Answer:
[tex]1.25\ \mu\text{g/m}^3[/tex]
Explanation:
v = Velocity of the breeze = 4 m/s
w = Width of the valley = 5000 m
h = Height of the valley = 1000 m
Volumetric flow rate is given by
[tex]\dot{V}=vwh\\\Rightarrow \dot{V}=4\times 5000\times 1000\\\Rightarrow \dot{V}=2\times10^{7}\ \text{m}^3/\text{s}[/tex]
[tex]\dot{m}[/tex] = Mass flow rate of pollutant = 25 g/s = [tex]25\times 10^6\ \mu\text{g/s}[/tex]
Concentration is given by
[tex]C=\dfrac{\dot{m}}{\dot{V}}\\\Rightarrow C=\dfrac{25\times 10^6}{2\times 10^7}\\\Rightarrow C=1.25\ \mu\text{g/m}^3[/tex]
The steady state concentration of pollutants in the valley, is [tex]1.25\ \mu\text{g/m}^3[/tex].
