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InterviewSolution
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A mixture contains `8g He` and `14 g N_(2)` in a vessel at `300K`. How much heat is required to increase the rms speed of these molecules to double their value. Also calculate the final temperatures. |
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Answer» Mole of `He = (8)/(4) = 2` Mole of `N_(2) = (14)/(28) = 0.5` `KE` of `He` at `300 K = (3)/(2) xx2xx 8.314 xx 300 = 7.4833 xx 10^(3)J` `KE` of `N_(2)` at `300K = (3)/(2) xx 0.5 xx 8.314 xx 300 = 1.873 xx 10^(3)J` Now, since `KE = (1)/(2) M (U_(rms))^(2)` On doubling `U_(rms)`, kinetic enegry will becomes four times of initial value. `:.` New `KE of He = 4 xx 7.483 xx 10^(3) J = 2.99 xx 10^(4)J` New `KE of N_(2) = 4 xx 1.871 xx 10^(3)J = 7.484 xx 10^(3)J` Thus, for `He, 4 xx 7.483 xx 10^(3) = (3)/(2)RT xxn (T` is new temperature) ltbRgt `= (3)/(2) xx2xx8.314 xxT` `T =1200K` Now, heat given at constant volume to gases to heat them from `300K` to `1200K` is `DeltaH_(He) = n xx C_(V) xxDeltaT` `= nxx (3)/(2) R xx DeltaT` `= 2 xx(3)/(2) xx 8.314 xx (1200 - 300) [C_(V)` for `He = (3)/(2)R]` `= 22447.8 J` `DeltaH_(N_(2)) = 0.5 xx (5)/(2) xx 8.314 xx (1200 - 300) = 9353.25J` `[C_(V)` for `N_(2) = (5)/(2)R]` `:.` Heat provded `= 22447.8 + 9353.25 = 31801.05J = 3.18 xx 10^(4)J` |
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