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An open vessel at 27^@C is heated until three fifth of the air in it has been expelled। Assuming volume of vessel constant find the temperature to which the vessel has been heated |
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Answer» Solution :When an open vessel is HEATED, its volume and PRESSURE may be regarded as constant। Suppose the pressure and volume in the vessel are P and V respectively। If `n_1` represents the number of moles of the gas in the vessel before heating and `n_2` after heating, we have (according to the ideal gas equation), `PV=n_1 RT_1 "" ...(i)` and `PV = n_2 RT_2"" ...(ii)` where `T_1`is the temperature of the gas before heating and `T_2` after heating. Dividing Eqs. (i) by (ii), we have `1 = (n_1 T_1)/(n_2 T_2) "or" n_2/n_2 = T_1/T_1""..(iii)` In the present case, `T_1= 273 + 27 = 300 K " and " T_2` = ? Since on heating three fifth air is EXPELLED, therefore `n_2 = n_1 - (n_1 xx 3/5 ) = 2/5 n_1` Substituting the values is Eq। (iii), we have `(2/5n_1)/n_1 =300/T_2` `:. "" T_2 = 750K` Hence, the vessel has been heated to `750 - 273 = 477^@C`. |
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