(1)

(a)

$\displaystyle Q_1=mRT_2\ln \frac{V_3}{V_2}$

(b)

$\displaystyle |Q_2|=mRT_1\ln \frac{V_1}{V_4}$

(c)

$(P, V, T)$ が $(P+\Delta P, V+\Delta V, T + \Delta T)$ に変化したとき，ボイルシャルルの法則より，

$PV = mRT$

$(P+\Delta P)(V+\Delta V)=mR(T + \Delta T)$

よって， $\Delta P \Delta V$ の項を無視すると，

$P\Delta V + V\Delta P = mR\Delta T$

等温条件での熱力学第一法則より，

$\displaystyle -PdV = mC_vdT = \frac{mC_v}{mR}(PdV + VdP)$

$\displaystyle \frac{dP}{P} + \frac{C_v+R}{C_v}\frac{dV}{V}=0$

$\displaystyle \frac{dP}{P} + \kappa \frac{dV}{V}=0$

この微分方程式を解くと，

$PV^\kappa =$ （一定）

(d)

ポアソン法則より，

$T_1V_1^{\kappa -1} = T_2V_2^{\kappa -1}$

$T_3V_3^{\kappa -1} = T_4V_4^{\kappa -1}$

$T_1=T_4, T_2=T_3$ より，

$\displaystyle \frac{T_1}{T_2} = \frac{V_2}{V_1} = \frac{V_3}{V_4}$

(e)

$\displaystyle \eta = 1 -\frac{Q_1}{|Q_2|}$

(d)より， $\displaystyle \frac{V_3}{V_2} = \frac{V_1}{V_4}$ だから， $\displaystyle \eta = 1- \frac{T_2}{T_1}$

(f)

$\displaystyle \frac{|Q_2|}{Q_1-|Q_2|} = \frac{T_2}{T_1-T_2}$

(2)

(a)

$\displaystyle T_2 = T_1\varepsilon^{\kappa -1}$

$T_3 = T_2\sigma = T_1\varepsilon^{\kappa-1}\sigma$

$\displaystyle T_4 = \left(\frac{V_3}{V_4}\right)^{\kappa -1 }T_3 = \left(\frac{\sigma}{\varepsilon}\right)^{\kappa - 1}T_1\varepsilon^{\kappa -1 } \sigma = T_1\sigma^\kappa$

(b)

$Q_1=mC_v\kappa\varepsilon^{\kappa -1}(\sigma - 1)T_1$

$|Q_2| = mC_v(\sigma^\kappa -1)T_1$

(c)

$W = -mC_v\varepsilon^{\kappa -1}(\sigma -1)T_1$

$+ mC_v(\kappa -1)(\varepsilon^{\kappa -1}-1)T_1$

$+ mC_v(\sigma^{\kappa} - \varepsilon^{\kappa -1}\sigma)T_1$

(d)

$\displaystyle \eta_d = 1- \frac{\sigma^\kappa -1 }{\kappa\varepsilon^{\kappa -1 }(\sigma -1 )}$

(e)

$\displaystyle \eta_c = 1 -\frac{mC_v(T_4-T_1)}{mC_v(T_3-T_2)} = \frac{T_1\left( \displaystyle \frac{T_4}{T_1}-1\right)}{T_2\left( \displaystyle \frac{T_3}{T_2} -1 \right)}$