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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. triatomicB. diatomicC. polyatomicD. monoatomic |
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Answer» Correct Answer - D `C_(V)` (heat absorbed per deg.rise per mole) `=(41.75 J)/(0.1 mol xx20^(@))=20.875 JK^(-1)mol^(-1)` `C_(P)=C_(V)+R=20.875+8.314JK^(-1) mol^(-1)` `=29.189 JK^(-1)mol^(-1)` `(C_(P))/(C_(V))=(29.189)/(20.875)=1.40` `:.` The gas is monoatomic |
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