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प्रश्न
Consider an ideal gas with following distribution of speeds.
| Speed (m/s) | % of molecules |
| 200 | 10 |
| 400 | 20 |
| 600 | 40 |
| 800 | 20 |
| 1000 | 10 |
Calculate Vrms and hence T. (m = 3.0 × 10−26 kg)
उत्तर
`V_(rms)^2 = (n_1v_1^2 + n_2v_2^2 + .... + n_nv_n^2)/(n_1 + n_2 + n_3 ..... n_n)`
`v_(rms)^2 = (10 xx (200)^2 + 20(400)^2 + 40(600)^2 + 20(800)^2 + 10(1000)^2)/(10 + 20 + 40 + 20 + 10)`
`v_(rms)^2 = (10^5 [4 + 32 + 144 + 128 + 100])/100`
= `10^3 [408]`
`v_(rms) = sqrt(10^4 xx 40.8)`
= `10^2 xx 6.39 ms^-1`
`1/2 mv_(rms)^2 = 3/2 K_BT`
`T = (mv_(rms)^2)/(3K_B)`
= `(3 xx 10^-26 xx 10^5 xx 4.08)/(3 xx 1.38 xx 10^-23)`
= `(204 xx 10^-23 xx 10^2)/(69 xx 10^-23)`
T = `2.96 xx 10^2` = 296 K
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