Question

From \[PV = \dfrac{1}{3}mN\langle {v^2}\rangle \] , show that the kinetic energy for a mole of an ideal gas is \[{E_k} = \dfrac{3}{2}RT\]?

Answer 1

Hint: We know about ideal gas equation whereas the given \[PV = \dfrac{1}{3}mN\langle {v^2}\rangle \] is not exact ideal gas equation which seems same to original ideal gas equation. \[PV = \dfrac{1}{3}mN\langle {v^2}\rangle \] is actually the ideal gas equation in term of kinetic energy per molecule of gas whenever \[N\] is Avogadro number and \[\langle {v^2}\rangle \] is the root mean square velocity and \[m\] is the mass. This question is actually related to kinetic theory of gases and we will prove it according to the principles of kinetic theory of gases or internal energy of ideal gas. We have to know that kinetic theory of gases is assuming molecules to be very small relative to distance between molecules.

Complete step-by-step answer:
Here our aim is to prove kinetic energy for a mole of an ideal gas is \[{E_k} = \dfrac{3}{2}RT\] from the equation \[PV = \dfrac{1}{3}mN\langle {v^2}\rangle \]. Here the problem is we will be having no idea about what is \[{E_k}\] . So, then depending on the units we need let us assume \[{E_k}\] be the average kinetic energy per mole. This will also make us assume that the molecule is not subjected to an external potential.
Next let us write down the given equation at once.
\[PV = \dfrac{1}{3}mN\langle {v^2}\rangle \]
This equation is actually from a derivation based on kinetic theory of gases which an ideal das possess whereas \[m\] is the mass in kilograms and \[N\] is the number of molecules.
Here one main fact we have to keep in mind is actually the question asks for average squared speed but not average speed squared which means \[\langle {v^2}\rangle \ne {\langle v\rangle ^2}\] unless the variance is found to be zero. Also, \[{v_{RMS}} = {\langle {v^2}\rangle ^{\dfrac{1}{2}}} \ne \langle v\rangle \]
Next, we have to look on to the translational energy of a free particle. It is given as follows, \[{E_{tr}} = {K_{tr}} = \dfrac{1}{2}m{v^2}\]
So, next we need the average translational energy. To get it, we have to divide the above equation by number of molecules which is \[N\] . It will be as follows,
\[{\langle \varepsilon \rangle _{tr}} \equiv \dfrac{{{E_{tr}}}}{N} = \dfrac{1}{2}m\langle {v^2}\rangle \]
Therefore, from the above equation \[\langle {v^2}\rangle = \dfrac{2}{m}{\langle \varepsilon \rangle _{tr}}\]. So let us substitute the value of \[\langle {v^2}\rangle \] from the above equation in \[PV = \dfrac{1}{3}mN\langle {v^2}\rangle \] . Then the equation will be as follows,
\[PV = \dfrac{1}{3}mN(\dfrac{2}{m}{\langle \varepsilon \rangle _{tr}})\]
\[ = \dfrac{2}{3}N{\langle \varepsilon \rangle _{tr}}\]
Next, let us go to the equation of ideal gas which is \[PV = nRT\] . Next let us compare this ideal gas equation with the above equation. Then, it will be as follows,
\[nRT = \dfrac{2}{3}N{\langle \varepsilon \rangle _{tr}} = \dfrac{2}{3}{E_{tr}}\]
On further calculation, we will get the equation as follows,
\[{E_{tr}} = \dfrac{3}{2}nRT\] \[J\]
We have to know that since \[nR = N{k_B}\] , therefore,
\[{E_{tr}} = \dfrac{3}{2}N{k_B}T\] \[J\]
We know that we have determined the energy in joules which is the same we have known from equipartition theorem in case of ideal gas molecules which is in absence of rotation as well as vibration and there are three degrees of freedom for translation.
Now, let us write the equation in terms of molar energy:
\[{\bar E_{tr}} = \dfrac{{{E_{tr}}}}{n} = \dfrac{3}{2}RT\]\[J.mo{l^{ - 1}}\]
Next, in terms of molecular energy, it will be as follows,
\[{\langle \varepsilon \rangle _{tr}} = \dfrac{{{E_{tr}}}}{N} = \dfrac{3}{2}{k_B}T\] \[J.{\left( {molecule} \right)^{ - 1}}\]

Hence, we show that the kinetic energy for a mole of an ideal gas is \[{E_k} = \dfrac{3}{2}RT\] from
\[PV = \dfrac{1}{3}mN\langle {v^2}\rangle \]

Note: We have to note that by determining temperature we can directly figure out the average kinetic energy of a gas molecule. It does not matter what gas we are considering unless and until it is ideal gas. By considering macroscopic parameters of gas like pressure, volume, temperature, we can accurately calculate microscopic parameters like momentum, velocity, internal energy, kinetic energy, thermal energy. Here, while dealing with the solution part, we will be having a chance of misunderstanding that is to assume average speed squared which means \[\langle {v^2}\rangle \ne {\langle v\rangle ^2}\] instead of average squared speed which is the right one.