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For a monatomic gas, kinetic energy `= E`. The relation with `rms` velocity isA. `u = ((2E)/(m))^(1//2)`B. `u = ((E)/(2m))^(1//2)`C. `u = ((3E)/(2m))^(1//2)`D. `u = ((E)/(3m))^(1//2)` |
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Answer» Correct Answer - A According to the kinetic gas eqauation , `pV = (1)/(3) mnc^(2)` For a single molecule of mass `m , n = 1` . Thus , `u_(rms) = sqrt((3 pV)/(m))` (1) According to gas equation , for one mole of gas , we have `pV = RT` For a single molecule , we have `pV = (R )/(N_(0))T = kT` (2) lt brgt where `k` is called the Boltzmann constant. Average kinetic energy per molecule `(E)` is given as `E = (3)/(2) kT` or `kT = (2)/(3) E` (3) Combining Eqs. (1),(2), and (3) , we have ` u_(rms) = sqrt((3 xx (2)/(3)E)/(m)) = sqrt((2E)/(m))` |
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