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.
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
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
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.
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.
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).
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 )
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...