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What is the diffusion coefficient of a gas with a molar mass of 16 g/mol in air at 25°C, assuming the gas follows Graham's law?

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Graham's law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. To find the diffusion coefficient  D  of a gas, we can use the following equation:D =  D * M  / MWhere:D = diffusion coefficient of the reference gas  air in this case M = molar mass of the reference gas  air M = molar mass of the given gasFirst, we need to find the diffusion coefficient of air. For this, we can use the Chapman-Enskog equation:D =  3.0  10  * T/ /  P *  *  Where:T = temperature in Kelvin  25C = 298.15 K P = pressure in atm  assuming 1 atm  = collision diameter  for air, approximately 3.62  10 m  = collision integral  for air, approximately 1.0 Plugging in the values:D =  3.0  10  *  298.15 / /  1 *  3.62  10  * 1.0 D  2.0  10 m/sNow, we can find the diffusion coefficient of the given gas. The molar mass of air is approximately 29 g/mol. The molar mass of the given gas is 16 g/mol.D =  2.0  10 * 29  / 16D   2.0  10 * 29  / 4D  1.45  10 m/sSo, the diffusion coefficient of the gas with a molar mass of 16 g/mol in air at 25C is approximately 1.45  10 m/s, assuming the gas follows Graham's law.

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