Speed of sound in different media

Given:

$$\beta_{\mathrm{Hg}}=3\cdot10^{-5}\,\mathrm{atm}^{-1},\qquad \rho_{\mathrm{Hg}}=13.6\cdot10^3\,\mathrm{kg/m^3}$$

$$\beta_{\mathrm{water}}=5\cdot10^{-5}\,\mathrm{atm}^{-1},\qquad \rho_{\mathrm{water}}=1.0\cdot10^3\,\mathrm{kg/m^3}$$

$$\beta_{\mathrm{air}}=0.71\,\mathrm{atm}^{-1},\qquad \rho_{\mathrm{air}}=1.2\,\mathrm{kg/m^3}$$

Main formula:

$$v=\sqrt{\frac{1}{\beta\rho}}$$

where $\beta$ is the compressibility of the medium, and $\rho$ is the density of the medium.

We convert the compressibility from $\mathrm{atm}^{-1}$ to $\mathrm{Pa}^{-1}$.

$$1\,\mathrm{atm}=101325\,\mathrm{Pa}$$

Therefore,

$$1\,\mathrm{atm}^{-1}=\frac{1}{101325}\,\mathrm{Pa}^{-1}$$

1. Mercury

$$\beta_{\mathrm{Hg}}= \frac{3\cdot10^{-5}}{101325} \approx2.96\cdot10^{-10}\,\mathrm{Pa}^{-1}$$

We use the formula:

$$v_{\mathrm{Hg}}= \sqrt{\frac{1}{\beta_{\mathrm{Hg}}\rho_{\mathrm{Hg}}}}$$

Substituting the values:

$$v_{\mathrm{Hg}}= \sqrt{\frac{1} {(2.96\cdot10^{-10})(13.6\cdot10^3)}}$$

First, we calculate the denominator:

$$(2.96\cdot10^{-10})(13.6\cdot10^3) \approx4.0256\cdot10^{-6}$$

Therefore:

$$v_{\mathrm{Hg}}= \sqrt{\frac{1}{4.0256\cdot10^{-6}}}$$

$$v_{\mathrm{Hg}}\approx498\,\mathrm{m/s}$$

Answer for mercury:

$$\boxed{v_{\mathrm{Hg}}\approx500\,\mathrm{m/s}}$$

2. Water

$$\beta_{\mathrm{water}}= \frac{5\cdot10^{-5}}{101325} \approx4.93\cdot10^{-10}\,\mathrm{Pa}^{-1}$$

We use the formula:

$$v_{\mathrm{water}}= \sqrt{\frac{1}{\beta_{\mathrm{water}}\rho_{\mathrm{water}}}}$$

Substituting the values:

$$v_{\mathrm{water}}= \sqrt{\frac{1} {(4.93\cdot10^{-10})(1.0\cdot10^3)}}$$

We calculate the denominator:

$$(4.93\cdot10^{-10})(1.0\cdot10^3) =4.93\cdot10^{-7}$$

Therefore:

$$v_{\mathrm{water}}= \sqrt{\frac{1}{4.93\cdot10^{-7}}}$$

$$v_{\mathrm{water}}\approx1424\,\mathrm{m/s}$$

Answer for water:

$$\boxed{v_{\mathrm{water}}\approx1420\,\mathrm{m/s}}$$

3. Air

$$\beta_{\mathrm{air}}= \frac{0.71}{101325} \approx7.01\cdot10^{-6}\,\mathrm{Pa}^{-1}$$

We use the formula:

$$v_{\mathrm{air}}= \sqrt{\frac{1}{\beta_{\mathrm{air}}\rho_{\mathrm{air}}}}$$

Substituting the values:

$$v_{\mathrm{air}}= \sqrt{\frac{1} {(7.01\cdot10^{-6})(1.2)}}$$

We calculate the denominator:

$$(7.01\cdot10^{-6})(1.2) =8.412\cdot10^{-6}$$

Therefore:

$$v_{\mathrm{air}}= \sqrt{\frac{1}{8.412\cdot10^{-6}}}$$

$$v_{\mathrm{air}}\approx345\,\mathrm{m/s}$$

Answer for air:

$$\boxed{v_{\mathrm{air}}\approx345\,\mathrm{m/s}}$$

Final answer:

$$\boxed{ v_{\mathrm{Hg}}\approx500\,\mathrm{m/s},\qquad v_{\mathrm{water}}\approx1420\,\mathrm{m/s},\qquad v_{\mathrm{air}}\approx345\,\mathrm{m/s} }$$