Answer
$β_{}=0.00381\ K^{-1}$
$α_{}=0.00482\ K^{-1}$
Work Step by Step
The volume expansivity and isothermal compressibility are expressed as $$
\beta=\frac{1}{v}\left(\frac{\partial v}{\partial T}\right)_P \text { and } \alpha=-\frac{1}{v}\left(\frac{\partial v}{\partial P}\right)_T
$$ Approximating differentials by differences about the specified state, $$
\begin{aligned}
\beta & \cong \frac{1}{v}\left(\frac{\Delta v}{\Delta T}\right)_{P=200 \mathrm{kPa}}=\frac{1}{v}\left(\frac{\boldsymbol{v}_{40^{\circ} \mathrm{C}}-\boldsymbol{v}_{20^{\circ} \mathrm{C}}}{(40-20)^{\circ} \mathrm{C}}\right)_{P=200 \mathrm{kPa}} \\
& =\frac{1}{0.11874 \mathrm{~m}^3 / \mathrm{kg}}\left(\frac{(0.12322-0.11418) \mathrm{m}^3 / \mathrm{kg}}{20 \mathrm{~K}}\right) \\
& =\mathbf{0 . 0 0 3 8 1\ \mathrm { K } ^ { - 1 }}
\end{aligned}
$$ and $$
\begin{aligned}
\alpha & \cong-\frac{1}{v}\left(\frac{\Delta v}{\Delta P}\right)_{T=30^{\circ} \mathrm{C}}=-\frac{1}{v}\left(\frac{v_{240 \mathrm{ka}}-v_{180 \mathrm{kPa}}}{(240-180) \mathrm{kPa}}\right)_{T=30^{\circ} \mathrm{C}} \\
& =-\frac{1}{0.11874 \mathrm{~m}^3 / \mathrm{kg}}\left(\frac{(0.09812-0.13248) \mathrm{m}^3 / \mathrm{kg}}{60 \mathrm{kPa}}\right) \\
& =\mathbf{0 . 0 0 4 8 2}\ \mathbf{k P a}^{-1}
\end{aligned}
$$
