Assuming the radius of diatomic molecules is approximately 1.0×10 -10 m for what
ID: 1489706 • Letter: A
Question
Assuming the radius of diatomic molecules is approximately 1.0×10-10 m for what pressure will the mean free path in room-temperature (20°C) nitrogen be 8.8 m? The Boltzmann constant is 1.38 × 10-23 J/K, Avogadro's number is 6.02 × 1023 molecules/mole, and the ideal gas constant is R = 8.314 J/mol·K = 0.0821 L·atm/mol·K.
Pleas show all work, and please specify the correct answer choice at the bottom. Also, please do not present answers that are not one of the choices!!! Thank you!
A. 1.1×107 atm B. 3.6×108 atm C. 2.6×108 atm D. 8.0×108 atmExplanation / Answer
Mean free path,
L = Kb*T / (sqrt(2) *pi* d^2*p)
kB is the Boltzmann constant in J/K,
T is the temperature in K,
p is pressure in Pascals,
and d is the diameter
L = Kb*T / (sqrt(2) *pi* d^2*p)
8.8= 1.38*10^-23 * 293 / sqrt(2)*pi*(2*10^-10)^2 * p
p = 2.6*10^-3 Pa
= 2.6*10^-3 / (1*10^-5) atm
= 2.6*10^-8 atm
Answer: C
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