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Heat and Thermodynamics question

2012 · Shift 1 · Q44
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  5. /2012 · Shift 1 · Q44

Heat and Thermodynamics question

2012 · Shift 1 · Q44

JEE AdvancedPhysicsHeat and ThermodynamicsMCQ+3 / −0.75
A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds (vrms(helium)vrms(arg⁡on))\left( {{{{v_{rms}}\left( {helium} \right)} \over {{v_{rms}}\left( {\arg on} \right)}}} \right)(vrms​(argon)vrms​(helium)​) is
  1. A
    0.32
  2. B
    0.45
  3. C
    2.24
  4. D
    3.16
View written solutionFree

Correct answer: D

  1. For any ideal gas, the rms speed is

vrms=3RTMv_{rms} = \sqrt{\frac{3RT}{M}}vrms​=M3RT​​

where MMM is the molar mass.

  1. Since both gases are at the same temperature T=300 KT = 300\,\text{K}T=300K, the ratio of rms speeds depends only on molar masses:

vrms(He)vrms(Ar)=MArMHe\frac{v_{rms}(\text{He})}{v_{rms}(\text{Ar})} = \sqrt{\frac{M_{\text{Ar}}}{M_{\text{He}}}}vrms​(Ar)vrms​(He)​=MHe​MAr​​​

  1. Given:
  • Helium atomic mass =4 amu= 4\,\text{amu}=4amu
  • Argon atomic mass =40 amu= 40\,\text{amu}=40amu

So,

vrms(He)vrms(Ar)=404=10\frac{v_{rms}(\text{He})}{v_{rms}(\text{Ar})} = \sqrt{\frac{40}{4}} = \sqrt{10}vrms​(Ar)vrms​(He)​=440​​=10​

  1. Numerically,

10≈3.16\sqrt{10} \approx 3.1610​≈3.16

  1. Hence the correct option is:

D: 3.16\boxed{\text{D: } 3.16}D: 3.16​

  1. Note: The number of moles in the mixture does not affect the rms speed ratio, because rms speed of each gas depends only on temperature and molar mass.
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