Answer
$−3.735\text{ kJ/kg}$
$−3.721\text{ kJ/kg}$
Work Step by Step
Analysis The saturation temperature of R-134a at $280 \mathrm{~kPa}$ is $-1.25^{\circ} \mathrm{C}$ (Table A-12). $$
g_f=h_f-T s_f=50.16 \mathrm{~kJ} / \mathrm{kg}-(-1.25+273.15 \mathrm{~K})(0.19822 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})=-3.735 \mathrm{~kJ} / \mathrm{kg}
$$ For the vapor phase, $$
g_g=h_g-T s_g=249.77 \mathrm{~kJ} / \mathrm{kg}-(-1.25+273.15 \mathrm{~K})(0.93228 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})=-3.721 \mathrm{~kJ} / \mathrm{kg}
$$ The results agree and demonstrate that phase equilibrium exists.
