College Physics (4th Edition)

(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.
(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.