For a spacecraft or a molecule to leave the moon, it must reach the escape velocity (speed) of the moon, which is 2.37 km/s. The average day

Question

For a spacecraft or a molecule to leave the moon, it must reach the escape velocity (speed) of the moon, which is 2.37 km/s. The average daytime temperature of the moon’s surface is 365 K. What is the rms speed (in m/s) of an oxygen molecule at this temperature?rms speed = m/s

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Trung Dũng 4 years 2021-07-27T19:43:23+00:00 2 Answers 83 views 0

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    0
    2021-07-27T19:44:51+00:00

    Answer:

    Vrms = 533 m/s

    Explanation:

    The formula for the root mean square molecular speed is given as;

    V_rms=√(3RT/M)

    Where;

    M is the molar mass of the molecule

    T is Temperature

    R is gas constant which has a value of 8.31 J/mol.k

    V_rms is root mean square speed

    Now let’s calculate the molar mass of the molecule whuch in this case is oxygen.

    Oxygen has a formula of O_2

    Now, molar mass of one atom is 16 g/mol

    Thus, molar mass of the 2 atoms would be; 2 x 16 = 32 g/mol = 0.032 kg/mol

    T = 365K

    So, plugging in the relevant values, we obtain;

    V_rms = √[(3×8.31×365 )/0.032]

    Vrms = 533 m/s

    0
    2021-07-27T19:44:55+00:00

    Answer:

    Vrms = 291 m/s

    Explanation:

    The root mean square velocity or vrms is the square root of the average square velocity and is. vrms=√3RTM. Where M is equal to the molar mass of the molecule in kg/mol.

    Temperature = 365 K

    Root mean square velocity = ?

    molar mass of oxygen = 16 g/mol.

    But xygen gas (O2) is comprised of two oxygen atoms bonded together. Therefore:

       molar mass of O2 = 2 x 16

       molar mass of O2 = 32 g/mol

       Convert this to kg/mol:

       molar mass of O2 = 32 g/mol x 1 kg/1000 g

       molar mass of O2 = 3.2 x 10-2 kg/mol

    Molar mass of Oxygen = 3.2 x 10-2 kg/mol

    Vrms = √[3(8.3145 (kg·m2/sec2)/K·mol)(365 K)/3.2 x 10-2 kg/mol]

    Vrms = 291 m/s

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