Download Thermodynamics of Ideal Gases: Work, Heat, and Internal Energy and more Study notes Physical Chemistry in PDF only on Docsity! or for infinitesimal changes • we defined work when studying mechanics as q = heat (discussed later) and w = work • where is the displacement against the opposing force • using our model system of an ideal gas inside a cylinder with a movable piston, we calculate the work based on the displacement of the piston which results in a change in volume of the ideal gas • work of expansion/compression (aka pV work) Ideal Gas • for infinitesimal changes sign convention Work of Reversible Isothermal Expansion • In this transformation the gas expands as weight is slowly removed from the piston. Since the temperature decreases due to the expansion, a thermostatic heat sink supplies heat to the system such that the temperature remains constant p(T, V) = nRT/V remove weight Heat Sink Δh Heat Sink Work of irreversible expansion against a constant external pressure • this occurs if we heat the gas in the cylinder very quickly • in this case the work is trivial to calculate • note: if the transformation is at constant volume then • note2: the external pressure must be > 0 for work to be done. • consider the transformation from {T, V1, p1} to {T, V2, p2} along the irreversible path below. • graphically this work is denoted by the shaded areaStep I Step II • consider a more direct transformation Along this path p = nRT/V • now the work of this reversible isothermal expansion Step III • the energy that flows from a body of temperature Th to a body of temperature Tc if Th > Tc • energy in the form of heat might also flow without a temperature change such as in a phase transition or isothermal expansion • heat transformations are best illustrated by example and require the definitions of the thermal coefficients (units: J/K) Heat capacity at constant pressure Heat capacity at constant volume Latent heat of a pressure change Latent heat of a volume change •Heat calculations and the first law (dU = δq + δw) constant volume (dV = 0) Thermodynamic definition of the heat capacity at constant V •To determine the heat released or absorbed at constant V if we assume CV is independent of T •If we know the heat capacity then we can measure the temperature change and determine the change in energy (which is equal to heat). This measurement is called bomb calorimetry. •We have implicitly imposed a sign convention on the heat (fortunately the correct one). A positive sign implies heat is absorbed by the system. •To determine the heat released or absorbed at constant p if we assume Cp is independent of T •If we know the heat capacity then we can measure the temperature change and determine the change in enthalpy (which is equal to heat). This measurement is called constant-pressure calorimetry (often done using a styrofoamcup) • We now have all of the background to calculate the work, the heat, the change in internal energy ΔU and the change in enthalpy ΔH for transformations in our model system. In addition, dU is an exact differential and U is a state function (a very important property; more on this later)By the way it is defined, dH must also be exact and H is a state function • Before doing an example we need to remember an important concept from the kinetic theory of gases – the energy of an ideal gas is proportional to the temperature.- The change in internal energy of an ideal gas in zero for an isothermal transformation. • Heat Capacity and • Enthalpy (H) and and • for an ideal gas the change in internal energy constconst • points 1 → 2 → 3 → 1 p1=p2=16.6 barV2=V3= 25 LT1=T3=p1V1/nR = 79.9 KT2=p2V2/nR = 2000 KCV,m= 20.79 J mol-1 K-1Cp,m= CV,m +R Step I (constant p expansion)
AU = nCy AT
dp = AH = NCp AT
V(X) w = AU — gp
AT = (T> —T,;) = 1920 K
AU = (2.50 mol)(20.79 Jmol-1K~1)(1920 K) = 99.8 kJ
gp = AH = (2.50 mol)(29.10 Jmol-*K—1)(1920 K) = 140 kj
w = AU — qp = —40.2 kJ
Step II (constant V cooling) constant VNow for the enthalpyThus,
