Two key definitions before we begin:
- thermodynamics = the study of energy and its interconversions.
- 1st Law of Thermodynamics = the total energy of the universe is constant.
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The Internal Energy of a System (E)
- think of the "system" as the "chemical reaction."
The Internal Energy of a system - the sum of the kinetic and potential energies of all the "particles" in the system (the reaction).
➞ this internal energy (E) can be changed by a flow of work (w), heat (q), or both:
ΔE = q + w
As the image above suggests, let's examine the specifics of heat (q) and work (w)...
... the signs (+ or -) of heat (q) and work (w) are identified from the point of view of the system:
A. if the reaction is endothermic, heat flows into the system and thus heat (q) is positive (+) ➞ q > 0
B. if the reaction is exothermic, heat flows out of the system and thus heat (q) is negative (-) ➞ q < 0
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C. if the reaction (system) does work (w) on the surroundings, energy flows out of the system, so work (w) is negative (-) ➞ w < 0
D. if the surroundings do work (w) on the system, energy flows into the system, so work (w) is positive (+) ➞ w > 0
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ex: Calculate ΔE for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1400 J of work is done on the system.
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How to Calculate Work (w)
The "work" associated with chemical processes is usually work done by gases (through expansion) or work done to gases (through compression).
Let's examine 2 CASES (below) and the following relationship between work, external pressure, and volume: w = - PΔV
- external pressure (P) = usually a constant in these types of problems, so w depends on ΔV
CASE 1 ➞ If a gas expands, the volume of the system increases, so the change in volume (ΔV) is positive (+).
Because pressure (P) is always positive (+), then the work (w) must be negative (-).
➞ this corelates well with "C" above because the reaction is doing work on the surroundings in order to expand the volume.
-- so, w < 0 (-) because energy is flowing out of the system.
CASE 2 ➞ If a gas gets compressed, the volume of the system decreases, so the change in volume (ΔV) is negative (-).
Because pressure (P) is always positive (+), then according to "w = -PΔV", the work (w) must be positive (+).
➞ this corelates well with "D" above because the surroundings are doing work on the system in order to compress the volume.
-- so, w > 0 (+) because energy is flowing into the system.
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ex: Calculate the work associated with the expansion of a gas from 46L to 64L at a constant external pressure of 15atm.
This is an "expansion" (CASE 1).
Don't think of the (-) minus sign as a "negative number," but instead think of the (-) sign as a direction.
➞ since the gas expands, work is negative (case 1).
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ex: A balloon is being inflated by heating the air inside it. The volume of the balloon increases from 4.00 x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J of energy as heat.
Assuming that the balloon expands against a constant pressure of 1.0atm, calculate ΔE for this process. (note: 1 L.atm = 101.3 J )
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Next, we'll talk about how "heat" (q) is measured experimentally using a concept called Calorimetry.
So, stick around for my next post on SECTION 6 - Thermochemistry,
where we'll discuss Calorimetry and Specific Heat Capacity...