaplan book and it states "Work is done by a gas on its surroundings (positive work) increases the volume of a gas. Work done on a gas (negative work) decreases the volume of the gas". They also give the equation for pressure volume work as:
W = P deltaV

Heres my question. Isn't all that I stated above old convention? When a system does work on its surroundings, shouldn't the sign of the work be negative (the same way it would be if we were talking about heat)? In physical chemistry we always said:
W = -PdeltaV
This would seem to make more sense when considering:
deltaE(sys) = q + w

If anyone could give me a clear explanation of this, that would be much appreciated. Also, if anyone knows which is the convention used on the MCAT, that would be helpful information as well.

The way I remember this work sign crap is this:

First, I always think of it in terms of work BY the gas. If the Q asks for work done ON the gas, I solve it first for work BY the gas, and just flip the sign on the answer.

Anyway, if a gas is being compressed, what's happening? W=Fd, and in this case, the displacement of the gas is in the OPPOSITE direction of the force exerted BY the gas. I.e. the gas is pushing up against the piston trying to compress it. Therefore, work done BY the gas as it's being compressed is (-).

Conversely, if a gas is expanding against some outside force, the force BY the gas is in the same direction as its direction of displacement. Therefore, work BY the gas as it's expanding is (+).

So, in terms of formulas, if you work in terms of work BY the gas, we have deltaEint = q - Wby.

And for PV diagrams, as you're moving from L --> R (increasing V), you can see that Wby the gas would be (+) (alternatively, work ON the gas moving from L --> R would be (-).

Did that help?

 

[treeclimbingmonkey]

The way I remember this work sign crap is this:

First, I always think of it in terms of work BY the gas. If the Q asks for work done ON the gas, I solve it first for work BY the gas, and just flip the sign on the answer.

Anyway, if a gas is being compressed, what's happening? W=Fd, and in this case, the displacement of the gas is in the OPPOSITE direction of the force exerted BY the gas. I.e. the gas is pushing up against the piston trying to compress it. Therefore, work done BY the gas as it's being compressed is (-).

Conversely, if a gas is expanding against some outside force, the force BY the gas is in the same direction as its direction of displacement. Therefore, work BY the gas as it's expanding is (+).

So, in terms of formulas, if you work in terms of work BY the gas, we have deltaEint = q - Wby.

And for PV diagrams, as you're moving from L --> R (increasing V), you can see that Wby the gas would be (+) (alternatively, work ON the gas moving from L --> R would be (-).

Did that help?

Yes!!