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Q. The average kinetic energy of an ideal gas, per molecule at $ {{25}^{o}}C $ will be

Punjab PMETPunjab PMET 1999

Solution:

Here: Temperature $ (T)={{25}^{o}}C=298\,\,K $ . Average kinetic energy permolecule $ =\frac{3RT}{2n}=\frac{3\times 8.314\times 298}{2\times (6.02\times {{10}^{23}})} $ $ =6.17\times {{10}^{-21}}J $ (where $ R=8.314 $ )