Answer
$p = 18.699 \frac{g}{cm^{3}}$
Work Step by Step
Given:
Gold at room temperature (20°C) density is $19.320 g/cm^{3}$
volume coefficient $(α_{v})= 3 α_{l}$
Required:
density $p$ of gold at 800°C if only thermal expansion is considered
Solution:
Using Equation 19.4 and rearranging it:
$ΔV= α_{v}ΔT{V_{0}}$
Using the basis of 1 $cm^{3}$ of material at 20°C, mass is 19.320 g, and the mass will remain constant upon heating to 800°C. Also, $α_{v} = 3α_{l}$, and from Table 19.1, $α_{l} = 14.2 \times 10^{-6} (°C^{-1}) $
Substituting,
$V = V_{0} + ΔV = V_{0} + (α_{v}ΔT{V_{0}}) = V_{0}(1 + 3α_{l}ΔT)= (1 cm^{3})(1 + (3)[14.2 \times 10^{-6} (°C^{-1})] (800°C - 20°C) = 1.03323 cm^{3}$
$p = \frac{19.320 g}{1.03323 cm^{3} } = 18.699 \frac{g}{cm^{3}}$