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Valter edited Aug 25, 2026, 6:51 PM

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		+	### 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>
		+
		+	#### Answer
		+	$m = 305 \text{ g}$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### 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>

			#### Answer
			$m = 305 \text{ g}$

			#### Answer

			[Insert a concise answer or boxed result]