Question

Let $v,v_{rms}and{v}_p$ respectively denote the mean speed, the root-mean-square speed and the most probable speed of the molecules in an ideal monoatomic gas at absolute temperature $T$. The mass of a molecule is $m$
(i) No molecule can have speed $>v_{rms}$
(ii) No molecule can have speed less than $v_p$
(iii) $v_p<v<v_{rms}$
(iv) The average K.E of a molecule is $\frac{3}{4}m{v}_p^2$

• A. (iii), (iv)
• B. (ii), (iii)
• C. (i), (iv)
• D. (i), (ii)

Answer 1

The correct option is A (iii), (iv)

We know that: $v_{rms}=\sqrt{\frac{3kT}{m}},v_p=\sqrt{\frac{2kT}{m}},v=\sqrt{\frac{8kT}{\pi m}}$

Root mean square speed
$v_{rms}=\sqrt{\frac{3kT}{m}}$
Most probable speed
$p=\sqrt{\frac{2kT}{m}}$
Aveage speed
$v=\sqrt{\frac{8kT}{\pi m}}$

$v_p<v<v_{rms}$

Now
$\overline{K}=\frac{3}{2}kT$

Average KE of a molecule is given by
$\overline{K}=\frac{3}{2}kT...\left(i\right)$

Most probable speed is given by
$v_p=\sqrt{\frac{2kT}{m}}$
$kT=\frac{1}{2}m{v}_p^2...\left(ii\right)$

From equation (𝑖) and (𝑖𝑖)

$\overline{K}=\frac{3}{2}\times \frac{1}{2}m{v}_p^2=\frac{3}{4}m{v}_p^2$
FINAL ANSWER: (c)