Chapter6LEcture

# Chemists usually speak not of e itself but of h which

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Chemists usually speak not of E itself, but of H, which is a modified energy. H = E + pV Generally, pV is a small amount compared to E If you add pV to E, you get H H is enthalpy. It is what chemists use. Every reaction has ΔH = H(products) H(reactants) If ΔH < 0, then the products have less H than the reactants. The difference goes into heating the surroundings. This is often the case for reactions that “go”, but not always. It was true for Na thrown into H 2 O!!

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6 Copyright: 2010 Prof. Magde Chapter 6: Work in Chemistry Doing work can be mechanical or electrical, mostly. Electrical is important in chemistry, both for its own reasons and for applications like batteries. That comes in 6C. We are interested now in mechanical work. Mechanical work is Force × distance = F•Δx Pressure is force per unit area. P = F/A So F = p•A So Work = p A Δx = p ΔV This has units of energy.

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6 Copyright: 2010 Prof. Magde Chapter 6: Work So work = W = p• V with a minus sign! This pV must have units of energy, because work changes E. Positive work increases the E of the system (and the H). To get more stored potential energy into a system, say the air in a tire, you need to compress the gas to a smaller volume. So reducing V increases E. So to increase W we need a negative V. Hence the minus sign above. A big problem is units of pressure. To work with joules, kg, m, you need Pascals. From the equation above, 1 Pascal = 1 J / m 3 But the p you know are 1 atmosphere and perhaps psi (pounds per square inch). 1 atm = 10 5 Pascals = about 14 psi.
6 Copyright: 2010 Prof. Magde Chapter 6: Work Continuing with W = p• V 1 atm = 10 5 Pascals = about 14 psi. We need the absolute pressure of the gas; but the pressures you know about for tires are all pressures above 1 atm. So a bicycle tire at 98 psi would be about 7 atm above 1 atm, or 8 atm absolute, or 8 10 5 Pascals, as intensive variable in the system Suppose we have a bicycle tire pump for which 1 stroke is 20 cm 4 cm 2 of volume change and suppose the p doesn’t change much so we can take it to be constant. Then E = W = p• V = 8 10 5 P 80 cm 3 10 -6 m 3 / cm 3 = 64 J Is 1 J a lot of work? 1 J can warm 1 mL of water less than 0.25 C. 4.18 J = 1 calorie, and 1000 Cal = 1 food calorie or 1 kCal 1 jelly donut is about 10 6 J or 1 MJ

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6 Copyright: 2010 Prof. Magde Chapter 6: Work By the way: We no longer use atm. The new unit is bar. One bar is close to one atm. This is better, because 1 bar = 10 5 P exactly. Suppose we start with a vacuum and put some gas in. How much work does that require? This is called a free expansion. The gas will expand freely, by itself, into the volume. As long as there is so little gas that the pressure remains essentially zero, no work is needed. And that makes sense. This means that no W is done and E stays constant (if there is no Q either).
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