$(1)$ %=image:/media/2015/02/02/142287293421096500.png: (I) \[ v_1= \frac{RT_1}{p_1}= \frac {286.99 \times 288}{0.1013 \times 10^6} =0.81592 \\ \therefore 0.8159 \ \rm{m^3/kg} \] (II) \[ p_2 = p_1 \left( \frac{v_1}{v_2} \right)^\kappa = 101.3 \times 14^{1.4}=4075.5 \\ \therefore 4076 \ \rm{kPa} \] (III) \[ \frac{v_1}{v_2}=14 \ \Longrightarrow \ v_2= \frac{v_1}{14}=0.05828 \\ \therefore 0.05828 \ \rm{m^3/kg} \] (IV) \[ T_2= \frac{p_2 v_2}{R}=\frac{4075.5 \times 10^3 \times0.05828}{0.28699 \times 10^3}=827.6 \\ \therefore 827.6 \ \rm{K} \] (V) \[ s_2 = s_1 =0 \\ \therefore 0 \ \rm{kJ/(kg\cdot K)} \] (VI) \[ p_3 = p_2 = 4076 \\ \therefore 4076 \ \rm{kPa} \] (VII) \[ q_{23} = c_p (T_3 - T_2) \\ \Longrightarrow \ T_3= T_2 + \frac{q_{23}}{c_p}=827.6 + \frac{2000 \times 10^3}{1.006 \times 10^3} = 2815.7 \\ \therefore 2816 \ \rm{K} \] (VIII) \[ v_2= \frac{RT_3}{p_3} = \frac{0.28699 \times 10^3 \times 2815.7}{4075.5 \times 10^3} = 0.19828 \\ \therefore 0.1983 \ \rm{m^3/kg} \] (IX) \[ s_2= c_p \ln \frac{T_3}{T_2} +s_2 = 1.006 \times \ln \frac{2815.7}{827.6} =1.2318 \\ \therefore 1.232 \ \rm{kJ/(kg \cdot K)} \] (X) \[ s_4 = s_3 = 1.232 \\ \therefore 1.232 \ \rm{kJ/(kg\cdot K)} \] (XI) \[ v_4 = v_1 = 0.8159 \\ \therefore 0.8159 \ \rm{m^3/kg} \] (XII) \[ p_4 = p_3 \left( \frac{v_3}{v_4} \right)^\kappa =4075.5 \times \left( \frac{0.19828}{0.81592} \right)^{1.4} =562.42 \\ \therefore 562.4 \ \rm{kPa} \] (XIII) \[ T_4 = \frac{p_4 v_4}{R} = \frac{562.4 \times 0.81592}{0.28699} = 1598.9 \\ \therefore 1599 \ \rm{K} \] $(2)$ \[ \begin{align} q_{41}=c_v (T_4 - T_1) &=0.719 \times (1599-288) \\ \therefore &=942.6 \ \rm{kJ/kg} \end{align} \] $(3)$ \[ \eta= \frac{q_{23}-q_{41}}{q_{23}}= \frac{2060-942.6}{2000}=0.5287 \\ \therefore \ 52.9 \% \] $(4)$ %=image:/media/2015/02/03/142289671307867300.png: