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Work: Sign Convention Problem 
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#1
Apr909, 03:28 PM

P: 25

1. The problem statement, all variables and given/known data
One mole of an ideal gas at 300K expands isothermally from a pressure of 2x10^6 Nm^2 to a pressure of 2x10^5 Nm^2. Calculate the work done ON the gas. 2. Relevant equations dW = PdV 3. The attempt at a solution Following through the integral, it becomes clear that work is either plus or minus 5.74kJ. My teacher has said he uses the sign convention that work done ON a system is always negative. I'm trying to understand what this means. Imagine a piston, if you push the piston you will do work, but if you pull the piston are you not also doing work too? According to the convention does this mean you will do negative work on the gas if you pull OR push? The reason I'm getting so confused is that in the mark scheme, the answer is given as 5.74kJ. This means the work done ON the gas in an expansion is negative. But later on, I'm told, using the same sign convention, compressing a gas means negative work has been done ON the gas. Surely that is a contradiction? Has my teacher confused himself?? 


#2
Apr909, 03:39 PM

P: 3,015

I belive that generally in Chemistry, Work done BY the system is negative, i.e., it LOSES energy.
In engineering generally work done ON the system is negative (since we want to design systems that DO work) It really is just a convention. So use whatever your prof uses. Hope that helps a little. I know it can be confusing. 


#3
Apr909, 04:08 PM

P: 25

Ah yes but I guess the main thing that is confusing me is: do you do work ON the system when you PUSH the piston or is it when you PULL the piston and why? The only convention I've been told is "work done ON the system is negative".



#4
Apr1009, 05:33 PM

Mentor
P: 12,068

Work: Sign Convention Problem
It all has to do with the basic definition of work,
W = F·ΔxWork is positive if F and Δx are in the same direction, and negative if they are in opposite directions. Imagine you are pushing on a piston containing a gas. You are exerting a force on it (the "system"), and it is exerting an equalbutopposite force back on you (the "surroundings") If the gas is allowed to expand as you push on it, the force on the system (i.e. the gas) is opposite to the displacement, and so there is negative work done on the system. Also, the force on the surroundings (i.e. you) is in the same direction as displacement, so positive work is done on the surroundings. In other words, the system does work on the surroundings, or positive work is done by the system rather than on the system. If a gas compresses, things are the other way around. In thermodynamics, one adopts a definition of "work" and specifies whether it is positive for work done by the system or on the system. As far as I'm aware, it's common to say work done by the system is positive, or in other words: dW = P dVHope that helps. 


#5
Apr1009, 07:19 PM

P: 25

So using the convention work done BY the system is positive, is it correct to say if a piston is pushed out by a gas, the gas has done positive work on the surroundings. Does this mean work done by the surroundings on gas is negative? Are the two statements equivalent?
BUT looking at it from the other way, the convention implies that if work done BY the system is positive, then work done ON the system is negative. So you push the piston, you are doing work ON the gas, which by definition is negative. So in both cases the surroundings is doing negative work but in the first case the gas expands and in the second the gas contracts. Surely that is a contradiction!? 


#6
Apr1009, 08:13 PM

Mentor
P: 12,068

The convention implies that IF work is done BY the system, then W is positive. And if work is done ON the system, then W is negative. If you push the piston, work is done ON the system, and W is negative. Or put another way: If the gas expands, (positive) work is done BY the system and W is positive. If the gas compresses, (positive) work is done ON the system and W is negative. Sorry if anything in my earlier post confused the issue. It can be tough to get a handle on these things. 


#7
Nov1909, 09:34 AM

P: 15

I think it is a good idea to refer to the internal energy of a system and then fix the sign of work and heat. Usually, a system changes from one state to another state by exchanging heat and work (two different forms of energy) with the surroundings.
Heat supplied to the system increases the internal energy (U) of the system. Similarly work supplied (that is, work done on the system) also increases U. On the other hand, heat lost and work done by the system result in a decrease in U. Thus, it is convenient to regard heat supplied and work done on the system as positive since they cause an increase in U. The above convention leads to the convenient expression for first law of thermodynamics: dU = q + w (Increase in energy = sum of energies supplied) where dU = increase in internal energy, q = heat supplied, w= work done on the system. 


#8
Dec2410, 08:52 AM

P: 6

mpkannan makes it clear you must look at everything in context before deciding the signs on everything. A question like "what is the sign on work" is too vague.
Two statements of the First Law of Thermo seen in textbooks: dU = Q + W and dU = Q  W W in the first statement is "work on the gas" W in the second statement is "work by the gas" And so, for example, when a gas expands W would be negative in the first statement (work done on the gas is negative since positive work is being done on the surroundings). BUT when a gas expands W would be positive in the second statement (work done by the gas is positive since work is being done on the surroundings). The fact that work is negative in one case but positive in the other (both for the same scenario) is not a contradiction, since in one case we're talking about work ON and in the other we're talking about work BY. Textbooks should present BOTH statements of this first law, CLEARLY defining that the first statement is work ON and the second is work BY, in my opinion. 


#9
Oct1712, 01:35 AM

P: 1

work done on the system is ve and by the system is +ve.



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