Question

SID: 973610-18-20DQ

The temperature of an ideal monatomic gas is increased from 25°C to 50°C. Does the average translational kinetic energy of each gas atom double? Explain. If your answer is no, what would the final temperature be if the average translational kinetic energy was doubled?

Short answer

• No. The temperature increased from 25°C to 50°C does not double the average translational kinetic energy of each gas atom.
• The final temperature would be$595.65K$, if the average translational kinetic energy was doubled.

Step-by-step solution

About the average kinetic energy and its equation  

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.

Temperature conversion and  temperature doubling calculations

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.

Calculation of temperature for doubled average 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.