Edits to “Statement”, “Answer”
en/5.5.26.md
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| @@ -1,11 +1,5 @@ | |||
| ### Statement | |||
| − | $5.5.26.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | ### Statement | ||
| − | |||
| $5.5.26^*.$ To measure the mass of water in fog droplets, a sample of air at a pressure $p_1 = 100 \text{ kPa}$ and a temperature $t_1 = 0 ^\circ\text{C}$ is hermetically sealed in a vessel with transparent walls, heated to a temperature at which the fog in the sample disappears, and the pressure at this temperature is measured. Estimate the mass of the fog in a sample volume $V = 1 \text{ m}^3$ if the fog disappearance temperature is $t_2 = 82 ^\circ\text{C}$, and the pressure in the vessel at this temperature is $p_2 = 180 \text{ kPa}$. | |||
| ### Solution | |||
| The idea of the problem is as follows: as the temperature increases, the fog droplets evaporate, and the partial pressure of water vapor is added to the air pressure. | |||
| First, it is necessary to convert the temperatures to the Kelvin scale: | |||
| $$T_1 = 273.15 + t_1$$ | |||
| $$T_2 = 273.15 + t_2$$ | |||
| Let's write down the equations for the initial and final states: | |||
| $$p_1 V = \nu_{\text{air}} R T_1$$ | |||
| $$p_2 V = \nu_{\text{air}} R T_2 + \frac{m_w}{M_w} R T_2$$ | |||
| Solving these equations, we obtain: | |||
| $$m_w = \frac{M_w V}{R T_2} \left( p_2 - p_1 \frac{T_2}{T_1} \right) \approx 0.305 \text{ kg}$$ | |||
| <br> | |||
| <i>Note: The answer in Savchenko's book is apparently incorrect.</i> | |||
| @@ -28,7 +22,3 @@Solution | |||
| #### Answer | |||
| $m = 305 \text{ g}$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,11 +1,5 @@ | |||
| ### Statement | ### Statement | ||
| $5.5.26.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $5.5.26^*.$ To measure the mass of water in fog droplets, a sample of air at a pressure $p_1 = 100 \text{ kPa}$ and a temperature $t_1 = 0 ^\circ\text{C}$ is hermetically sealed in a vessel with transparent walls, heated to a temperature at which the fog in the sample disappears, and the pressure at this temperature is measured. Estimate the mass of the fog in a sample volume $V = 1 \text{ m}^3$ if the fog disappearance temperature is $t_2 = 82 ^\circ\text{C}$, and the pressure in the vessel at this temperature is $p_2 = 180 \text{ kPa}$. | $5.5.26^*.$ To measure the mass of water in fog droplets, a sample of air at a pressure $p_1 = 100 \text{ kPa}$ and a temperature $t_1 = 0 ^\circ\text{C}$ is hermetically sealed in a vessel with transparent walls, heated to a temperature at which the fog in the sample disappears, and the pressure at this temperature is measured. Estimate the mass of the fog in a sample volume $V = 1 \text{ m}^3$ if the fog disappearance temperature is $t_2 = 82 ^\circ\text{C}$, and the pressure in the vessel at this temperature is $p_2 = 180 \text{ kPa}$. | ||
| ### Solution | ### Solution | ||
| The idea of the problem is as follows: as the temperature increases, the fog droplets evaporate, and the partial pressure of water vapor is added to the air pressure. | The idea of the problem is as follows: as the temperature increases, the fog droplets evaporate, and the partial pressure of water vapor is added to the air pressure. | ||
| First, it is necessary to convert the temperatures to the Kelvin scale: | First, it is necessary to convert the temperatures to the Kelvin scale: | ||
| $$T_1 = 273.15 + t_1$$ | $$T_1 = 273.15 + t_1$$ | ||
| $$T_2 = 273.15 + t_2$$ | $$T_2 = 273.15 + t_2$$ | ||
| Let's write down the equations for the initial and final states: | Let's write down the equations for the initial and final states: | ||
| $$p_1 V = \nu_{\text{air}} R T_1$$ | $$p_1 V = \nu_{\text{air}} R T_1$$ | ||
| $$p_2 V = \nu_{\text{air}} R T_2 + \frac{m_w}{M_w} R T_2$$ | $$p_2 V = \nu_{\text{air}} R T_2 + \frac{m_w}{M_w} R T_2$$ | ||
| Solving these equations, we obtain: | Solving these equations, we obtain: | ||
| $$m_w = \frac{M_w V}{R T_2} \left( p_2 - p_1 \frac{T_2}{T_1} \right) \approx 0.305 \text{ kg}$$ | $$m_w = \frac{M_w V}{R T_2} \left( p_2 - p_1 \frac{T_2}{T_1} \right) \approx 0.305 \text{ kg}$$ | ||
| <br> | <br> | ||
| <i>Note: The answer in Savchenko's book is apparently incorrect.</i> | <i>Note: The answer in Savchenko's book is apparently incorrect.</i> | ||
| @@ -28,7 +22,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $m = 305 \text{ g}$ | $m = 305 \text{ g}$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||