Kinetic Energy Of Ideal Gas

The Kinetic Energy of Ideal Gas

In ideal gases, the molecules are considered point particles. The point particles can have only translational motion and thus only translational energy. So for an ideal gas, the internal energy can only be translational kinetic energy.

Hence kinetic energy (or internal energy) of n mole ideal gas

$E=\frac{1}{2}nM{v}_{ms}^2=\frac{1}{2}nM\times \frac{3RT}{M}=\frac{3}{2}nRT$

1. kinetic energy of 1 molecule

$E=\frac{3}{2}kT$

$\text{where}\mathrm{k}=\text{Boltzmann's constant and}k=1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$

I.e. Kinetic energy per molecule of gas does not depend upon the mass of the molecule but only depends upon the temperature of the gas.

2. kinetic energy of 1-mole ideal gas

$E=\frac{3}{2}RT$

i.e. Kinetic energy per mole of gas depends only upon the temperature of the gas.

3. At T = 0, E = 0 i.e. at absolute zero the molecular motion stops.

The relation Between Pressure and Kinetic energy

As we know

P=\frac{1}{3} \frac{m N}{V} v_{m s}^2=\frac{1}{3} \frac{M}{V} v_{m s}^2 \Rightarrow P=\frac{1}{3} \rho v_{m s}^2......(1)

And K.E. per unit volume=
$E=\frac{1}{2}\left(\frac{M}{V}\right)v_{ms}^2=\frac{1}{2}\rho v_{ms}^2$... (2)

So from equation (1) and (2), we can say that $P=\frac{2}{3}E$

i.e. the pressure exerted by an ideal gas is numerically equal to two-thirds of the mean kinetic energy of translation per unit volume of the gas.

Law of Equipartition of Energy

According to this law, for any system in thermal equilibrium, the total energy is equally distributed among its various degrees of freedom.

I.e Each degree of freedom is associated with energy $E=\frac{1}{2}kT$

1. At a given temperature T, all ideal gas molecules will have the same average translational kinetic energy as $\frac{3}{2}kT$

2. Different energies of a system of the degree of freedom f are as follows

(i) Total energy associated with each molecule $=\frac{f}{2}kT$
(ii) Total energy associated with $N$ molecules $=\frac{1}{2}NkT$
(iii) Total energy associated with 1 mole $=\frac{1}{2}RT$
(iv) Total energy associated with $n$ mole $=\frac{nf}{2}RT$

We can understand better through video.

Solved Examples Based on the Kinetic Energy of Ideal Gas

Q 1. A gas molecule of mass M at the surface of the Earth has kinetic energy equivalent to 0⁰C. If it were to go up straight without colliding with any other molecules, how high it would rise ? Assume that the height attained is much less than the radius of the earth. (k_B is Boltzmann constant)

1) 0
2) $\frac{273k_B}{2Mg}$
3) $\frac{546k_B}{3Mg}$
4) $\frac{819k_B}{2Mg}$

Solution:

The kinetic energy of gas due to translation per mole

$\begin{array}{rl}& E=\frac{3}{2}PV\\ & =\frac{3}{2}RT\end{array}$

wherein

R = Universal gas constant

T = temperature of gas

K.E at temperature $T=3/2K_{b}T$
$\mathrm{T}$ is in Kelvin
$\begin{array}{rl}& \mathrm{T}=273\mathrm{K}\\ & \text{Kinetic energy}=T=3/2K_b(273)\\ & =819K_b/2\\ & h=819K_B/2\mathrm{mg}\end{array}$

Hence, the answer is option (4).

Q 2. An ideal gas occupies a volume of 2m³ at a pressure of 3 x 10⁶ Pa. The energy (in Joule) of the gas is

1) 9000000

2) 60000

3) 300

4) 100000000

Solution:

The kinetic energy of gas due to translation per mole

$E=\frac{3}{2}PV=\frac{3}{2}RT$
wherein

R = Universal gas constant

T = temperature of gas

$\begin{array}{rl}& p=3\times {10}^6{p}_a\\ & v=2m^3\\ & E=\frac{f}{2}nRT=\frac{f}{2}pv\end{array}$

Let gas be monoatomic
$\begin{array}{rl}& f=3\\ & E=\frac{3}{2}\times 3\times {10}^6\times 2\\ & E=9\times {10}^6\mathrm{J}\end{array}$

Hence, the answer is option (1).

Example 3: Two kg of a monoatomic gas is at a pressure of $4\times {10}^4\mathrm{N}/{\mathrm{m}}^2$. The density of the gas is $8\mathrm{Kg}/{\mathrm{m}}^3$. What is the order of energy (in Joule) of the gas due to its thermal motion?

1) 10⁵

2) 10³

3) 10⁶

4) 10⁴

Solution:

$\begin{array}{rl}& \text{Thermal energy}=N\frac{3}{2}KT=\frac{N}{N_A}\frac{3}{2}RT=\frac{3}{2}nRT=\frac{3}{2}PV=\frac{3}{2}P\left(\frac{\mathrm{m}}{8}\right)\\ & \text{Thermal energy}=\frac{3}{2}\times 4\times {10}^4\times \left(\frac{2}{8}\right)=1.5\times {10}^4\end{array}$

Hence, the answer is the option (4).

Example 4: A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. Considering only translational and rotational modes, the total internal energy (in RT) of the system is:

1) 20

2) 12

3) 4

4) 15

Solution:

Total internal energy

$\begin{array}{rl}& U=\frac{f_o}{2}{n}_{o}RT+\frac{f_a}{2}{n}_{a}RT\\ & =\frac{5}{2}3RT+\frac{3}{2}5RT\\ & =15RT\end{array}$

Hence, the answer is option (4).

Example 5: An $\mathrm{HCl}$ molecule has rotational, translational and vibrational motions. If the RMS velocity of $\mathrm{HCl}$ molecules in its gaseous phase is $\overline{v}$, $m$ is its mass and $k_B$ is Boltzmann constant, then its temperature will be :

1) $\frac{m{\overline{v}}^2}{6k_B}$
2) $\frac{m{\overline{v}}^2}{3k_B}$
3) $\frac{m{\overline{v}}^2}{7k_B}$
4) $\frac{m{\overline{v}}^2}{5k_B}$

Solution:

The kinetic energy of gas due to translation per mole

$\begin{array}{rl}& E=\frac{3}{2}PV\\ & =\frac{3}{2}RT\end{array}$
wherein

R = Universal gas constant

T = temperature of gas

HCl has 3 translational, 2 rotational, and 1 vibrational degree of freedom

$\begin{array}{rl}& \text{Translational Kinetic energy}=K.E_{\text{Translation}}=\frac{3}{2}{K}_{B}T\\ & \frac{1}{2}m{\overline{v}}^2=\frac{3}{2}{K}_{B}T\\ & T=\frac{m\overline{v^2}}{3k_B}\end{array}$

Hence, the answer is the option (2).

Summary

An important aspect of motion and interaction among gas particles is the kinetic energy for an ideal gas; hence it becomes necessary to get acquainted with the concept of kinetic energy for an ideal gas. According to this concept, the higher the temperature of a given volume of gas, the more intense its kinetic movements would become. This correlation accounts for a number of properties displayed by gases including their pressure-volume relationships and temperatures thereby making it stand out in describing thermodynamic processes.

Frequently Asked Questions (FAQs):

Q 1. What are real-life applications of the pressure of Ideal Gas?

Ans: Pressure is used in many everyday situations such as when inflating a balloon or filling a tire with air.

Q 2. What is the pressure at absolute zero?

Ans: If the temperature could be reduced to absolute zero, the pressure of the ideal gas would be 0, implying that the gas would exert no force on the walls of its container.

Q 3. Can the pressure of an ideal gas be zero?

Ans: No

Q4: How does temperature affect the kinetic energy of an ideal gas?

Ans: The kinetic energy of an ideal gas is directly proportional to its temperature. As the temperature increases, the average kinetic energy of the gas molecules increases, meaning the molecules move faster.

Q5: Why is kinetic energy important for understanding ideal gases?

Ans: Kinetic energy is crucial for understanding ideal gases because it helps explain how temperature and pressure are related to the motion of gas molecules. It forms the basis for the kinetic theory of gases, which provides a microscopic explanation for macroscopic gas properties.