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The Mean Free Path of a Molecule Is 2.25×107 m2.25 \times 10^{-7} \mathrm{~m}

Question 38

Multiple Choice

The mean free path of a molecule is 2.25×107 m2.25 \times 10^{-7} \mathrm{~m} in an ideal gas at standard temperature and pressure (0C\left(0^{\circ} \mathrm{C}\right. and 1 atm) 1 \mathrm{~atm}) . What is the mean free path if the pressure of the gas is doubled and the temperature is tripled?


A) 0.75×107 m0.75 \times 10^{-7} \mathrm{~m}
B) 3.38×107 m3.38 \times 10^{-7} \mathrm{~m}
C) 3.75×107 m3.75 \times 10^{-7} \mathrm{~m}
D) 2.25×107 m2.25 \times 10^{-7} \mathrm{~m}
E) 1.13×107 m1.13 \times 10^{-7} \mathrm{~m}

Correct Answer:

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