Ask Question
11 April, 02:56

If he (g) has an average kinetic energy of 5610 j/mol under certain conditions, what is the root mean square speed of o2 (g) molecules under the same conditions?

+2
Answers (1)
  1. 11 April, 05:29
    0
    Kinetic energy is the energy possessed by a particle or a body when in motion.

    It is calculated by 1/2 mv², where m is the mass and v is the velocity.

    But from the Grahams law of diffusion

    V (he) / V (o2) = √ (M (o2) / M (he)

    = √ (32/4)

    = 2.83

    This means that Helium diffuses at a rate of 2.83 times faster than oxygen.

    Since the conditions are the same then the two gases will have equal kinetic energy,

    Hence; (He = 4g/mol, O2 = 32g/mol

    5610 = 0.5 * 4/1000 * v²

    v² = 2805000

    v = 1674.8 m/s for He

    Therefore, since He diffuses at a rate of 2.83 times faster than oxygen

    Then, the velocity of Oxygen will be;

    1674.8/2.83

    = 591.8 m/s

    Thus, the velocity of oxygen molecules is 591.8 m/s
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “If he (g) has an average kinetic energy of 5610 j/mol under certain conditions, what is the root mean square speed of o2 (g) molecules ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers