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Heat and Thermodynamics question

2023 · 15 Apr · Shift 1 · Q61
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  5. /2023 · 15 Apr · Shift 1 · Q61

Heat and Thermodynamics question

2023 · 15 Apr · Shift 1 · Q61

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A flask contains Hydrogen and Argon in the ratio 2:12: 12:1 by mass. The temperature of the mixture is 30∘C30^{\circ} \mathrm{C}30∘C. The ratio of average kinetic energy per molecule of the two gases ( K\mathrm{K}K argon/K hydrogen) is : (Given: Atomic Weight of Ar=39.9\mathrm{Ar}=39.9Ar=39.9 )
  1. A
    39.92\frac{39.9}{2}239.9​
  2. B
    2
  3. C
    39.9
  4. D
    1
View written solutionFree

Correct answer: D

  1. Key concept: Average kinetic energy per molecule

For any ideal gas, the average translational kinetic energy per molecule depends only on temperature:

K‾=32kT\overline{K} = \frac{3}{2}kTK=23​kT

where:

  • kkk = Boltzmann constant
  • TTT = absolute temperature

So, for all gases at the same temperature, the average kinetic energy per molecule is the same, regardless of molecular mass.

  1. Apply to Hydrogen and Argon

Both Hydrogen and Argon are present in the same flask, so they are at the same temperature:

T=30∘C=303 KT = 30^\circ C = 303\,KT=30∘C=303K

Thus,

K‾Ar=32kT\overline{K}_{\text{Ar}} = \frac{3}{2}kTKAr​=23​kT

and

K‾H2=32kT\overline{K}_{\text{H}_2} = \frac{3}{2}kTKH2​​=23​kT
  1. Find the required ratio
KargonKhydrogen=32kT32kT=1\frac{K_{\text{argon}}}{K_{\text{hydrogen}}} = \frac{\frac{3}{2}kT}{\frac{3}{2}kT} = 1Khydrogen​Kargon​​=23​kT23​kT​=1
  1. Check options
  • A: 39.92\dfrac{39.9}{2}239.9​ ❌
  • B: 222 ❌
  • C: 39.939.939.9 ❌
  • D: 111 ✅

Hence, the correct option is D.

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