Question

A fluid system undergoes a non-flow frictionless process following the pressure-volume relation as =5⁄+1.5where p is in bar and V is in m3. During the process, the volume changes from 0.15 m3 to 0.05 m3 and the system rejects 45 kJ of heat.
Determine:
4.1change in internal energy;
4.2change in enthalpy; and
4.3change in entropy.

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Answer 1

Answer:

Given = pressure volume relation = P =5 /V +1.5

             pressure in bar and volume in m³

             volume changes from i.e.ΔV= 0.15 m3 to 0.05 m3

             system rejects i.e ΔH = 45 kJ of heat.

to find =

4.1change in internal energy;  (ΔU)

4.2change in enthalpy; and (ΔH)

4.3change in entropy. (ΔS)

Solution =   apply the 1st law of thermodynamics,

                   ΔQ = ΔU + ΔW

                  -45 = ΔU - 564  

                   ΔU = 519KJ

           

change in enthalpy  (ΔH) = H(product) -  H(rectants)    

                                  (ΔH) =   45    -   519

                                (ΔH)  =   - 474