Problem 1

New and well-forgotten old refrigerants. Refrigerants are compared thermodynamically in a model refrigeration cycle: liquid refrigerant boils at constant pressure

p_1

and temperature

T_1

(absorbing heat

Q

), the vapour is compressed reversibly and adiabatically to

T_2

(work

W

), then cooled at constant pressure

p_2

and returned to the initial state. For 1 mole of refrigerant with

T_1 = 280

K and

T_2 = 380

K, ideal vapour, the data are:

\ce{NH3}

—

\Delta H_{vap} = 21.3

kJ mol

^{-1}

C_V(\text{gas}) = 26.7

J K

^{-1}

mol

^{-1}

;

\ce{CHF2Cl}

—

20.0

kJ mol

^{-1}

and

48.8

J K

^{-1}

mol

^{-1}

;

\ce{CF3CH2F}

—

22.1

kJ mol

^{-1}

and

79

J K

^{-1}

mol

^{-1}

;

\ce{CF3CF=CH2}

—

19.1

kJ mol

^{-1}

and

120

J K

^{-1}

mol

^{-1}

. (a) Calculate

Q

and

W

for

\ce{NH3}

and

\ce{CHF2Cl}

. (b) The coefficient of performance is

COP = Q/W

; calculate it for both refrigerants. (c) Calculate the COP for

\ce{CF3CH2F}

and

\ce{CF3CF=CH2}

— did efficiency improve relative to

\ce{CHF2Cl}

Step 2 of 4: Work of adiabatic compression

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W = C_V(\text{gas})(T_2 - T_1): \quad W(\ce{NH3}) = 26.7 \times 100 = 2.67\ \text{kJ},\quad W(\ce{CHF2Cl}) = 48.8 \times 100 = 4.88\ \text{kJ}

Analysis

For an ideal gas in an adiabatic step, $q = 0$ so the compression work equals the internal-energy rise: $W = C_V(T_2 - T_1) = C_V \times 100$ K, giving 2.67 kJ ( $\ce{NH3}$ ) and 4.88 kJ ( $\ce{CHF2Cl}$ ).