Question

When pressure remaining constant, at what temperature will the r.m.s. Speed of a gas molecules increase by 10% of r.m.s. Speed at NTP

Answer 1

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

Answer 2

Hello friend

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 = 273 k at NTP

Hence,

(T₂/T₁) = 1.1² = 1.21

=> T₂ = 1.21 x 273 = 330.33 kelvin

Then 330.33- 273 = 57.33°C

Thank you