| 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. | | 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. | | Determine the speed of sound in these media. |
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| Given: | | Given: |
| \[ | | \[ |
| \beta_{\mathrm{Hg}}=3\cdot10^{-5}\,\mathrm{atm}^{-1},\qquad \rho_{\mathrm{Hg}}=13.6\cdot10^3\,\mathrm{kg/m^3} | | \beta_{\mathrm{Hg}}=3\cdot10^{-5}\,\mathrm{atm}^{-1},\qquad \rho_{\mathrm{Hg}}=13.6\cdot10^3\,\mathrm{kg/m^3} |
| \] | | \] |
| \[ | | \[ |
| \beta_{\mathrm{H_2O}}=5\cdot10^{-5}\,\mathrm{atm}^{-1},\qquad \rho_{\mathrm{H_2O}}=1.0\cdot10^3\,\mathrm{kg/m^3} | | \beta_{\mathrm{H_2O}}=5\cdot10^{-5}\,\mathrm{atm}^{-1},\qquad \rho_{\mathrm{H_2O}}=1.0\cdot10^3\,\mathrm{kg/m^3} |
| \] | | \] |
| \[ | | \[ |
| \beta_{\text{air}}=0.71\,\mathrm{atm}^{-1},\qquad \rho_{\text{air}}=1.2\,\mathrm{kg/m^3} | | \beta_{\text{air}}=0.71\,\mathrm{atm}^{-1},\qquad \rho_{\text{air}}=1.2\,\mathrm{kg/m^3} |
| \] | | \] |
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| Main formula: | | Main formula: |
| \[ | | \[ |
| v=\sqrt{\frac{1}{\beta\rho}} | | v=\sqrt{\frac{1}{\beta\rho}} |
| \] | | \] |
| where $\beta$ is the compressibility of the medium, and $\rho$ is the density of the medium. | | where $\beta$ is the compressibility of the medium, and $\rho$ is the density of the medium. |
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| We convert the compressibility from $\mathrm{atm}^{-1}$ to $\mathrm{Pa}^{-1}$: | | We convert the compressibility from $\mathrm{atm}^{-1}$ to $\mathrm{Pa}^{-1}$: |
| \[ | | \[ |
| 1\,\mathrm{atm}=101325\,\mathrm{Pa} \implies 1\,\mathrm{atm}^{-1}=\frac{1}{101325}\,\mathrm{Pa}^{-1} | | 1\,\mathrm{atm}=101325\,\mathrm{Pa} \implies 1\,\mathrm{atm}^{-1}=\frac{1}{101325}\,\mathrm{Pa}^{-1} |
| \] | | \] |
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| 1. Mercury | | 1. Mercury |
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| \[ | | \[ |
| \beta_{\mathrm{Hg}}=\frac{3\cdot10^{-5}}{101325} \approx 2.96\cdot10^{-10}\,\mathrm{Pa}^{-1} | | \beta_{\mathrm{Hg}}=\frac{3\cdot10^{-5}}{101325} \approx 2.96\cdot10^{-10}\,\mathrm{Pa}^{-1} |
| \] | | \] |
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| We use the formula: | | We use the formula: |
| \[ | | \[ |
| v_{\mathrm{Hg}}=\sqrt{\frac{1}{\beta_{\mathrm{Hg}}\rho_{\mathrm{Hg}}}} | | v_{\mathrm{Hg}}=\sqrt{\frac{1}{\beta_{\mathrm{Hg}}\rho_{\mathrm{Hg}}}} |
| \] | | \] |
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| Substituting the values: | | Substituting the values: |
| \[ | | \[ |
| v_{\mathrm{Hg}}=\sqrt{\frac{1}{(2.96\cdot10^{-10})(13.6\cdot10^3)}} | | v_{\mathrm{Hg}}=\sqrt{\frac{1}{(2.96\cdot10^{-10})(13.6\cdot10^3)}} |
| \] | | \] |
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| We calculate the denominator: | | We calculate the denominator: |
| \[ | | \[ |
| (2.96\cdot10^{-10})(13.6\cdot10^3) \approx 4.0256\cdot10^{-6} | | (2.96\cdot10^{-10})(13.6\cdot10^3) \approx 4.0256\cdot10^{-6} |
| \] | | \] |
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| Therefore: | | Therefore: |
| \[ | | \[ |
| v_{\mathrm{Hg}}=\sqrt{\frac{1}{4.0256\cdot10^{-6}}} \approx 498\,\mathrm{m/s} | | v_{\mathrm{Hg}}=\sqrt{\frac{1}{4.0256\cdot10^{-6}}} \approx 498\,\mathrm{m/s} |
| \] | | \] |
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| Answer for mercury: | | Answer for mercury: |
| \[ | | \[ |
| v_{\mathrm{Hg}}\approx 500\,\mathrm{m/s} | | v_{\mathrm{Hg}}\approx 500\,\mathrm{m/s} |
| \] | | \] |
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| 2. Water ($\mathrm{H_2O}$) | | 2. Water ($\mathrm{H_2O}$) |
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| \[ | | \[ |
| \beta_{\mathrm{H_2O}}=\frac{5\cdot10^{-5}}{101325} \approx 4.93\cdot10^{-10}\,\mathrm{Pa}^{-1} | | \beta_{\mathrm{H_2O}}=\frac{5\cdot10^{-5}}{101325} \approx 4.93\cdot10^{-10}\,\mathrm{Pa}^{-1} |
| \] | | \] |
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| We use the formula: | | We use the formula: |
| \[ | | \[ |
| v_{\mathrm{H_2O}}=\sqrt{\frac{1}{\beta_{\mathrm{H_2O}}\rho_{\mathrm{H_2O}}}} | | v_{\mathrm{H_2O}}=\sqrt{\frac{1}{\beta_{\mathrm{H_2O}}\rho_{\mathrm{H_2O}}}} |
| \] | | \] |
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| Substituting the values: | | Substituting the values: |
| \[ | | \[ |
| v_{\mathrm{H_2O}}=\sqrt{\frac{1}{(4.93\cdot10^{-10})(1.0\cdot10^3)}} | | v_{\mathrm{H_2O}}=\sqrt{\frac{1}{(4.93\cdot10^{-10})(1.0\cdot10^3)}} |
| \] | | \] |
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| We calculate the denominator: | | We calculate the denominator: |
| \[ | | \[ |
| (4.93\cdot10^{-10})(1.0\cdot10^3) = 4.93\cdot10^{-7} | | (4.93\cdot10^{-10})(1.0\cdot10^3) = 4.93\cdot10^{-7} |
| \] | | \] |
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| Therefore: | | Therefore: |
| \[ | | \[ |
| v_{\mathrm{H_2O}}=\sqrt{\frac{1}{4.93\cdot10^{-7}}} \approx 1424\,\mathrm{m/s} | | v_{\mathrm{H_2O}}=\sqrt{\frac{1}{4.93\cdot10^{-7}}} \approx 1424\,\mathrm{m/s} |
| \] | | \] |
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| Answer for water: | | Answer for water: |
| \[ | | \[ |
| v_{\mathrm{H_2O}}\approx 1420\,\mathrm{m/s} | | v_{\mathrm{H_2O}}\approx 1420\,\mathrm{m/s} |
| \] | | \] |
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| 3. Air | | 3. Air |
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| \[ | | \[ |
| \beta_{\text{air}}=\frac{0.71}{101325} \approx 7.01\cdot10^{-6}\,\mathrm{Pa}^{-1} | | \beta_{\text{air}}=\frac{0.71}{101325} \approx 7.01\cdot10^{-6}\,\mathrm{Pa}^{-1} |
| \] | | \] |
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| We use the formula: | | We use the formula: |
| \[ | | \[ |
| v_{\text{air}}=\sqrt{\frac{1}{\beta_{\text{air}}\rho_{\text{air}}}} | | v_{\text{air}}=\sqrt{\frac{1}{\beta_{\text{air}}\rho_{\text{air}}}} |
| \] | | \] |
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| Substituting the values: | | Substituting the values: |
| \[ | | \[ |
| v_{\text{air}}=\sqrt{\frac{1}{(7.01\cdot10^{-6})(1.2)}} | | v_{\text{air}}=\sqrt{\frac{1}{(7.01\cdot10^{-6})(1.2)}} |
| \] | | \] |
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| We calculate the denominator: | | We calculate the denominator: |
| \[ | | \[ |
| (7.01\cdot10^{-6})(1.2) = 8.412\cdot10^{-6} | | (7.01\cdot10^{-6})(1.2) = 8.412\cdot10^{-6} |
| \] | | \] |
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| Therefore: | | Therefore: |
| \[ | | \[ |
| v_{\text{air}}=\sqrt{\frac{1}{8.412\cdot10^{-6}}} \approx 345\,\mathrm{m/s} | | v_{\text{air}}=\sqrt{\frac{1}{8.412\cdot10^{-6}}} \approx 345\,\mathrm{m/s} |
| \] | | \] |
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| Answer for air: | | Answer for air: |
| \[ | | \[ |
| v_{\text{air}}\approx 345\,\mathrm{m/s} | | v_{\text{air}}\approx 345\,\mathrm{m/s} |
| \] | | \] |
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| Answer: | | Answer: |
| \[ | | \[ |
| v_{\mathrm{Hg}}\approx500\,\mathrm{m/s},\qquad v_{\mathrm{H_2O}}\approx1420\,\mathrm{m/s},\qquad v_{\text{air}}\approx345\,\mathrm{m/s} | | v_{\mathrm{Hg}}\approx500\,\mathrm{m/s},\qquad v_{\mathrm{H_2O}}\approx1420\,\mathrm{m/s},\qquad v_{\text{air}}\approx345\,\mathrm{m/s} |
| \] | | \] |
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| #### Answer | | #### Answer |
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| [Insert a concise answer or boxed result] | | [Insert a concise answer or boxed result] |
