Edits to “Statement”, “Solution”, “Note on the displaced volume”
en/5.5.5.md
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| @@ -1,11 +1,34 @@ | |||
| ### Statement | |||
| − | $5.5.5.$ [Insert the problem statement] | ||
| + | $5.5.5.$ A gas cylinder with a volume of $50 \text{ L}$ was filled with air at $27^\circ\text{C}$ to a pressure of $10 \text{ MPa}$. What volume of water can be displaced from a submarine's ballast tank by the air from this cylinder, if the displacement occurs at a depth of $40 \text{ m}$? The temperature of the air after expansion is $3^\circ\text{C}$. | ||
| ### Solution | |||
| − | Statement$5.5.5.$ A gas cylinder with a volume of $50 \text{ L}$ was filled with air at $27^\circ\text{C}$ to a pressure of $10 \text{ MPa}$. What volume of water can be displaced from a submarine's ballast tank by the air from this cylinder, if the displacement occurs at a depth of $40 \text{ m}$? The temperature of the air after expansion is $3^\circ\text{C}$.SolutionLet's write down the initial parameters of the air inside the cylinder:Initial volume: $V_1 = 50 \text{ L}$Initial pressure: $P_1 = 10 \text{ MPa} = 10^7 \text{ Pa}$Initial temperature: $T_1 = 27^\circ\text{C} = 300 \text{ K}$After the air is released into the ballast tank at a depth $h = 40 \text{ m}$, it reaches a new state. The external pressure at this depth consists of the atmospheric pressure $P_{\text{atm}}$ and the hydrostatic pressure of the water:$$P_2 = P_{\text{atm}} + \rho gh$$where $\rho = 10^3 \text{ kg/m}^3$ is the density of water, and we take standard atmospheric pressure as $P_{\text{atm}} \approx 10^5 \text{ Pa}$.The final temperature of the air after expansion is $T_2 = 3^\circ\text{C} = 276 \text{ K}$.Since the mass of the air remains constant, we can apply the combined gas law:$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$From this, we can express the final total volume $V_2$ occupied by the air:$$V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}$$$$V_2 = \frac{P_1 V_1 T_2}{T_1 (P_{\text{atm}} + \rho gh)}$$Note on the displaced volume: The total volume of the air after expansion is $V_2$. However, the cylinder itself already holds a volume $V_1$. The volume of water actually displaced from the tank is only the additional volume the gas takes up as it leaves the cylinder. Therefore, the displaced volume of water $\Delta V$ is the difference between the final and initial gas volumes.$$\Delta V = V_2 - V_1 = \frac{P_1 V_1 T_2}{T_1 (P_{\text{atm}} + \rho gh)} - V_1$$Let's substitute the numerical values into the equation. For convenience, we can keep $V_1$ in liters, which will give us the result for $\Delta V$ directly in liters:$$\Delta V = \frac{10^7 \cdot 50 \cdot 276}{300 \cdot (10^5 + 10^3 \cdot 9.8 \cdot 40)} - 50$$$$\Delta V = \frac{1.38 \cdot 10^{11}}{300 \cdot (100000 + 392000)} - 50$$$$\Delta V = \frac{1.38 \cdot 10^{11}}{147600000} - 50 \approx 935 - 50 = 885 \text{ L}$$Answer$$\Delta V = 885 \text{ L}$$ | ||
| + | Let's write down the initial parameters of the air inside the cylinder: | ||
| + | 1. Initial volume: $V_1 = 50 \text{ L}$ | ||
| + | 2. Initial pressure: $P_1 = 10 \text{ MPa} = 10^7 \text{ Pa}$ | ||
| + | 3. Initial temperature: $T_1 = 27^\circ\text{C} = 300 \text{ K}$ | ||
| + | After the air is released into the ballast tank at a depth $h = 40 \text{ m}$, it reaches a new state. The external pressure at this depth consists of the atmospheric pressure $P_{\text{atm}}$ and the hydrostatic pressure of the water: | ||
| + | $$P_2 = P_{\text{atm}} + \rho gh$$ | ||
| + | where $\rho = 10^3 \text{ kg/m}^3$ is the density of water, and we take standard atmospheric pressure as $P_{\text{atm}} \approx 10^5 \text{ Pa}$. | ||
| + | The final temperature of the air after expansion is $T_2 = 3^\circ\text{C} = 276 \text{ K}$. | ||
| + | |||
| + | Since the mass of the air remains constant, we can apply the combined gas law: | ||
| + | $$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$ | ||
| + | |||
| + | From this, we can express the final total volume $V_2$ occupied by the air: | ||
| + | $$V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}$$ | ||
| + | $$V_2 = \frac{P_1 V_1 T_2}{T_1 (P_{\text{atm}} + \rho gh)}$$ | ||
| + | |||
| + | #### Note on the displaced volume: | ||
| + | The total volume of the air after expansion is $V_2$. However, the cylinder itself already holds a volume $V_1$. The volume of water actually displaced from the tank is only the "additional" volume the gas takes up as it leaves the cylinder. Therefore, the displaced volume of water $\Delta V$ is the difference between the final and initial gas volumes. | ||
| + | |||
| + | $$\Delta V = V_2 - V_1 = \frac{P_1 V_1 T_2}{T_1 (P_{\text{atm}} + \rho gh)} - V_1$$ | ||
| + | |||
| + | Let's substitute the numerical values into the equation. For convenience, we can keep $V_1$ in liters, which will give us the result for $\Delta V$ directly in liters: | ||
| + | $$\Delta V = \frac{10^7 \cdot 50 \cdot 276}{300 \cdot (10^5 + 10^3 \cdot 9.8 \cdot 40)} - 50 = 885 \text{ L}$$ | ||
| + | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $$\Delta V = 885 \text{ L}$$ | ||
| @@ -1,11 +1,34 @@ | |||
| ### Statement | ### Statement | ||
| $5.5.5.$ [Insert the problem statement] | $5.5.5.$ A gas cylinder with a volume of $50 \text{ L}$ was filled with air at $27^\circ\text{C}$ to a pressure of $10 \text{ MPa}$. What volume of water can be displaced from a submarine's ballast tank by the air from this cylinder, if the displacement occurs at a depth of $40 \text{ m}$? The temperature of the air after expansion is $3^\circ\text{C}$. | ||
| ### Solution | ### Solution | ||
| Statement$5.5.5.$ A gas cylinder with a volume of $50 \text{ L}$ was filled with air at $27^\circ\text{C}$ to a pressure of $10 \text{ MPa}$. What volume of water can be displaced from a submarine's ballast tank by the air from this cylinder, if the displacement occurs at a depth of $40 \text{ m}$? The temperature of the air after expansion is $3^\circ\text{C}$.SolutionLet's write down the initial parameters of the air inside the cylinder:Initial volume: $V_1 = 50 \text{ L}$Initial pressure: $P_1 = 10 \text{ MPa} = 10^7 \text{ Pa}$Initial temperature: $T_1 = 27^\circ\text{C} = 300 \text{ K}$After the air is released into the ballast tank at a depth $h = 40 \text{ m}$, it reaches a new state. The external pressure at this depth consists of the atmospheric pressure $P_{\text{atm}}$ and the hydrostatic pressure of the water:$$P_2 = P_{\text{atm}} + \rho gh$$where $\rho = 10^3 \text{ kg/m}^3$ is the density of water, and we take standard atmospheric pressure as $P_{\text{atm}} \approx 10^5 \text{ Pa}$.The final temperature of the air after expansion is $T_2 = 3^\circ\text{C} = 276 \text{ K}$.Since the mass of the air remains constant, we can apply the combined gas law:$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$From this, we can express the final total volume $V_2$ occupied by the air:$$V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}$$$$V_2 = \frac{P_1 V_1 T_2}{T_1 (P_{\text{atm}} + \rho gh)}$$Note on the displaced volume: The total volume of the air after expansion is $V_2$. However, the cylinder itself already holds a volume $V_1$. The volume of water actually displaced from the tank is only the additional volume the gas takes up as it leaves the cylinder. Therefore, the displaced volume of water $\Delta V$ is the difference between the final and initial gas volumes.$$\Delta V = V_2 - V_1 = \frac{P_1 V_1 T_2}{T_1 (P_{\text{atm}} + \rho gh)} - V_1$$Let's substitute the numerical values into the equation. For convenience, we can keep $V_1$ in liters, which will give us the result for $\Delta V$ directly in liters:$$\Delta V = \frac{10^7 \cdot 50 \cdot 276}{300 \cdot (10^5 + 10^3 \cdot 9.8 \cdot 40)} - 50$$$$\Delta V = \frac{1.38 \cdot 10^{11}}{300 \cdot (100000 + 392000)} - 50$$$$\Delta V = \frac{1.38 \cdot 10^{11}}{147600000} - 50 \approx 935 - 50 = 885 \text{ L}$$Answer$$\Delta V = 885 \text{ L}$$ | Let's write down the initial parameters of the air inside the cylinder: | ||
| 1. Initial volume: $V_1 = 50 \text{ L}$ | |||
| 2. Initial pressure: $P_1 = 10 \text{ MPa} = 10^7 \text{ Pa}$ | |||
| 3. Initial temperature: $T_1 = 27^\circ\text{C} = 300 \text{ K}$ | |||
| After the air is released into the ballast tank at a depth $h = 40 \text{ m}$, it reaches a new state. The external pressure at this depth consists of the atmospheric pressure $P_{\text{atm}}$ and the hydrostatic pressure of the water: | |||
| $$P_2 = P_{\text{atm}} + \rho gh$$ | |||
| where $\rho = 10^3 \text{ kg/m}^3$ is the density of water, and we take standard atmospheric pressure as $P_{\text{atm}} \approx 10^5 \text{ Pa}$. | |||
| The final temperature of the air after expansion is $T_2 = 3^\circ\text{C} = 276 \text{ K}$. | |||
| Since the mass of the air remains constant, we can apply the combined gas law: | |||
| $$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$ | |||
| From this, we can express the final total volume $V_2$ occupied by the air: | |||
| $$V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}$$ | |||
| $$V_2 = \frac{P_1 V_1 T_2}{T_1 (P_{\text{atm}} + \rho gh)}$$ | |||
| #### Note on the displaced volume: | |||
| The total volume of the air after expansion is $V_2$. However, the cylinder itself already holds a volume $V_1$. The volume of water actually displaced from the tank is only the "additional" volume the gas takes up as it leaves the cylinder. Therefore, the displaced volume of water $\Delta V$ is the difference between the final and initial gas volumes. | |||
| $$\Delta V = V_2 - V_1 = \frac{P_1 V_1 T_2}{T_1 (P_{\text{atm}} + \rho gh)} - V_1$$ | |||
| Let's substitute the numerical values into the equation. For convenience, we can keep $V_1$ in liters, which will give us the result for $\Delta V$ directly in liters: | |||
| $$\Delta V = \frac{10^7 \cdot 50 \cdot 276}{300 \cdot (10^5 + 10^3 \cdot 9.8 \cdot 40)} - 50 = 885 \text{ L}$$ | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $$\Delta V = 885 \text{ L}$$ | ||