Chapter 4: Applications of the First Law Different types of work: Configuration work: (reversible process) Dissipative work: (irreversible process) Adiabatic work: (independent of path) Work: not a system property, i.e. not a state variable Conventional sign of work: on (-) or by (+) the system Expansivity and isothermal compressibility

4.2 Mayer’s equation Heat capacity: limiting ratio of …… ( not an exact differential !!) Heat capacity depends on the conditions at which heat transfer takes place Specific heat capacity c v and c p For the ideal gas system: c p – c v = R Calculate the internal energy of an ideal gas system based on the definition of c v

If c v is independent of temperature, then u = u 0 + c v (T – T 0 )

4.3 Enthalpy and Heats of Transformation The heat of transformation is the heat transfer accompanying a phase change. Phase change is an isothermal and isobaric process. Phase change only entails a change of volume. thus the work (only configuration work) equals: w = P(v 2 – v 1 )

Variation in the internal energy is expressed as: du = dq - Pdv for a finite change: u 2 – u 1 = l – P(v 2 -v 1 ) where l represents the latent heat of transformation. thus l = ( u 2 + Pv 2 ) – (u 1 + Pv 1 ) At this point, introducing h = u + Pv (the small h denotes the specific enthalpy) Therefore, the latent heat of transformation is equal to the difference in enthalpies of the two phases.

Conventional notation: 1 denotes a solid, 2 a liquid and 3 a vapor, i.e. h’ represents the enthalpy of solid, h’’ is the enthalpy of liquid, …. l 12 = h’’ – h’ represents solid to liquid transformation (fusion). l 23 = h’’’ – h’’ represents liquid to vapor transformation (evaporation). Enthalpy is a state function, i.e. integration around a closed cycle produces 0!! (see Fig. 4.2)

4.4 Relationships involving enthalpy The natural choice in the variable h is h = h(T, P) The analysis can be proceeded in the same way as to the internal energy u

As Thus: Since: then

For an ideal gas Then Since for ideal gas h depends on T only,

4.5 Comparison of u And h See Table 4.2 in the textbook

4.6 Work done in an adiabatic process In adiabatic process: dq = 0. The equation dq = c p dT – vdP can be rearranged into vdP = c p dT Similarly, one gets Pdv = -c v dT Dividing the above two equation:

Assuming The integration of the above equation leads to where K is a constant Similarly, one gets

The work done in the adiabatic process is For a reversible adiabatic process: w = u 1 – u 2 = c v (T 1 – T 2 )

Example: An ideal monatomic gas is enclosed in an insulated chamber with a movable piston. The initial values of the state variables are P 1 = 8atm, V 1 = 4 m 3 and T 1 = 400K. The final pressure after the expansion is P 2 = 1 atm. Calculate V 2, T 2, W and ∆U. Solution: For an ideal monatomic gas, the ratio of specific heats γ = 5/3 since P 1 V 1 γ = P 2 V 2 γ V 2 = V 1 (P 1 /P 2 ) 1/γ thus V 2 = 13.9m 3 According to ideal gas law: PV = nRT T 2 can be easily calculated as 174K The work done by the system is w = 2.74 x 10 6 J For the adiabatic process: ∆U is equal to the work done on the system thus is x 10 6 J

Chapter 5: Consequences of the First Law

5.1 The Gay-Lussac-Joule Experiments