Question

average kinetic energy and rms speed of gas molecule
consider a gas container at room temperature $t_1(25.0^{circ}c)$ and find the average speed of the gas molecule at $t_2(55^{circ}c)$?
$e_{int}=\frac{3}{2}nk_{b}t$ : mono - atomic
$e_{int}=\frac{f}{2}nk_{b}t$ : any type
thermal
fluctuations
$f
ightarrow$ degrees of freedom
$overline{u}_{rms}=sqrt{\frac{f k_{b}t}{m}}$ : one atom
or
molecule.
$overline{u}_{rms}proptosqrt{t}$
$t
ightarrow 0k$
zero thermal
flux.
rms = root - mean - square
$overline{u}_{rms}=sqrt{\frac{f k_{b}t}{m}}$
$overline{u}_{rms}proptosqrt{t}$
$overline{u}_{rms}(1)proptosqrt{t_1}$
$overline{u}_{rms}(2)proptosqrt{t_2}$
$\frac{u_1}{u_2}=sqrt{\frac{t_1}{t_2}}$
$u_2 = u_1sqrt{\frac{t_2}{t_1}}$
$=u_1sqrt{\frac{273 + 55}{273+25}}$
$u_2 = u_1$?
$u_2>u_1$

Explanation:

Step1: Convert temperatures to Kelvin

$T_1 = 25 + 273 = 298K$, $T_2=55 + 273 = 328K$

Step2: Use the relation $\frac{u_{rms1}}{u_{rms2}}=\sqrt{\frac{T_1}{T_2}}$

Let the r - ms speed at $T_1$ be $u_{rms1}$ and at $T_2$ be $u_{rms2}$. Then $u_{rms2}=u_{rms1}\sqrt{\frac{T_2}{T_1}}=u_{rms1}\sqrt{\frac{328}{298}}$

Answer:

The r - ms speed of the gas molecule at $55^{\circ}C$ is $u_{rms1}\sqrt{\frac{328}{298}}$ times the r - ms speed at $25^{\circ}C$, where $u_{rms1}$ is the r - ms speed at $25^{\circ}C$.