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Chemistry
The average kinetic energy of an ideal gas, per molecule at 25oC will be
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Q. The average kinetic energy of an ideal gas, per molecule at $ {{25}^{o}}C $ will be
Punjab PMET
Punjab PMET 1999
A
$ 6.17\times {{10}^{-21}}J $
B
$ 6.17\times {{10}^{-20}}J $
C
$ 61.7\times {{10}^{-21}}J $
D
$ 7.16\times {{10}^{-20}}J $
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 $ )