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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Kinetic Theory of Gases

NTA Abhyas 2020 · Physics · Kinetic Theory of Gases

The temperature at which a & molecule of nitrogen will have the...

The temperature at which a & molecule of nitrogen will have the same rms velocity as a & molecule of oxygen at 127°C127 ° C& is&

Options

  1. A457℃457℃
  2. B273℃273℃
  3. C350℃350℃
  4. D77℃77℃

Correct answer

D. 77℃77℃

Step-by-step solution

Root mean square velocity

vrms=3RTM{v}_{rms} = \sqrt{\dfrac{3RT}{M}}

where RR is gas constant, TT the temperature and MM molecular weight.

Given, MN2=28,MO2=32,TO2=127℃=127+273=400{MN}_{2} = 28, {MO}_{2} = 32, {TO}_{2} = 127℃ = 127 + 273 = 400 K

∴vO2vN2=TO2MO2×MN2TN2=40032×28TN2=1∴ \dfrac{{{v}_{O}}_{2}}{{{v}_{N}}_{2}} = \sqrt{\dfrac{{{T}_{O}}_{2}}{{{M}_{O}}_{2}} \times \dfrac{{{M}_{N}}_{2}}{{{T}_{N}}_{2}}} = \sqrt{\dfrac{400}{32} \times \dfrac{28}{{{T}_{N}}_{2}}} = 1

⇒TN2=350K=77℃⇒ {TN}_{2} = 350K = 77℃.
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