Question

Problem #4: We will later develop a formula based on the kinetic theory of gases to predict the self-diffusivity of a single-species ideal gas as: $D_{gas} = \sqrt{\frac{1}{6M \pi d^2 P N_A}(RT)^{3/2}}$ A simplified version of this formula, using non-SI units is often given as: $D_{gas} = 1.61 \times 10^{-3} \sqrt{\frac{1}{M d^2 P}(T)^{3/2}}$ where T is in Kelvin, P is in atm, d is in Å, and M is in grams per mole. Derive how the pre-factor of 1.61 x 10?³ is obtained. Clearly show all your work.

          Problem #4: We will later develop a formula based on the kinetic theory of gases to predict the
self-diffusivity of a single-species ideal gas as:
$D_{gas} = \sqrt{\frac{1}{6M \pi d^2 P N_A}(RT)^{3/2}}$
A simplified version of this formula, using non-SI units is often given as:
$D_{gas} = 1.61 \times 10^{-3} \sqrt{\frac{1}{M d^2 P}(T)^{3/2}}$
where T is in Kelvin, P is in atm, d is in Å, and M is in grams per mole. Derive how the pre-factor of
1.61 x 10?³ is obtained. Clearly show all your work.

Problem #4: We will later develop a formula based on the kinetic theory of gases to predict the
self-diffusivity of a single-species ideal gas as:
Dgas = √((1)/(6M π d^2 P NA)(RT)^3/2)
A simplified version of this formula, using non-SI units is often given as:
Dgas = 1.61 × 10^-3√((1)/(M d^2 P)(T)^3/2)
where T is in Kelvin, P is in atm, d is in Å, and M is in grams per mole. Derive how the pre-factor of
1.61 x 10?³ is obtained. Clearly show all your work.

Added by Charlotte M.

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