Solution

Condition:

The compressibility of mercury, water, and air is equal to $3 \cdot 10^{-5}$, $5 \cdot 10^{-5}$, and $0.71\,\mathrm{atm}^{-1}$, respectively, and their density is $13.6 \cdot 10^3$, $1 \cdot 10^3$, and $1.2\,\mathrm{kg/m}^3$, respectively.
Determine the speed of sound in these media.

Given:
$$\beta_{\mathrm{Hg}}=3\cdot10^{-5}\,\mathrm{atm}^{-1},\qquad \rho_{\mathrm{Hg}}=13.6\cdot10^3\,\mathrm{kg/m^3}$$
$$\beta_{\mathrm{H_2O}}=5\cdot10^{-5}\,\mathrm{atm}^{-1},\qquad \rho_{\mathrm{H_2O}}=1.0\cdot10^3\,\mathrm{kg/m^3}$$
$$\beta_{\text{air}}=0.71\,\mathrm{atm}^{-1},\qquad \rho_{\text{air}}=1.2\,\mathrm{kg/m^3}$$

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

1. Mercury

$$\beta_{\mathrm{Hg}}=\frac{3\cdot10^{-5}}{101325} \approx 2.96\cdot10^{-10}\,\mathrm{Pa}^{-1}$$

We use the formula:
$$v_{\mathrm{Hg}}=\sqrt{\frac{1}{\beta_{\mathrm{Hg}}\rho_{\mathrm{Hg}}}}$$

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

We calculate the denominator:
$$(2.96\cdot10^{-10})(13.6\cdot10^3) \approx 4.0256\cdot10^{-6}$$

Therefore:
$$v_{\mathrm{Hg}}=\sqrt{\frac{1}{4.0256\cdot10^{-6}}} \approx 498\,\mathrm{m/s}$$

Answer for mercury:
$$v_{\mathrm{Hg}}\approx 500\,\mathrm{m/s}$$

2. Water ($\mathrm{H_2O}$)

$$\beta_{\mathrm{H_2O}}=\frac{5\cdot10^{-5}}{101325} \approx 4.93\cdot10^{-10}\,\mathrm{Pa}^{-1}$$

We use the formula:
$$v_{\mathrm{H_2O}}=\sqrt{\frac{1}{\beta_{\mathrm{H_2O}}\rho_{\mathrm{H_2O}}}}$$

Substituting the values:
$$v_{\mathrm{H_2O}}=\sqrt{\frac{1}{(4.93\cdot10^{-10})(1.0\cdot10^3)}}$$

We calculate the denominator:
$$(4.93\cdot10^{-10})(1.0\cdot10^3) = 4.93\cdot10^{-7}$$

Therefore:
$$v_{\mathrm{H_2O}}=\sqrt{\frac{1}{4.93\cdot10^{-7}}} \approx 1424\,\mathrm{m/s}$$

Answer for water:
$$v_{\mathrm{H_2O}}\approx 1420\,\mathrm{m/s}$$

3. Air

$$\beta_{\text{air}}=\frac{0.71}{101325} \approx 7.01\cdot10^{-6}\,\mathrm{Pa}^{-1}$$

We use the formula:
$$v_{\text{air}}=\sqrt{\frac{1}{\beta_{\text{air}}\rho_{\text{air}}}}$$

Substituting the values:
$$v_{\text{air}}=\sqrt{\frac{1}{(7.01\cdot10^{-6})(1.2)}}$$

We calculate the denominator:
$$(7.01\cdot10^{-6})(1.2) = 8.412\cdot10^{-6}$$

Therefore:
$$v_{\text{air}}=\sqrt{\frac{1}{8.412\cdot10^{-6}}} \approx 345\,\mathrm{m/s}$$

Answer for air:
$$v_{\text{air}}\approx 345\,\mathrm{m/s}$$

Answer:
$$v_{\mathrm{Hg}}\approx500\,\mathrm{m/s},\qquad v_{\mathrm{H_2O}}\approx1420\,\mathrm{m/s},\qquad v_{\text{air}}\approx345\,\mathrm{m/s}$$