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Heat and Thermodynamics question

2021 · 1 Sep · Shift 2 · Q70
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Heat and Thermodynamics question

2021 · 1 Sep · Shift 2 · Q70

JEE MainPhysicsHeat and ThermodynamicsNumerical+4 / −1
The average translational kinetic energy of N2 gas molecules at ............. ∘^\circ∘ C becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt. (Given kB = 1.38 ×\times× 10 −-− 23 J/K) (Fill the nearest integer).
Numerical answer
View written solutionFree

Correct answer: 500

  1. Average translational kinetic energy of a gas molecule

For one molecule of an ideal gas,

⟨K⟩=32kBT\langle K \rangle = \frac{3}{2}k_B T⟨K⟩=23​kB​T

where kB=1.38×10−23 J/Kk_B = 1.38 \times 10^{-23}\,\text{J/K}kB​=1.38×10−23J/K.

  1. Kinetic energy of the electron

An electron accelerated through a potential difference VVV gains kinetic energy

K=eVK = eVK=eV

Here,

V=0.1 V,e=1.6×10−19 CV = 0.1\,\text{V}, \qquad e = 1.6 \times 10^{-19}\,\text{C}V=0.1V,e=1.6×10−19C

So,

K=(1.6×10−19)(0.1)=1.6×10−20 JK = (1.6 \times 10^{-19})(0.1) = 1.6 \times 10^{-20}\,\text{J}K=(1.6×10−19)(0.1)=1.6×10−20J
  1. Equate the two energies

Given that the average translational kinetic energy of an N2N_2N2​ molecule equals the electron kinetic energy,

32kBT=1.6×10−20\frac{3}{2}k_B T = 1.6 \times 10^{-20}23​kB​T=1.6×10−20

Substituting kBk_BkB​,

32(1.38×10−23)T=1.6×10−20\frac{3}{2}(1.38 \times 10^{-23})T = 1.6 \times 10^{-20}23​(1.38×10−23)T=1.6×10−20 2.07×10−23 T=1.6×10−202.07 \times 10^{-23} \, T = 1.6 \times 10^{-20}2.07×10−23T=1.6×10−20

Thus,

T=1.6×10−202.07×10−23T = \frac{1.6 \times 10^{-20}}{2.07 \times 10^{-23}}T=2.07×10−231.6×10−20​ T≈7.73×102=773 KT \approx 7.73 \times 10^2 = 773\,\text{K}T≈7.73×102=773K
  1. Convert to degree Celsius
T(∘C)=773−273=500∘CT(^\circ C) = 773 - 273 = 500^\circ CT(∘C)=773−273=500∘C
  1. Nearest integer
500\boxed{500}500​
  1. Comparison with stored correct answer

Stored correct answer = 500500500

Our derived answer = 500500500

So, the derived answer agrees with the stored correct answer.

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