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What is the rate of diffusion of nitrogen gas in air at 25°C and 1 atm pressure, if the molecular weight of nitrogen is 28 g/mol and the average velocity of the gas molecules is 500 m/s?

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To calculate the rate of diffusion of nitrogen gas in air, we can use Graham's law of diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. However, we need to find the molar mass of air first.Air is composed of approximately 78% nitrogen  N2 , 21% oxygen  O2 , and 1% argon  Ar  by volume. The molecular weights of N2, O2, and Ar are 28 g/mol, 32 g/mol, and 40 g/mol, respectively.To find the average molar mass of air, we can use the following equation:M_air =  fraction of N2  M_N2  +  fraction of O2  M_O2  +  fraction of Ar  M_Ar M_air =  0.78  28 g/mol  +  0.21  32 g/mol  +  0.01  40 g/mol M_air  28.97 g/molNow, we can use Graham's law of diffusion to find the rate of diffusion of nitrogen gas in air:Rate_N2 / Rate_air =  M_air / M_N2 Rate_N2 / Rate_air =  28.97 g/mol / 28 g/mol Rate_N2 / Rate_air  1.018Since we are given the average velocity of nitrogen gas molecules  500 m/s , we can assume that this is the rate of diffusion of nitrogen in air:Rate_N2 = 500 m/sNow, we can find the rate of diffusion of air:Rate_air = Rate_N2 / 1.018Rate_air  491.16 m/sSo, the rate of diffusion of nitrogen gas in air at 25C and 1 atm pressure is approximately 500 m/s, while the rate of diffusion of air is approximately 491.16 m/s.

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