Question

Assuming the radius of diatomic molecules is approximately 2.3 ×10-10 m for what pressure in Pa will the mean free path in room-temperature (20°C) nitrogen be 0.027 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 = 8.315 J/mol•K = 0.0821 L ∙ atm/mol ∙ K.

Answer 1

For a radius of diatomic molecules is approximately 2.3 ×10-10 m, the  pressure in Pa is mathematically given as

P=0.134Pa

What pressure in Pa will the mean free path in room-temperature (20°C) nitrogen be 0.027 m?

Generally, the equation for the ideal gas  is mathematically given as

PV=nRT

Therefore

$P=\frac{1kT}{\sqrt{2}\pi(p)d^2}$

$P=\frac{1.38*10^-23*293}{\sqrt{2}\pi*(2*1.3*10^{-10})^2)*0.1}$

P=0.134Pa

In conclusion, Pressure

P=0.134Pa

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