In this problem are internal change and enthalpy change same when first heated a

Question

In this problem are internal change and enthalpy change same when first heated and cooled later? xample 2.8 alculate the internal energy and enthalpy changes resulting if air changes from an ini- tial state of 5°C and 10 bar, where its molar volume is 2.312 x 10 3 m3 mol.to a final state of 60 C and 1 bar. Assume also that air remains a gas for which PVIT is constant and that Cv 20.785 and Cp 29.100 Jmol.K- Solution 2.8 Because property changes e independent of process, calculations may be based on any process that accomplishes the change. Here. we choose a two-step. mechan- ically reversible process wherein 1 mol of air is (a) cooled at constant volume to the final pressure, and (b) heated at constant pressure to the final temperature. Of course, other paths could be chosen, and would yield the same result. T.-: 5 + 273. 15-. 278. I 5 K T2 60+273.15 333.15 K With PV- kT, the ratio T/P is constant for step (a). The intermediate temperature between the two steps is therefore: T (278.15) (1/10)27.82 K and the temperature changes for the two steps are: ATa 27.82-278.15-250.33 K = 333.15-27.82 305.33 K For step (a), by Eqs. (2.17) and (2.14), U, = Cv Ta (20.785)(-250.33) =-52031 J -5203.1+2.312 x 10-3 m3x (-9x 105) Pa-7283.9J For step (b), the final volume of the air is: ½=ViP2 2=2.312 × 10-3(10x333|5)=2.769x 10-2 m P1T2 Tx 278.15 2.769 x0-2m3 By Eqs. (2.21) and (2.14), = c,ATb = (29. 100) (305.33) = 8885.1 J 8885.1- (1 x 105)(0.02769 0.00231) 6347.1 J For the two steps together, AU-5203.16347.11144.0J H -7283.9 + 8885. 1 = 1 601 .2 J These values would be the same for any process that results in th e same change of state

Explanation / Answer

Yes the total change in internal energy or enthalpy change will be same if it is heated first and then cooled later.

Reason is that these are the properties and properties do not differ by adopting a different process, however the initials and final conditions must be same.

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