Question

A gas mixture consists of 2 moles of \[{{O}_{2}}\] and 4 moles of \[Ar\] at temperature \[T\]. Neglecting all vibrational modes, the total internal energy of the system is
\[\begin{align}
  & A)11RT \\
 & B)4RT \\
 & C)15RT \\
 & D)9RT \\
\end{align}\]

Answer 1

Hint: We will need to know the degree of freedom without the vibrational modes. We will find internal energy of both gases individually by using the formula which includes the degree of freedom. Then we will add these internal energies to find the total internal energy of the mixture. The internal energy formula must be known.

Formula used:
\[U=\mu \left( \dfrac{D.F}{2}\times RT \right)\]

Complete step by step answer:
Here, we have a mixture of diatomic and monatomic gases. Oxygen (\[{{O}_{2}}\]) is a diatomic gas in which two atoms of the same form are the molecules which create it. That is oxygen as a diatomic gas, containing two oxygen atoms. If we take the case of argon (\[Ar\]), it is a monatomic gas which consists of single \[Ar\] atom ions.
Now, if we take the equation given for finding internal energy,
\[U=\mu \left( \dfrac{D.F}{2}\times RT \right)\]
We have \[D.F\], Which is the degree of freedom of the gas. We know oxygen is a diatomic gas and the degree of freedom without vibrational modes for a diatomic gas is \[D.F=5\]. Argon, as a monatomic gas, has a degree of freedom as \[D.F=3\].
Now, we will find the internal energy of 2 moles of oxygen by using the formula
\[\Rightarrow {{U}_{{{O}_{2}}}}=\mu \left( \dfrac{5}{2}\times RT \right)=2\left( \dfrac{5}{2}\times RT \right)\]
\[{{U}_{{{O}_{2}}}}=5RT\]
Therefore, internal energy of oxygen is \[5RT\].
Now, internal energy of 4 moles of argon is,
\[\Rightarrow {{U}_{Ar}}=\mu \left( \dfrac{3}{2}\times RT \right)=4\left( \dfrac{3}{2}\times RT \right)\]
\[{{U}_{Ar}}=6RT\]
The internal energy of argon is \[6RT\].
Now, to get total internal energy, we need to add both these energies.
\[\Rightarrow {{U}_{total}}={{U}_{{{O}_{2}}}}+{{U}_{Ar}}\]
\[{{U}_{total}}=5RT+6RT=11RT\]
So, the total internal energy of the mixture is \[11RT\].

Hence, the correct option is A.

Note:
For solving this question, it is important to remember the properties of monoatomic and diatomic gases. Here we also saw how the internal energy of these gases changes according to change in degree of freedom. Here, we have neglected the vibrational modes so we are not considering vibrational modes of degree of freedom.