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Heat and Thermodynamics question

2023 · 11 Apr · Shift 2 · Q60
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  5. /2023 · 11 Apr · Shift 2 · Q60

Heat and Thermodynamics question

2023 · 11 Apr · Shift 2 · Q60

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The root mean square speed of molecules of nitrogen gas at 27∘C27^{\circ} \mathrm{C}27∘C is approximately : (Given mass of a nitrogen molecule =4.6×10−26 kg=4.6 \times 10^{-26} \mathrm{~kg}=4.6×10−26 kg and take Boltzmann constant kB=1.4×10−23JK−1\mathrm{k}_{\mathrm{B}}=1.4 \times 10^{-23} \mathrm{JK}^{-1}kB​=1.4×10−23JK−1 )
  1. A
    91 m/s
  2. B
    1260 m/s
  3. C
    27.4 m/s
  4. D
    523 m/s
View written solutionFree

Correct answer: D

  1. Use the formula for RMS speed

For a gas molecule, vrms=3kBTmv_{\mathrm{rms}}=\sqrt{\frac{3k_B T}{m}}vrms​=m3kB​T​​

where:

  • kB=1.4×10−23 J K−1k_B=1.4\times 10^{-23}\,\mathrm{J\,K^{-1}}kB​=1.4×10−23JK−1
  • T=27∘C=300 KT=27^\circ\mathrm{C}=300\,\mathrm{K}T=27∘C=300K
  • m=4.6×10−26 kgm=4.6\times 10^{-26}\,\mathrm{kg}m=4.6×10−26kg
  1. Substitute the values

vrms=3×1.4×10−23×3004.6×10−26v_{\mathrm{rms}}=\sqrt{\frac{3\times 1.4\times 10^{-23}\times 300}{4.6\times 10^{-26}}}vrms​=4.6×10−263×1.4×10−23×300​​

  1. Simplify the numerator

3×1.4×300=12603\times 1.4\times 300 = 12603×1.4×300=1260

So, vrms=1260×10−234.6×10−26v_{\mathrm{rms}}=\sqrt{\frac{1260\times 10^{-23}}{4.6\times 10^{-26}}}vrms​=4.6×10−261260×10−23​​

=12604.6×103=\sqrt{\frac{1260}{4.6}\times 10^3}=4.61260​×103​

Now, 12604.6≈273.9\frac{1260}{4.6}\approx 273.94.61260​≈273.9

Thus, vrms=273.9×103v_{\mathrm{rms}}=\sqrt{273.9\times 10^3}vrms​=273.9×103​

=273900=\sqrt{273900}=273900​

  1. Calculate the square root

273900≈523 m/s\sqrt{273900}\approx 523\,\mathrm{m/s}273900​≈523m/s

  1. Match with the options
  • A: 91 m/s91\,\mathrm{m/s}91m/s
  • B: 1260 m/s1260\,\mathrm{m/s}1260m/s
  • C: 27.4 m/s27.4\,\mathrm{m/s}27.4m/s
  • D: 523 m/s523\,\mathrm{m/s}523m/s

Hence, the correct option is: D   523 m/s\boxed{\text{D }\; 523\,\mathrm{m/s}}D 523m/s​

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