InterviewSolution
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InterviewSolution
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When 0.1 mol of a gas absorbs `41.75 J` of heat at constant volume, the rise in temperature occurs equal to `20^(@)C`. The gas must beA. monoatomicB. diatomicC. triatomicD. polyatomic |
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Answer» Correct Answer - B Atomicity of gas is deiced by the ratio `C_(P)//C_(V)`. `q = nC_(V) Delta T` Thus molar heat capacity at constant volume `C_(V) = (q)/(n Delta T)` `= (41.75 J)/((0.1 mol)(20 K))` `20.88 J K^(-1) mol^(-1)` For an ideal gas, `C_(P) - C_(V) = R` or `C_(P) = R + C_(V)` `= (8.314) + (20.88)` `= 29.19 J K^(-1) mol^(-1)` Finally, `(C_(P))/(C_(V)) = (29.19)/(20.88) = 1.40` This implies than the gas is diatomic. |
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