%=image:/media/2015/02/02/142287524866311400.png: $(1)$ まず，比熱比$\kappa$を求める． $\kappa = \frac{c_v + R}{c_v} = \frac{0.72 + 0.29}{0.72} = 1.4028 \\$ 次に，各々の状態変化の過程を考慮して， \[ v_1 = \frac{RT_1}{p_1} = \frac {0.29 \times 300}{100} = 0.37 \\ T_2 = T_1 + 300 \\ v_2 + \frac {v_1}{12} = 0.0725 \\ p_2 + \frac {RT_2}{v_2} = \frac{0.29 \times 300}{0.0725}=1200 \\ s_2 - s_1 = - R\ln \frac{p_2}{p_1} \ \Longrightarrow \ s_1 = s_2+ R\ln \frac{p_2}{p_1} = 0.29\ln 12 = 0.7206 \\ s_2 - s_1 = 12\ln \frac{v_2}{v_1} \ \Longrightarrow \ s_1 = s_2+ R\ln \frac{v_2}{v_1} = -0.29\ln \frac{1}{12} = 0.7206 \\ s_3 = s_1 =0.07206 \\ v_3 = v_2 = 0.0725 \\ p_3 = p_1 \left( \frac{v_1}{v_3} \right)^\kappa = 100 \times 12 ^{1.4028} = 3264.9…=3265 \\ T_3 = \frac{p_3 V_3}{R} = 816.2 \] $(2)$ \[ q_{23} = c_v (T_3 -T_1) = 0.72 (816.6 - 300) = 371.75 =372 \\ q_{12} = \int_1^2 pdv = RT_1 \int_1^2 \frac {dv}{v} = RT_1\ln \frac{v_2}{v_1} = 0.29 \times 300 \times \ln \frac{1}{12} = -216.2 \\ \eta = \frac {q_{23}+q_{12}}{q_{23}} = 0.4187 = 41.9\% \] $(3)$ %=image:/media/2015/02/03/142289714382818900.png: