Question

Calculate the ratio of kinetic energies of 3g of hydrogen and 4g of oxygen at a given temperature. The mean kinetic energy of a molecule at ${{0}^{0}}C$ is $5.621\times {{10}^{-14}}erg$ . Calculate Boltzmann’s constant if the value of $R=8.314\times {{10}^{7}}erg$ and also calculate the no. of molecules present in mole of gas.

Answer 1

Hint: The answer is dependent on the concept of kinetic energy which is given by the formula, $K.E.=\dfrac{3}{2}nRT$. Find out the number of moles of hydrogen and oxygen which is at constant temperature and then the Boltzmann’s constant $K$.

Complete answer:
In the chapters which deal with the basic concepts of chemistry, we have dealt with the several calculations such as calculating the kinetic energy of the molecules, calculation of vibrational and rotational energies and also several other parameters.
Let us now see the calculation of the ratios of kinetic energies of the hydrogen and oxygen molecules and also the Boltzmann’s constant $K$.
- Kinetic energy of a molecule in chemistry is defined as the average kinetic energy of the gas particles is directly proportional to the absolute temperature of the gas.
- The expression for kinetic energy is $K.E.=\dfrac{3}{2}nRT$
Now, since the temperature is constant, the ratios of kinetic energies for both the atoms will be,
\[\dfrac{{{(K.E.)}_{{{H}_{2}}}}}{{{(K.E.)}_{{{O}_{2}}}}}=\dfrac{{{n}_{{{H}_{2}}}}}{{{n}_{{{O}_{2}}}}}\]
Now, ${{n}_{{{H}_{2}}}}=\dfrac{mass\,of\,{{H}_{2}}}{molar\,mass}=\dfrac{3}{2}$ . Similarly, ${{n}_{{{O}_{2}}}}=\dfrac{mass\,of\,{{O}_{2}}}{molar\,mass}=\dfrac{4}{16}$
Thus, the ratios of these two terms according to the above equation will be,
\[\dfrac{{{(K.E.)}_{{{H}_{2}}}}}{{{(K.E.)}_{{{O}_{2}}}}}=\dfrac{{{n}_{{{H}_{2}}}}}{{{n}_{{{O}_{2}}}}}=\dfrac{{3}/{2}\;}{{4}/{16}\;}\]
Or, this can also be written as, ${{n}_{{{H}_{2}}}}:{{n}_{{{O}_{2}}}}=\dfrac{3}{2}:\dfrac{4}{16}$
By the cross multiplication rule and further simplification will give, ${{n}_{{{H}_{2}}}}:{{n}_{{{O}_{2}}}}=6:1$
Now, we know that average kinetic energy expression is given in terms of Boltzmann’s constant as,
\[Average\,K.E.=\dfrac{3}{2}KT\]
where $K$ is Boltzmann's constant.
Thus, the above equation can be written for $K$ and after substituting the values, we get the equation as,
\[K=\dfrac{5.621\times {{10}^{-14}}\times 2}{3\times 273}=1.372\times {{10}^{-16}}erg\,molecul{{e}^{-1}}{{K}^{-1}}\]
Therefore, the total number of molecules present in one mole of a gas will be given by the Avogadro number which is given by, $A=\dfrac{R}{K}=\dfrac{8.314\times {{10}^{7}}}{1.372\times {{10}^{-16}}}=6.059\times {{10}^{23}}$

Note:
Note that the Boltzmann’s constant is nothing but the proportionality factor that relates the average relative kinetic energy of the particles in a gas with the thermodynamic temperature of the gas.