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InterviewSolution
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Calculate the number of molecules is 2 c.c of the perfect gas at `27^(@)C` at a pressure of 10 cm. of mercury. Mean kinetic energy of each molecule at `27^(@)C = 7 xx 10^(-4) erg` and acceleration due to gravity `= 980 cm s^(-2)`. |
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Answer» Correct Answer - `5.71 xx 10^(8)` K.E of the given gas at `27^(@)C` `=1/2 MC^(2) = 3/2 PV` `=3/2 xx (10 xx 13.6 xx 980) xx 2 ergs` `:.` No. Of molecules in the given volume `=(K.E. "of given gas")/(K.E."per molecule of a gas")` `(3/2 xx 10 xx 13.6 xx 980 xx2)/(7 xx 10^(-4)` `=5.71 xx 10^(8)` |
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