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In a `2.5` L flask at `27^(@)C` temperature, the pressure of a gas was found to be 8 atm. If ` 41 xx 10^(23)` molecules of the same gas are introduced into the container, the temperature changed to `T_(2)`. The pressure of gas is found to be 10 atm. Find out the value of `T_(2)`.A. 253 KB. 347 KC. 230 KD. 370 K |
Answer» `6.023 xx 10^(23)` molecules correspond to 1 mole ` 2.41 xx 10^(23) ` molecules correspond to ? ` = (2.41 xx 10^(23) ) / (6.023 xx 10^(23)) = 0.4` moles Case (i) `._(1)V_(1) = n_(1).R_(1)." "T_(1)` ` 8.(2.5) = n_(1)(0.08).(300)` `n_(1) = (8(2.5))/((0.08)(300))` ` n_(1) = 0.833` moles Total moles, `n_(2) = 0.833 + 0.4` ` n_(2) = 1.233` Case (ii) ` P_(2)V_(2) = n_(2)R T_(2)` ` 10(2.5) = 1.233 (0.08). T_(2)` `T_(2) = 253.4` K |
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