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SOLUTION TO EXERCISE 13
SOLUTION TO EXERCISE 13
At S.T.P., one mole of gas occupies 0.0224 m^3. Thus, the number density is
6.02 * 10^23
n = -------------- = 2.69 * 10^25 particles per m^3. Since 1 m = 100 cm,
0.0224 m^3
(1 m)^3 = (100 cm)^3 = 10^6 cm^3; so the number density is also
6.02 * 10^23
n = -------------- = 2.69 * 10^19 particles per cm^3.
22400 cm^3
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