11/12.3 Internal Energy, Enthalpy, and Cp/Cv

Chad breaks down Internal Energy, heat, work, and Enthalpy and their relationships to Constant Volume and Constant Pressure Heat Capacities.
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[@melissastan6884]

Hi Chad, In my textbook, it stated that the equation is delta U = Q - W, but you wrote delta U = Q + W. Is it supposed to be + or - W. If it is -W could you please explain why we are subtracting W from Q. Thanks. Awesome videos by the way.

[@ChadsPrep]

Hello Melissa! You'll see this presented two ways depending upon the textbook. dU = Q + W where W is defined as -PdV and therefore dU = Q - PdV. The second way is dU = Q - W where W is defined as +PdV and therefore dU = Q - PdV. So they both lead to the same equation but differ in how work is defined. In the first (the one I use) work is defined relative to the system. When the system gains energy work is positive and when the system loses energy work is negative. But in the second work is defined as work that can be performed by the system. When work can be extracted from the system (and performed on the surroundings) it is defined as positive, but when work is done on the system by the surroundings it is defined as negative. So one focuses on the system itself whereas the second focuses on what can be done by the system. My apologies for the confusion but I didn't create it, we're just left to deal with it. Happy Studying!

[@gerardo5211]

DH= DU + P(DV), P should be constant so no delta is neded

[@ChadsPrep]

Correct Gerardo! H = U + PV so DH = DU + D(PV) This is often expanded to DH = DU + PDV + VDP But we often consider examples carried out at constant pressure in which case DP = 0 and therefore DH = DU + PDV You can actually expand this out further by substituting in for DU (DU = q+w) to show that DH = q at constant pressure. Happy Studying!

[@gerardo5211]

Chad's Prep ok I get it, you explain very good, thank you for your videos!