Chapter 15, Problem 32SP

The density of a particular sample of gold is 19.30 ${\text{g/cm}}^3$ at 20.0 °C, and the coefficient of linear expansion is $14.3\times {10}^{-6}°{\text{C}}^{-1}$. Compute the density of that sample at 90.0 °C. [Hint: Take a look at Problem 15.9.]

To determine

The density of gold sample at $90.0°\text{C}$ if its density is $19.30{\text{ g/cm}}^3$ at $20°\text{C}$.

Answer to Problem 32SP

Solution:

$19.2{\text{ g/cm}}^{\text{3}}$

Explanation of Solution

Given data:

The density of a particular sample of gold is $19.30{\text{ g/cm}}^3$ at $20°\text{C}$.

The coefficient of linear expansion of gold is $14.3\times {10}^{-6}°{\text{C}}^{-1}$.

Formula used:

Mass of the sample does not change with the temperature, thus it can be written as

$\begin{array}{c}{m}_o=m\\ {\rho }_o{V}_o=\rho V\end{array}$

Here, $\rho$ and $V$ are the density and volume of the sample at the temperature $T$.

But

$\begin{array}{c}\rho =\frac{{\rho }_o{V}_o}{V}\\ \rho =\frac{{\rho }_o{V}_o}{V_o+\Delta V}\\ \rho =\frac{{\rho }_o{V}_o}{1+\frac{\Delta V}{V_o}}\end{array}$

And

$\Delta V=V_o\beta \Delta T$

Therefore,

$\begin{array}{c}\rho =\frac{{\rho }_o{V}_o}{1+\beta \Delta T}\\ \left(\rho -{\rho }_o\right)=-\rho \beta \Delta T\end{array}$

Practically, $\rho \approx {\rho }_o$. Then, write the expression for linear expansion of solids in terms of density:

$\rho -{\rho }_0=-{\rho }_0\beta \left(T-T_0\right)$

Here, $\rho$ is the density at temperature $T$, $\beta$ is the coefficient of linear expansion, ${\rho }_0$ is the initial density of solid at temperature $T_0$, and $T$ is the final temperature.

Write the expression for coefficient of volumetric expansion in terms of coefficient of linear expansion:

$\beta =3\alpha$

Here, $\beta$ is the coefficient of volumetric expansion and $\alpha$ is the coefficient of linear expansion.

Explanation:

Recall the expression for coefficient of volumetric expansion in terms of coefficient of linear expansion for gold:

$\beta =3\alpha$

Substitute $14.3\times {10}^{-6}°{\text{C}}^{-1}$ for $\alpha$

$\begin{array}{c}\beta =3\left(14.3\times {10}^{-6}°{\text{C}}^{-1}\right)\\ =4.29\times {10}^{-5}°{\text{C}}^{-1}\end{array}$

Recall the expression for linear expansion of gold in terms of density:

$\rho -{\rho }_0=-{\rho }_0\beta \left(T-T_0\right)$

Substitute $19.30{\text{ g/cm}}^3$ for ${\rho }_0$, $4.29\times {10}^{-5}°{\text{C}}^{-1}$ for $\beta$, $90.0°\text{C}$ for $T$, and $20.0°\text{C}$ for $T_0$

$\rho -\left(19.30{\text{ g/cm}}^3\right)=-\left(19.30{\text{ g/cm}}^3\right)\left(4.29\times {10}^{-5}°{\text{C}}^{-1}\right)\left(90.0°\text{C}-20.0°\text{C}\right)$

Solve for $\rho$

$\begin{array}{c}\rho =\left(19.30{\text{ g/cm}}^3\right)-\left(19.30{\text{ g/cm}}^3\right)\left(4.29\times {10}^{-5}°{\text{C}}^{-1}\right)\left(90.0°\text{C}-20.0°\text{C}\right)\\ =\left(19.30{\text{ g/cm}}^3\right)-\left(19.30{\text{ g/cm}}^3\right)\left(4.29\times {10}^{-5}°{\text{C}}^{-1}\right)\left(70.0°\text{C}\right)\\ =\left(19.30{\text{ g/cm}}^3\right)-\left(0.0579{\text{ g/cm}}^3\right)\\ \approx 19.2{\text{ g/cm}}^3\end{array}$

Conclusion:

The density of the gold sample at $90.0°\text{C}$ is $19.2{\text{ g/cm}}^3$.