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Form the property table of R - 134 a refrigerant at P = 2bar and T = 20 degree c, U_1 = 244.99 KJ/log S_1 = 1.0206 KJ/kg k U_1 = 0.11394 m^3/kg and at T = - 22 degree c, and there is no work so (V initial = v final) therefore V_1 = V_2 U_1 = U_2 { m = 1 kg} = d 0.11394 = 0.0007326 + n (0.1590 - 0.0007326) x = 0.7152 so V_2 = U_b2 + xUtg_2 = 61(21.68) + [0.7152 times(214.70 - 21.66)] U_2 = 159.727 S_2 = Sf_2 + Sfg_2 = 0.0897 + 0.7152 times(0.9351 - 0.0897) S_2 = 0.69433 KJ/kg k from 1^st law triangle q = DU + DW triangle q = DU { DW = 0}triangle q = (U_2 - U_10) = (159.727 - 244.99) d q = - 85.263 KJ and every = DU - To DS = - 85.263 - 293 times (0.69433 - 1.0206 every = 10.33411 KJ
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