Issue by KC: The environmental protection company (EPA) has proposed a new common for microparticulates in air:?
A, For particles up to 2.five μm in diameter, the highest allowable sum is 50 μg/m3. If your eight.two ft × twelve.two ft × 9.9 ft dorm space just meets the new EPA common, how numerous of these particles are in your place? (Suppose the particles are spheres of 2.five μm diameter, and manufactured mainly of soot, a kind of carbon with a density of two.5 g/cm3.) Part B, How numerous of these particles are in each six.47×10-1 L breath you acquire?

Greatest solution:

Reply by nealjking
a) Place:
Volume: 8.two*twelve.two*nine.9 = 990.4 (ft^three) = 28 (m^three)

Allowable mass:
50e-six (g) * 28 = one.4e-3 (g)

Mass per particle:
four*pi*r^three/3 * two.5 = pi*d^three/six * two.five = pi*(two.5e-4)^3 *two.5 = 1.23e-10 (g)

Amount = (1.4e-three)/(one.23e-10) = one.14e7

b) Number in a breath:
The quantity density is one.14e7/28 = four.07e5/m^three

Given that 1 Liter = 1e-3 * m^3, the range density is four.07e8/L, so the amount in a breath of 6.47e-one (L) is
6.47e-one * four.07e8 = six.47*four.07 * e7 = two.63e8

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