Answer
(a) $\frac{\Delta V}{V_0} = \beta~\Delta T$, where $\beta = \frac{1}{T_0}$
(b) $\beta = 3411~\times 10^{-6}~K^{-1}$
This value is very close to the value of $3400\times 10^{-6}~K^{-1}$ listed in the table.
Work Step by Step
(a) We can find an expression for the original volume:
$P~V_0 = nRT_0$
$V_0 = \frac{nRT_0}{P}$
We can find an expression for the final volume:
$P~V_2 = nRT_2$
$V_2 = \frac{nRT_2}{P}$
We can divide $V_2$ by $V_1$:
$\frac{V_2}{V_1} = \frac{\frac{nRT_2}{P}}{\frac{nRT_0}{P}}$
$\frac{V_2}{V_0} = \frac{T_2}{T_0}$
$\frac{V_0+\Delta V}{V_0} = \frac{T_0+\Delta T}{T_0}$
$1+\frac{\Delta V}{V_0} = 1+\frac{\Delta T}{T_0}$
$\frac{\Delta V}{V_0} = \frac{\Delta T}{T_0}$
$\frac{\Delta V}{V_0} = \beta~\Delta T$, where $\beta = \frac{1}{T_0}$
(b) We can find $\beta$ when $T_0 = 293.15~K$:
$\beta = \frac{1}{293.15~K} = 3411~\times 10^{-6}~K^{-1}$
Note that this value is very close to the value of $3400\times 10^{-6}~K^{-1}$ listed in the table.
