91Ó°ÊÓ

The average translational kinetic energyof a monoatomic gas is proportional to the absolute temperature.

So, doubling the absolute temperature can double the energy too.

The average translational kinetic energy of a monoatomic gas is expressed as $E_{av}=\frac{3}{2}{k}_{B}T$where $E_{av}$is the average translational kinetic energy, $k_B=1.380649\text{'}{10}^{-23}{m}^{2}kg{s}^{-2}{K}^{-1}$is the Boltzmann’s constant, $T$is the absolute temperature.

The absolute temperature is to be in kelvin units. Convert the temperature given in degrees to kelvin by using the relation$K=°C+273$.

Given that, the temperature is increased from $25°Cto50°C$. Thus, in terms of kelvin, $25°C$becomes, $K=25°C+273$which is $298K$, and $50°C$becomes $K=50°C+273$which is $323K$.

Thus, increasing temperature from $25°C$to $50°C$is such as increasing the temperature from $298K$to $323K$which means temperature itself is not doubled. Then, we should not expect the kinetic energy to be doubled.

Calculating the average translational kinetic energies for $298K$and $323K$, by substituting the values of temperature and Boltzmann’s constant, in the expression $E_{av}=\frac{3}{2}{k}_{B}T$to check if$E_{av}$doubled or not.

For $298K$,

$$\begin{gathered} E_{av\left(298K\right)}=\frac{3}{2}*1.38\text{'}{10}^{-23}*298 \\ {E}_{av\left(298K\right)}=6.168*{10}^{-21}J \end{gathered}$$

For $323K$,

$$\begin{gathered} E_{av\left(323K\right)}=\frac{3}{2}*1.38\text{'}{10}^{-23}*323 \\ {E}_{av\left(323K\right)}=6.686*{10}^{-21}J \end{gathered}$$

Thus, the kinetic energy just changed from$E_{av\left(298K\right)}=6.168*{10}^{-21}J$to $E_{av\left(298K\right)}=6.686*{10}^{-21}J$, and is thus not doubled.

Hence, changing the absolute temperature from $298K$to $323K$does not double the average translational kinetic energy.

Given, the average kinetic energy is twice the initial value. Taking the energy as$E_{av\left(298K\right)}$, it becomes$2E_{av\left(298K\right)}$for the current case. From the obtainedlocalid="1663831113300" $E_{av\left(298K\right)}$for $298K$as, $6.168*{10}^{-21}J$, $2E_{av\left(298K\right)}$becomes, $2E_{av\left(298K\right)}=1.233*{10}^{-20}J$.

Substituting the values of$E_{av\left(298K\right)}$(now,$2E_{av\left(298K\right)}=1.233*{10}^{-20}J$), and Boltzmann’s constant the $k_B=1.38\text{'}{10}^{-23}{m}^{2}kg{s}^{-2}{K}^{-1}$in the expression of average kinetic energy, $E_{av}=\frac{3}{2}{k}_{B}T$,

$1.233*{10}^{-20}=\frac{3}{2}*1.38*{10}^{-23}*T$

Thus, the value of T is obtained as,

$$\begin{gathered} T=\frac{2*1.233*{10}^{-20}}{3*1.38*{10}^{-20}} \\ T=595.65K \end{gathered}$$

Hence, the final temperature is $595.65K$if the average translational kinetic energy was doubled.