Chemistry · Mole concept, molar mass, percentage composition, empirical and molecular formulae
What is the density of wet air with relative humidity at 1 atm and ? Given : vap
What is the density of wet air with \( 75 \% \) relative humidity at 1 atm and \( 300 \mathrm{K} \) ? Given : vapour pressure of \( H_{2} O \) is 30 torr and average molar mass of air is \( 29 g / m o l \)
- A. \( 1.614 g / L \)
- B. \( 0.96 g / L \)
- C. \( 1.06 g / L \)
- D. \( 1.164 g / L \)
Step-by-step solution
Given RH=75%, saturation vapor pressure of H2O is 30 torr, total pressure 1 atm=760 torr, temperature 300 K. Partial pressure of H2O = 0.75*30 = 22.5 torr. Partial pressure of dry air = 760-22.5 = 737.5 torr. Mole fractions: dry air = 737.5/760, H2O = 22.5/760. Average molar mass M_avg = (737.5/760)*29 + (22.5/760)*18 ≈ 28.674 g/mol. Using ideal gas law: density ρ = (P * M_avg)/(R*T) = (1 atm * 28.674 g/mol)/(0.0821 L·atm/(mol·K) * 300 K) ≈ 1.164 g/L.
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