One lb of Refrigerant 134a contained within a piston–cylinder assembly undergoes a process from a state where the temperature is 60F and the refrigerant is saturated liquid to a state where the pressure is 140 lbf/in 2 and quality is 50%. Determine the change in specific entropy of the refrigerant, in Btu/lbR. Can this process be accomplished adiabatically?

It is necessary to develop this problem to resort to the A-11E tables in English Units for R134a (since the problem requires it, if it were SI just to change to that table)

State 1 indicates that the refrigerant is at 60 ° F,

In the first table (attached image of the value taken) the value of the entropy is

[tex]S_f=S_1= 0.06675Btu/lbm.R[/tex]

For State 2 the refrigerant is at 50% quality and at a pressure of [tex]140lbf / in ^ 2[/tex]

In table 2 of the refrigerant (for the pressure values) we perform the reading and we have to

[tex]Sf= 0.09214[/tex]

[tex]Sg=0.21879[/tex]

We know that,

[tex]S_2 = Sf +X_{quality}(S_g-S_f)[/tex]

[tex]S_2 = 0.09214+0.5(0.21979-0.09214)[/tex]

[tex]S_2 = 0.155965BTU/Lb.R[/tex]

The change in enthalpy would be given by

[tex]\Delta S = S_2-S_1 = 0.155965BTU/Lb.R- 0.06675Btu/lbm.R[/tex]

[tex]\Delta S = 0.089215Btu/lbm.R[/tex]

The change in enthalpy is positive, so the process can be completed adiabatically