Question

Calculate the most probable speed, average speed and the root mean square speed for
gas molecules at 300 K. The mass of gas molecule is 5 × 10−26 kg and
kB = 1.38 × 10−23 JK−1
.

Answer 1

Answer on Question #66465, Physics / Molecular Physics | Thermodynamics

Calculate the most probable speed, average speed and the root mean square speed for gas molecules at 300 K. The mass of gas molecule is $5 \times 10^{-26} \mathrm{~kg}$ and $\mathrm{kB} = 1.38 \times 10^{-23} \mathrm{JK}^{-1}$

Find: $v_{\text{m.p.}} - ?v_{\text{a.}} - ?v_{\text{r.m.s.}} - ?$

Given:

T=300 K

$m_0 = 5 \times 10^{-26} \mathrm{~kg}$

$k = 1.38 \times 10^{-23} \mathrm{~J} \times \mathrm{K}^{-1}$

Solution:

The most probable speed: $v_{\text{m.p.}} = \sqrt{\frac{2kT}{m_0}}$ (1)

Of (1) $\Rightarrow v_{\text{m.p.}} = 406.9 \, \text{m/s}$

Average speed: $v_{\text{a.}} = \sqrt{\frac{8kT}{\pi m_0}}$ (2)

Of (2) $\Rightarrow v_{\text{a.}} = 459.3 \, \text{m/s}$

The root mean square speed:

$v_{\text{r.m.s.}} = \sqrt{\frac{3kT}{m_0}}$ (3)

Of (3) $\Rightarrow v_{\text{r.m.s.}} = 498.4 \, \text{m/s}$

Answer:

$v_{\text{m.p.}} = 406.9 \, \text{m/s}$

$v_{\text{a.}} = 459.3 \, \text{m/s}$

$v_{\text{r.m.s.}} = 498.4 \, \text{m/s}$