Question

Calculate the total and average kinetic energy of $32{{ g}}$ methane molecules at $27{{{ }}^ \circ }{{C}}$. $($ ${{R}}$ $ = $ $8.314$ ${{J }}{{{K}}^{ - 1}}{{mo}}{{{l}}^{ - 1}}$ $)$

Answer 1

Hint:Here, we are asked to calculate the total and average kinetic energies possessed by a given mass of methane gas molecules at a particular temperature. Total kinetic energy refers to the kinetic energy of the total number of moles of gas molecules whereas, average kinetic energy is the kinetic energy possessed by the maximum number of molecules.

Complete answer:
The total kinetic energy of molecules in a gas is calculated using the formula,
Total Kinetic Energy or Total K.E $ = $ $\dfrac{3}{2}{{nRT}}$
Where, ${{n}}$ is the number of moles of gas, ${{R}}$ is the universal gas constant and ${{T}}$ is the temperature at which the reaction is taking place.
Number of moles can be calculated by dividing mass of substance by molar mass of substance.
${{n = }}\dfrac{{{m}}}{{{M}}}$
Where, ${{n}}$ is the number of moles of gas, ${{m}}$ is the mass of the substance, and ${{M}}$ is the molar mass of the substance.
Here, the gas taken is methane with a molecular formula, ${{C}}{{{H}}_4}$ .
Mass of methane, ${{m}}$ $ = $ $32{{ g}}$
Molar mass of methane, ${{M}}$ $ = $ $16{{ g mo}}{{{l}}^{ - 1}}$
Therefore, the number of moles of methane, ${{n}}$ $ = $ $\dfrac{{32{{ g}}}}{{16{{ g mo}}{{{l}}^{ - 1}}}}$ $ = $ $2{{ mol}}$
In the question, the temperature is given as $27{{{ }}^ \circ }{{C}}$ .
${{T}}$$ = $ $27{{{ }}^ \circ }{{C}}$ $ = $ $27 + 273{{ K}}$
${{T}}$ $ = $ $300{{ K}}$
In order to calculate the total kinetic energy of methane molecules, we can substitute the values of ${{n}}$ , ${{R}}$ , and ${{T}}$ in the formula.
Total Kinetic Energy or Total K.E $ = $ $\dfrac{3}{2} \times {{ 2 mol }} \times {{ }}8.314{{ J }}{{{K}}^{ - 1}}{{mo}}{{{l}}^{ - 1}}{{ }} \times {{ 300 K}}$
Total Kinetic Energy or Total K.E $ = $ $7482.6{{ J}}$ $ = $ $7.482{{ kJ}}$

Hence, the total kinetic energy of methane molecules at $27{{{ }}^ \circ }{{C}}$ is $7.482{{ kJ}}$ .

Average kinetic energy $ = $ $\dfrac{3}{2}{{ }}\dfrac{{{{RT}}}}{{{{{N}}_{{A}}}}}$
Where, ${{R}}$ is the universal gas constant, ${{T}}$ is the temperature at which the reaction is taking place, and ${{{N}}_{{A}}}$ is the Avogadro Number.
In order to calculate the average kinetic energy of methane molecules, we can substitute the values of ${{R}}$ , ${{T}}$ and ${{{N}}_{{A}}}$ in the formula.
At a given temperature, different molecules in a sample of matter possess different kinetic energies or a range of kinetic energies. Most of the particles exhibit a kinetic energy in the middle range whereas some particles show a huge variation to the higher or lower sides. The kinetic energy possessed by the maximum number of molecules is calculated by taking the average kinetic energy. It is calculated using the formula:
Average kinetic energy $ = $ $\dfrac{3}{2}{{ }}\dfrac{{8.314{{ J }}{{{K}}^{ - 1}}{{mo}}{{{l}}^{ - 1}}{{ }} \times {{ 300 K}}}}{{6.023{{ }} \times {{ 1}}{{{0}}^{23}}}}$
On calculation, we get the value of average kinetic energy as,
Average kinetic energy $ = $$6.21{{ }} \times {{ 1}}{{{0}}^{ - 21}}{{ J}}$

Hence, the average kinetic energy of methane molecules at $27{{{ }}^ \circ }{{C}}$ is $6.21{{ }} \times {{ 1}}{{{0}}^{ - 21}}{{ J}}$ .

Note:
We know that kinetic energy is a temperature-dependent quantity. It is very important to remember the formulae of total kinetic energy and average kinetic energy. Total kinetic energy calculation requires the number of moles of gas taken whereas, average kinetic energy is independent of it. It only demands the knowledge of temperature.