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Kinetic Theory — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsKinetic Theory3 Marks
Question
You have the following group of particles, $n_i$ represents no. of molecules with speed $v_i(ms^{-1})$:
$n_i$ $2$ $4$ $8$ $6$ $3$
$v_i$ $1.0$ $2.0$ $3.0$ $4.0$ $5.0$
Calculate:
  1. The average speed.
  2. The $\text{r.m.s}$. speed.
  3. The most probable speed.
✓

Answer

$n_i$ = Number of molecules $v_i$ = Speed of molecules
$n_i$ $2$ $4$ $8$ $6$ $3$
$v_i$ $1.0$ $2.0$ $3.0$ $4.0$ $5.0$
  1. The average speed
$\text{v}_{\text{avg}}=\frac{\text{n}_1\text{v}_1+\text{n}_2\text{v}_2+\text{n}_3\text{v}_3+...+\text{n}_5\text{v}_5}{\text{n}_1+\text{n}_2+\text{n}_3+...+\text{n}_5}$
$=\frac{(2\times1)+(4\times2)+(8\times3)+(6\times4)+(3\times5)}{2+4+8+6+3}$
$=\frac{2+8+24+24+15}{23}=\frac{73}{23}=3.17\text{ms}^{-1}$
  1. $\text{v}_{\text{rms}}=\sqrt{\frac{\text{n}_1\text{v}_1^2+\text{n}_2\text{v}_2^2+\text{n}_3\text{v}_3^2+\text{n}_4\text{v}_4^2+\text{n}_5\text{v}_5^2}{\text{n}_1+\text{n}_2+\text{n}_3+\text{n}_4+\text{n}_5}}$
$=\sqrt{\frac{2(1)^2+4(2)^2+8(3)^2+6(4)^2+3(5)^2}{2+4+8+6+3}}$
$=\sqrt{\frac{2+16+72+96+75}{23}}$
$=\sqrt{\frac{261}{23}}=\sqrt{11.347}$
$=3.37\text{ms}^{-1}$
  1. The most probable sped $=3\text{ms}^{-1}$

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