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18 March, 01:44

A sample of argon gas (molar mass 40 g) is at four times the absolute temperature of a sample of hydrogen gas (molar mass 2 g). the ratio of the rms speed of the argon molecules to that of the hydrogen is

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  1. 18 March, 01:48
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    To solve this problem, let us recall that the formula for gases assuming ideal behaviour is given as:

    rms = sqrt (3 R T / M)

    where

    R = gas constant = 8.314 Pa m^3 / mol K

    T = temperature

    M = molar mass

    Now we get the ratios of rms of Argon (1) to hydrogen (2):

    rms1 / rms2 = sqrt (3 R T1 / M1) / sqrt (3 R T2 / M2)

    or

    rms1 / rms2 = sqrt ((T1 / M1) / (T2 / M2))

    rms1 / rms2 = sqrt (T1 M2 / T2 M1)

    Since T1 = 4 T2

    rms1 / rms2 = sqrt (4 T2 M2 / T2 M1)

    rms1 / rms2 = sqrt (4 M2 / M1)

    and M2 = 2 while M1 = 40

    rms1 / rms2 = sqrt (4 * 2 / 40)

    rms1 / rms2 = 0.447

    Therefore the ratio of rms is:

    rms_Argon / rms_Hydrogen = 0.45
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