InterviewSolution
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InterviewSolution
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The temperature of an open room of volume `30 m^(3)` increases from `17^(@)C to 27^(@)C` due to sunshine. The atmospheric pressure in the room remains `1 xx 10^(5) Pa`. If `n_(i) and n_(f)` are the number of molecules in the room before and after heating then `n_(f)` and `n_(i)` will beA. `2.5 xx 10^(25)`B. `-2.5 xx 10^(25)`C. `-1.61 xx 10^(23)`D. `1.38 xx 10^(23)` |
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Answer» Correct Answer - B Here, `V_(1) = 30m^(3), T_(1) = 17+273 = 290 K` `P_(1) = 1 xx 10^(5)Pa` `V_(2) = 30 m^(3) , T_(2) = 27+273 = 300K`, `P_(2) = 1 xx 10^(5) Pa`. Let `N_(1),N_(2)` be the no, of moles of a gas at temperature `T_(1) and T_(2)` respectively. then `N_(1) =(P_(1)V_(1))/(RT_(1)) = ((1 xx 10^(5)) xx 30)/(83 xx 290) = 1.24 xx 10^(3)` or `N_(2) = (P_(2)V_(2))/(RT_(2)) = ((1 xx 10^(5)) xx 30)/(83 xx 300) = 1.20 xx 10^(3)` ltbr. Change in the number of moles `N_(2)-N_(1) =(1.20-1.24) xx 10^(3) = -0.04 xx 10^(3)` Change in the number of molecules `n_(f)-n_(i) = (N_(2)-N_(1)) xx (6.023 xx 10^(23))` `= - (0.04 xx 10^(3)) xx (6.023 xx 10^(23))` =-2.5 xx 10^(25)`. |
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