- Physical Chemistry I
Spring 2012, Unique 52135
Lecture Summary, 27 and 30 January 2012
Law of Thermodynamics: heat + work = Internal
Heat and work are path functions, meaning that the amount of work done by a system or heat necessary to do it depends on the path the state takes from the initial to final states. In solving thermodynamic problems, it is always very important to define the path clearly; often the rest of the problem will be trivial. In the past couple days we have talked about a number of specific examples of the effect of path on work and heat.
--Free Expansion: expansion into a vacuum, where P(ext) = 0:
w = 0, deltaU = q
--Expansion against constant pressure, P(ext) = constant
w = -P(ext)detalV
--Isothermal reversible expansion, P(ext) = P(sys) at every point along the pathway:
w = -nRT ln (Vf/Vi)
At constant volume, no work can be done and the change in internal energy, deltaU, is entirely due to q:
deltaU = qv = CvdeltaT (this is subscript "v" meaning at constant volume
For an isothermal process, deltaT = 0 and deltaU = 0.
At constant pressure, the system can do work, and this is such an important case that we give this thermodynamic state another term, enthalpy:
deltaH = deltaU + delta(PV)
deltaH = qp = CpdeltaT (where subscript p indicates constant pressure).
Cp > Cv always. For an ideal gas, Cp - Cv = nR. You should know how to derive this.