Chemistry Labs

Problem 1

New and well-forgotten old refrigerants. Refrigerants are compared thermodynamically in a model refrigeration cycle: liquid refrigerant boils at constant pressure p1p_1 and temperature T1T_1 (absorbing heat QQ), the vapour is compressed reversibly and adiabatically to T2T_2 (work WW), then cooled at constant pressure p2p_2 and returned to the initial state. For 1 mole of refrigerant with T1=280T_1 = 280 K and T2=380T_2 = 380 K, ideal vapour, the data are: NHX3\ce{NH3} — ΔHvap=21.3\Delta H_{vap} = 21.3 kJ mol−1^{-1}, CV(gas)=26.7C_V(\text{gas}) = 26.7 J K−1^{-1} mol−1^{-1}; CHFX2Cl\ce{CHF2Cl} — 20.020.0 kJ mol−1^{-1} and 48.848.8 J K−1^{-1} mol−1^{-1}; CFX3CHX2F\ce{CF3CH2F} — 22.122.1 kJ mol−1^{-1} and 7979 J K−1^{-1} mol−1^{-1}; CFX3CF=CHX2\ce{CF3CF=CH2} — 19.119.1 kJ mol−1^{-1} and 120120 J K−1^{-1} mol−1^{-1}. (a) Calculate QQ and WW for NHX3\ce{NH3} and CHFX2Cl\ce{CHF2Cl}. (b) The coefficient of performance is COP=Q/WCOP = Q/W; calculate it for both refrigerants. (c) Calculate the COP for CFX3CHX2F\ce{CF3CH2F} and CFX3CF=CHX2\ce{CF3CF=CH2} — did efficiency improve relative to CHFX2Cl\ce{CHF2Cl}?
Step 4 of 4: COP of third- and fourth-generation refrigerants
COP(CFX3CHX2F)=22.179×100×10−3=2.80,COP(CFX3CF=CHX2)=19.1120×100×10−3=1.59⇒NoCOP(\ce{CF3CH2F}) = \dfrac{22.1}{79 \times 100 \times 10^{-3}} = 2.80,\quad COP(\ce{CF3CF=CH2}) = \dfrac{19.1}{120 \times 100 \times 10^{-3}} = 1.59 \Rightarrow \text{No}
Analysis

The same calculation gives COP=2.80COP = 2.80 for CFX3CHX2F\ce{CF3CH2F} and 1.591.59 for CFX3CF=CHX2\ce{CF3CF=CH2} — both below 4.10, so the newer generations did not improve energy efficiency versus CHFX2Cl\ce{CHF2Cl}; they were adopted for environmental (ozone, greenhouse) reasons instead.

Common pitfall. Use CVC_V in J K−1^{-1} mol−1^{-1} and convert to kJ: forgetting the factor 10−310^{-3} inflates WW by 1000 and destroys the COP.