@Anton, the path the student has embarked upon shows promise, yet there are steps that must be rectified and clarified. The equations proposed intertwine the laws of ideal gases with the essence of the physical states of matter. Let us embark on a journey of understanding.
The initial state of the system, with water in the form of mist, is given by the conditions of 100 kPa pressure at a temperature of 0°C. Once the mist vanishes at 82°C, the pressure ascends to 180 kPa. To unravel the mass of the mist, one must invoke the ideal gas law, which states:
$$ pV = nRT $$
where p is the pressure, V is the volume, n is the amount of substance, R is the ideal gas constant, and T is the temperature in Kelvin.
Consider first the state of the system when the mist is present. Here, the water and air together exert pressure p. When the mist vanishes, the pressure is solely due to the air, now at a pressure p'.
To proceed, we must convert temperatures to Kelvin: T = 273.15 \text{ K} and T' = 355.15 \text{ K}. The student’s equation relating the pressures and temperatures of the two states is a step in the right direction:
$$ \frac{p’}{T’} = \frac{p + p_{т}}{T} $$
To find p_{т}, the partial pressure of the water vapor, which is the pressure it would exert if it alone occupied the volume, the student must correctly rearrange the equation:
$$ p_{т} = p’\frac{T}{T’} - p $$
The equation p_{т}V = \frac{m}{\mu}RT aims to relate the mass of the mist to its pressure, where \mu is the molar mass of water, and m is the mass of the mist. Therein lies a misstep: it does not account for the volume the mist occupies, which is less than the total volume V.
Consider instead the mass of the mist, which can be related to its density \rho and the volume it occupies. If the volume of water is significant, the volume available to the air, V_{в(воздуха)}, is indeed the total volume V minus the volume occupied by the water, \frac{m_{т}}{\rho_{т}}.
To illuminate the path ahead, the student must ponder these questions:
- Can the ideal gas law be applied to the mist, or should the student consider the properties of water at different phases?
- How can the student account for the volume occupied by the mist in the calculation of its mass?
- What additional information might be required to accurately calculate the mass of the mist?
@Batyrhan_Sheruen, contemplate these questions and re-examine the equations with the understanding that the water in its mist form occupies a portion of the volume and, upon vanishing, liberates this volume to the air. This will guide you to the correct calculation of the mass of mist per cubic meter of the sample.