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Heat and Thermodynamics question

2018 · 15 Apr · Shift 2 · Q62
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Heat and Thermodynamics question

2018 · 15 Apr · Shift 2 · Q62

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The value closest to the thermal velocity of a Helium atom at room temperature (300 K) in ms-1 is : [kB =1.4 ×\times× 10-23 J/K; mHe = 7 ×\times× 10 -27 kg ]
  1. A
    1.3 ×\times× 104
  2. B
    1.3 ×\times× 103
  3. C
    1.3 ×\times× 105
  4. D
    1.3 ×\times× 102
View written solutionFree

Correct answer: B

  1. For the thermal speed of a gas atom, we use the order-of-magnitude relation

vth≈kBTmv_{th} \approx \sqrt{\frac{k_B T}{m}}vth​≈mkB​T​​

(Using 3kBT/m\sqrt{3k_BT/m}3kB​T/m​ would give the rms speed, but the closest option remains the same.)

  1. Substitute the given values:

kB=1.4×10−23 J/K,T=300 K,m=7×10−27 kgk_B = 1.4 \times 10^{-23}\,\text{J/K}, \quad T = 300\,\text{K}, \quad m = 7 \times 10^{-27}\,\text{kg}kB​=1.4×10−23J/K,T=300K,m=7×10−27kg

So,

vth=(1.4×10−23)(300)7×10−27v_{th} = \sqrt{\frac{(1.4 \times 10^{-23})(300)}{7 \times 10^{-27}}}vth​=7×10−27(1.4×10−23)(300)​​

  1. Simplify inside the square root:

1.4×300=4201.4 \times 300 = 4201.4×300=420

Hence,

(1.4×10−23)(300)7×10−27=420×10−237×10−27\frac{(1.4 \times 10^{-23})(300)}{7 \times 10^{-27}} = \frac{420 \times 10^{-23}}{7 \times 10^{-27}}7×10−27(1.4×10−23)(300)​=7×10−27420×10−23​

=60×104=6×105= 60 \times 10^4 = 6 \times 10^5=60×104=6×105

Therefore,

vth=6×105=6×102.5v_{th} = \sqrt{6 \times 10^5} = \sqrt{6} \times 10^{2.5}vth​=6×105​=6​×102.5

Now,

6≈2.45,102.5≈316\sqrt{6} \approx 2.45, \qquad 10^{2.5} \approx 3166​≈2.45,102.5≈316

So,

vth≈2.45×316≈774 m/sv_{th} \approx 2.45 \times 316 \approx 774\,\text{m/s}vth​≈2.45×316≈774m/s

  1. If we use rms speed,

vrms=3kBTm=3 vth≈1.732×774≈1340 m/sv_{rms} = \sqrt{\frac{3k_BT}{m}} = \sqrt{3}\,v_{th} \approx 1.732 \times 774 \approx 1340\,\text{m/s}vrms​=m3kB​T​​=3​vth​≈1.732×774≈1340m/s

This is

≈1.3×103 m/s\approx 1.3 \times 10^3\,\text{m/s}≈1.3×103m/s

  1. Evaluate the options:
  • A: 1.3×1041.3 \times 10^41.3×104 too large
  • B: 1.3×1031.3 \times 10^31.3×103 matches
  • C: 1.3×1051.3 \times 10^51.3×105 far too large
  • D: 1.3×1021.3 \times 10^21.3×102 too small

Hence, the closest value is Option B.

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