为解决的问题：
为何冬天比夏天干？夏天的水汽含量是冬天的多少倍？
冬天，华南地区（如广州）比武汉地区水汽含量多多少倍？
干空气与水汽的相对分子质量，分别用符号\(R_d\)和\(R_v\)来表示（下标\(_d\)表示dry，\(_v\)表示water）。
根据高中化学可知，\(M_d = 28.97 g/mol, M_v = 18 g/mol\)。带入上述公式，可得
对于干空气，\(R_d = R^* / M_d = 8.314 / 28.97 = 0.287\;(J ·g^{-1}K^{-1})\)
对于水汽，\(R_v = R^* / M_v = 8.314 / 18 = 0.4615 \;(J ·g^{-1}K^{-1})\)
\[ \epsilon = \frac{R_d}{R_v} = \frac{M_v}{M_d} ≈ 0.622 \\ \frac{R_v}{R_d} = \frac{1}{\epsilon} ≈ 1.608 \]
\[ R_v = \frac{1}{\epsilon} R_d \]
\[ \rho_v = \frac{e}{R_v T} = \frac{\epsilon e}{R_d T} \\ \]
\[ \rho_d = \frac{p - e}{R_d T} \\ \]
\[ \begin{align} \rho &= \rho_d + \rho_v \\ &= \frac{p - e}{R_d T} + \frac{e}{R_v T} \\ &= \frac{p - e}{R_d T} + \frac{\epsilon e}{R_d T} \\ &= \frac{p - (1 - \epsilon)e }{R_d T} \\ &= \frac{p}{R_d T} (1 - 0.378 \frac{ e }{p}) \end{align} \]
\[ \begin{align} q &= \frac{\rho_v}{\rho_d + \rho_v} \\ &= \frac{\epsilon e}{p - e + \epsilon e} \\ &= \frac{\epsilon e}{p - (1 - \epsilon)e} \end{align} \]
\[ qp = (1 - \epsilon ) e q + \epsilon e \\ e = \frac{qp}{ \epsilon + (1 - \epsilon) q } \]
转化的意义： - 已知饱和水汽压，求饱和比湿q - 已知相对湿度，求比湿（回答为何冬天干）
\[ \begin{align} w &= \frac{\rho_v}{\rho_d } \\ &= \frac{\epsilon e}{p - e} \\ \end{align} \]
\[ wp = we + \epsilon e \\ e = \frac{w p}{w + \epsilon} \]
\[ \begin{align} \frac{e}{P} &= \frac{\rho_v}{\rho} \frac{R_v}{R}, (R_d ≈ R) \\ &≈ \frac{\rho_v}{\epsilon \rho} \end{align} \]
\[ \rho_v = \epsilon \rho \frac{e}{P} \]
\[ \frac{e}{P - e} = \frac{\rho_v}{\rho_d} \frac{R_v}{R_d} \\ \frac{e}{P - e} = \frac{\rho_v}{\epsilon \rho_d} \]
\[ \frac{e}{P} = \frac{\rho_v}{\epsilon \rho_d + \rho_v} \\ \]
\[ \rho_v = \epsilon \rho_d \frac{e}{P - e} \]