when pressure remaining contant at what temperature will the r.m.s speed of a gas molecules increase by $10$ % of the r.m.s. speed at NTP?

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Solution

The temperature at which speed increases by 10% is 354.53 kelvin

We know that ,

V = √(3RT/M)

Hence,

V ∝ √T

=> V₂/V₁ = √(T₂/T₁)

since the velocity increases by 10%, so,

V₂/V₁ = 1.1 and

T₁ = 293 k at NTP

Hence,

(T₂/T₁) = 1.1² = 1.21

=> T₂ = 1.21 x 293 = 354.53 kelvin

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when pressure remaining contant at what temperature will the r.m.s speed of a gas molecules increase by 10 % of the r.m.s. speed at NTP?

The temperature at which speed increases by 10% is 354.53 kelvin We know that , V = √(3RT/M) Hence, V ∝ √T => V₂/V₁ = √(T₂/T₁) since the velocity increases by 10%, so, V₂/V₁ = 1.1 and T₁ = 293 k at NTP Hence, (T₂/T₁) = 1.1² = 1.21 => T₂ = 1.21 x 293 = 354.53 kelvin

when pressure remaining contant at what temperature will the r.m.s speed of a gas molecules increase by $10$ % of the r.m.s. speed at NTP?