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What is the sign convention for work in dE = q - w

  1. May 2, 2015 #1
    Why is it not just dE = q + w because I assume if you use general sign convention and take work to be negative when it is done by the system and positive when it is done on the system, you should get the right answer. Silly question but I'd appreciate some clarity on this formula please?
  2. jcsd
  3. May 2, 2015 #2

    Doc Al

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    Staff: Mentor

    I believe that the convention to take work done on the system as positive is catching on. (It's the one I prefer.) I believe the other convention is just historical. With a heat engine, you put heat in and get work out, so they thought it natural to make those two positive.

    As long as you are consistent, either convention works.
  4. May 10, 2015 #3

    James Pelezo

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    Gold Member

    In some textbooks on the 1st Law of Thermodynamics, ∆E = q + w while in some use ∆E = q - w. The difference is when using ∆E = q + w, one must determine if the process of interest is undergoing 'expansion' or undergoing 'compression' and assign a sign for endothermic work (+) or exothermic work (-). Using ∆E = q - w this is unnecessary. For example, consider 3H2(g) + N2(g) => 2NH3(g). Note there are 4 molar volumes of gas on the reactant side of the equation and 2 molar volumes of gas on the product side of the equation. The change in molar volume from reactants to products is a decrease in 2 Vm of gas. This is considered a 'compression' type reaction in which work is endothermic and a (+) sign must be assigned to work. Visualize ... squeezing a balloon ... the molecules of air (gas) push back on your hands indicating they have a higher energy;i.e., endothermic work. Some texts refer to this as the surroundings doing work on the system. So when using ∆E = q + w it becomes ∆E = q + (+w). For the above ammonia reaction, the heat of reaction ( forming 2 moles ammonia ) is q = -91.8 Kj and work is w = +5 Kj. The change in internal energy ∆E = q + w = (-91.8 Kj) + (+5.0 Kj) = -86.8 Kj net change in internal energy. Now, consider the evaporation of 1 mole of water: H2O(l) => H2O(g). This process is going from 0.00 molar volumes for liquid water (only gases have molar volumes) to 1.00 molar volumes for gas phase water. This process is considered an 'expansion' process and work is exothermic. One must assign a (-) sign to the work giving ∆E = q + (-w). The heat of vaporization of water is q = 40.7 Kj and work for expansion is w = - 2.5 Kj, giving ∆E = q + w = (+40.7 Kj) + (-2.5 Kj) = +38.2 Kj net change in internal energy. The assignment of the (-) sign is b/c expanding gases will lose work energy and hence be exothermic. Visualize the balloon experiment again, but this time visualize the balloon instantaneously disappearing, or popping the balloon. The gas expands rapidly and ( I like to think of the gas particles as relaxing ) loses a small amount work energy. This is referred to as the system doing work on the surroundings.

    Now the same results can be obtained using ∆E = q - w, but this time substitute the calculated change in molar volume along with the sign imposed by the calculation. That is, for the ammonia reaction the change in molar volume is the same decrease in molar volumes, or Vm = -2. The work of 2 molar volumes is ~ -5.0 kj. Including heat of reaction, the calculation is ∆E = q - w = (-91.8 Kj) - (-5.0 Kj) = (-91.8 Kj + 5.0 Kj) = -86.8 Kj. Same as the above calculation, but the sign of the decreasing volume is substituted along with the change in molar volume calculated. The water expansion is done by the same logic. Water(l) => Water(g) gives a 1 molar volume increase, or Vm = +1.00V m => work = ~ +2.5 Kj. Subbing the heat of vaporization q = +40.7 Kj, gives ∆E = q - w = (+40.7 Kj) - ( +2.5 Kj) = (+40.7 Kj - 2.5 Kj) = +38.2 Kj. Same as above evaporation problem.

    Please note, the calculation of work values is a quick estimation using w = 2.5(∆Vm). If using the ∆E = q + w the equation becomes
    ∆E = q + [2.5∆V m] in which you must sub an assigned sign => (-) for expansion and (+) for compression, but if using ∆E = q - w the equation becomes ∆E = q - [2.5∆V m] in which you simply calculate the change in volume and include the imposed sign of the change.

    Here's a link to some of my thermodynamic lectures. Use them if you wish. This is an open website, free for anyone interested in topics in general chemistry. Hope this helps. http://www.chemunlimited.com/vidsx312sp15.html
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