Question

The concentration of Ozone corresponding to a mixing ratio of 120 ppbv at pressure of 1 atmosphere and temperature of 25°C is ________ \(μg/m^3\) (rounded off to 1 decimal place).
Atomic weight of oxygen = 16; R= 8.314 J/K-g.mole.

Answer 1

Correct Answer: 234

Step 1: Convert temperature to Kelvin and calculate the molar mass of ozone. \[ T = 25°C + 273.15 = 298.15 \, K \] \[ M_{O_3} = 16 \times 3 = 48 \, \text{g/mol} \] Step 2: Use the ideal gas law to find the molar volume at STP. \[ V = \frac{nRT}{P} = \frac{1 \times 0.0821 \times 298.15}{1} = 24.467 \, \text{L} \] Step 3: Calculate the concentration of ozone in µg/m³ for 120 ppbv. \[ \text{Concentration} = \frac{48 \times 10^6 \text{ µg}}{24.467 \text{ m}^3} \times \frac{120}{10^9} = 234.0 \text{ µg/m}^3 \]