Homework Answers
Given that 0.456 * 10-6 g of NH3 is
present in 1 m3 of Air.
For Ideal gas,
PV = NRT
V = NRT / P
Here, N = 0.456 * 10-6 / (14 + (1.1 * 3))
= 0.026 * 10-6 mole
R = 0.0821 L atm / mol K = 8.21 * 10-5 m3 atm
/ mol K
Thus V = (0.026 * 10-6 * 8.21 * 10-5 *
273) / 1
= 5.83 * 10-10 m3
Thus, mixing ratio in ppbv = (5.83 * 10-10) *
109 / 1
= 0.583 m3 / m3
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