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NCERT Exemplar · Q28

Q.Consider an ideal gas with following distribution of speeds. Speed (m/s) | % of molecules
200 | 10
400 | 20
600 | 40
800 | 20
1000 | 10

(i) Calculate VrmsV_{rms} and hence TT. (m=3.0×10−26m = 3.0 \times 10^{-26} kg)
(ii) If all the molecules with speed 1000 m/s escape from the system, calculate new VrmsV_{rms} and hence TT.
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Weighting each speed by its percentage gives the mean-square speed, whose square root is VrmsV_{rms}; the temperature then follows from 12mVrms2=32kBT\tfrac12mV_{rms}^2=\tfrac32k_BT.

  1. Initially Vrms≈639 m/sV_{rms}\approx639\ \text{m/s} and T≈296 KT\approx296\ \text{K}.
  2. After the fastest group (1000 m/s) escapes, Vrms≈585 m/sV_{rms}\approx585\ \text{m/s} and T≈248 KT\approx248\ \text{K} -- removing the fastest molecules cools the remaining gas.

The method

v2‾=∑ifivi2,Vrms=v2‾,T=mVrms23kB\overline{v^2} = \sum_i f_i v_i^2, \qquad V_{rms}=\sqrt{\overline{v^2}}, \qquad T = \frac{mV_{rms}^2}{3k_B}

with m=3.0×10−26 kgm = 3.0\times10^{-26}\ \text{kg} and kB=1.38×10−23 J/Kk_B = 1.38\times10^{-23}\ \text{J/K}.

(i) Initial distribution

Speed viv_i (m/s)%fif_ifivi2f_iv_i^2 (m2^2/s2^2)
200100.104,000
400200.2032,000
600400.40144,000
800200.20128,000
1000100.10100,000
Total1001.00408,000

v2‾=408,000 m2/s2⇒Vrms=408,000≈638.7 m/s\overline{v^2} = 408{,}000\ \text{m}^2/\text{s}^2 \quad\Rightarrow\quad V_{rms} = \sqrt{408{,}000} \approx 638.7\ \text{m/s}

T=mVrms23kB=(3.0×10−26)(408,000)3(1.38×10−23)≈295.65 K≈296 KT = \frac{mV_{rms}^2}{3k_B} = \frac{(3.0\times10^{-26})(408{,}000)}{3(1.38\times10^{-23})} \approx 295.65\ \text{K} \approx 296\ \text{K}

(ii) After molecules with speed 1000 m/s escape

The 10% at 1000 m/s leave; the remaining 90% keep their speeds. Re-normalising by dividing by 0.90:

v2‾new=4,000+32,000+144,000+128,0000.90=308,0000.90≈342,222 m2/s2\overline{v^2}_{\text{new}} = \frac{4{,}000+32{,}000+144{,}000+128{,}000}{0.90} = \frac{308{,}000}{0.90} \approx 342{,}222\ \text{m}^2/\text{s}^2

Vrms,new=342,222≈585.0 m/sV_{rms,\text{new}} = \sqrt{342{,}222} \approx 585.0\ \text{m/s} …

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