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Heat and Thermodynamics question

2016 · Q156
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Heat and Thermodynamics question

2016 · Q156

NEETPhysicsHeat and ThermodynamicsMCQ+4 / −1
The molecules of a given mass of a gas have r.m.s. velocity of 2000 m s−-−1 at 27oC and 1.0 ×\times× 105 N m−-−2 pressure. When the temperature and pressure of the gas are respectively, 127oC and 0.05 ×\times× 105 N m−-−2, the r.m.s. velocity of its molecules in m s−-−1 is
  1. A
    10023{{100\sqrt 2 } \over 3}31002​​
  2. B
    1003{{100} \over 3}3100​
  3. C
    1002100\sqrt 21002​
  4. D
    4003{{400} \over {\sqrt 3 }}3​400​
View written solutionFree

Correct answer: D

It is observed that the rms velocity of molecule is directly proportional to temperature, so

vrms∝T{v_{rms}} \propto \sqrt T vrms​∝T​

⇒vrms′=vrmsT′T\Rightarrow v_{rms}^{'} = {v_{rms}}\sqrt {{{T'} \over T}}⇒vrms′​=vrms​TT′​​

Hence, vrms′=200×400300=400/3v_{rms}^{'} = 200 \times \sqrt {{{400} \over {300}}} = 400/\sqrt 3 vrms′​=200×300400​​=400/3​

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