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प्रश्न
The molecules of a given mass of a gas have root mean square speeds of 100 ms−1 at 27°C and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at 127°C and 2.0 atmospheric pressure?
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उत्तर
We know that for a given mass of a gas
`V_(rms) = sqrt((3RT)/M)`
Where R is gas constant
T is the temperature in Kelvin
M is the molar mass of the gas
Clearly, `V_(rms) ∝ sqrt(T)`
As R and M are constants,
`(V_(rms))_1/(V_(rms))_2 = sqrt(T_1/T_2)`
Given, `(V_(rms))_1` = 100 m/s
T1 = 27°C = 27 + 273 = 300 K
T2 = 127°C = 127 + 273 = 400 K
∴ From equation (i)
`100/((V_(rms))_2) = sqrt(300/400) = sqrt(3)/2`
⇒ `(V_(rms))_2 = (2 xx 100)/sqrt(3) = 200/sqrt(3)` m/s
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