Lessons 46–47 · 2 lessons · P. Habibullayev, A. Boydedayev, A. Bahromov, K. Suyarov, J. Usarov, M. Yuldasheva. Physics Grade 9, revised and expanded 3rd edition. G‘afur G‘ulom Publishing House, Tashkent, 2019
46
Problem solving
Textbook: p. 125
GoalSolve problems on air humidity: ρ, φ, p and p₀, the dew point, ρ = pM/(RT).
New words
saturated vapour pressure p₀ · to‘yingan bug‘ bosimi p₀vapour density · bug‘ zichligicalculating from the dew point · shudring nuqtasidan hisoblashmolar mass M · molyar massa M
Explanation
In humidity problems first identify the given quantities: φ, ρ or p, and the temperature. Take ρ₀ or p₀ at that temperature from the table. Since relative humidity is φ = ρ/ρ₀ = p/p₀, we have ρ = φρ₀ and p = φp₀. The vapour density can also be found from the pressure: from pV = (m/M)RT we get ρ = pM/(RT); for water vapour M = 0.018 kg/mol, R = 8.31 J/(mol·K), T = t + 273 K. If the dew point is given, the actual vapour pressure p in the air equals the saturated pressure at the dew point; then φ is p divided by p₀ at the actual air temperature. The mass of vapour in a room is m = ρV. Finally check the answer: φ cannot exceed 100 %.
Worked examples
At 20 °C with φ = 60 %: ρ = 0.6·17.3 ≈ 10.4 g/m³.
Air at 20 °C (p₀ = 2.33 kPa), dew point 8 °C (p = 1.06 kPa): φ = 1.06/2.33·100 % ≈ 45 %.
Class activity
Calculation task: measure the volume of your room (length·width·height). Measure the room temperature and take ρ₀ from the table. Assuming a relative humidity of 50 %, find the mass of water vapour in the room. How many drops would that much water make in a glass?
Practice
1
Why can φ not exceed 100 %?
Excess vapour condenses, so there cannot be more than saturated vapour.
2
In a 60 m³ room at 20 °C, φ = 50 %. What is the vapour mass? (ρ₀ = 17.3 g/m³)
m = 0.5·17.3·60 = 519 g
3
The saturated vapour pressure at 20 °C is p₀ = 2330 Pa. Find ρ₀ (M = 0.018, R = 8.31, T = 293 K).