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Estimate, to within an order of magnitude, on the basis of kinetic theory the heat conductivity of a gas in terms of its temperature, density, molecular weight, and heat capacity at constant volume. Make your own estimates of collision cross-sections and molecular mean free paths. You may restrict your attention to pressures near atmospheric, temperatures near room temperature and dimensions of the order of centimeters or meters. Do not concern yourself with heat transfer by convection. $\left(k=1.38 \times 10^{-16} \mathrm{erg} / \mathrm{K}\right)$.

   Estimate, to within an order of magnitude, on the basis of kinetic theory the heat conductivity of a gas in terms of its temperature, density, molecular weight, and heat capacity at constant volume. Make your own estimates of collision cross-sections and molecular mean free paths. You may restrict your attention to pressures near atmospheric, temperatures near room temperature and dimensions of the order of centimeters or meters. Do not concern yourself with heat transfer by convection. $\left(k=1.38 \times 10^{-16} \mathrm{erg} / \mathrm{K}\right)$.
 
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Problems and Solutions on Thermodynamics and Statistical Mechanics
Problems and Solutions on Thermodynamics and Statistical Mechanics
U.S.T. of China… 1st Edition
Chapter 2, Problem 200 ↓

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We need to estimate the heat conductivity \( k \) of a gas in terms of its temperature \( T \), density \( \rho \), molecular weight \( M \), and heat capacity at constant volume \( C_V \). The heat conductivity can be expressed using kinetic theory as: \[ k  Show more…

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Estimate, to within an order of magnitude, on the basis of kinetic theory the heat conductivity of a gas in terms of its temperature, density, molecular weight, and heat capacity at constant volume. Make your own estimates of collision cross-sections and molecular mean free paths. You may restrict your attention to pressures near atmospheric, temperatures near room temperature and dimensions of the order of centimeters or meters. Do not concern yourself with heat transfer by convection. $\left(k=1.38 \times 10^{-16} \mathrm{erg} / \mathrm{K}\right)$.
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