InterviewSolution
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InterviewSolution
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A flask contains argon and chlorine in the ratio 2:1 by mass. The temperature of the mixture is `27^(@)`. Obtain the ratio of (i) average kinetic energy per molecule, and (ii) root mean square speed of the molecules of two gases. Atomic mass of argon = 39.9 u, Molecular mass of chlorine = 70.9 u. |
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Answer» we know that average `K.E.//"molecure"` of any ideal gas `= 3.2 K_(B) T` . Where `K_(B)` is Boltzmann constant and T is temperature. It does not depend upon nature of the gas. (i) As argon and chlorine, both have the same temperature in flask, the ratio of average `K.E//"moluecule"` of the gases = `1:1`. (i) If m is mass of a molecule of the gas, then average `K.E//"molecule" =1/2 m upsilon_(rms)^(2)` `:. 1/2 m_(1)upsilon_(1)^(2) = 1/2 m_(2)upsion_(2)^(2) = 3/2 K_(B)T` or ` (upsion_(1)^(2))/(upsion_(2)^(2)) = (m_(2))/(m_1) = (M_2)/(M_1) = (70.9)/(39.9) = 1.77` where M represents the `"atomic" // "molecular"` mass of the gas (For argon, a molecule is just an atom of argon). `:. (upsilon_1)/(upsilon_2) = sqrt(1.77) = 1.33` . |
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