
One mole of oxygen at 273K and one mole of Sulphur dioxide at 546K are taken in two separate containers, then:
(A) Kinetic energy of ${{O}_{2}}$> Kinetic energy of $S{{O}_{2}}$
(B) Kinetic energy of ${{O}_{2}}$< Kinetic energy of $S{{O}_{2}}$
(C) Kinetic energy of both are equal
(D) None of the above
Answer
511.5k+ views
Hint: Try to recall the chapter kinetic theory of gases chapter. The kinetic energy of a molecule can be calculated from the formula given below.
Formula:$\text{K}\text{.E = }\dfrac{\text{f}}{2}{{\text{K}}_{b}}\text{ x T}$
Where,
K.E stands for kinetic energy of the molecule,
f stands for degrees of freedom,
T stands for temperature of the container,
${{\text{K}}_{b}}$ stands for the Boltzmann constant
Complete step by step answer:
The kinetic theory of gases is a simple model of the behavior of gases, based on which many thermodynamic concepts have been established.
The model describes a gas as a large number of identical microscopic particles (for atoms or molecules), all of which are in constant, Brownian motion. Their size is assumed to be much smaller than the average distance between the colliding particles.
The particles undergo random elastic collisions and with the walls of the container.
The three main conclusions of the kinetic theory of gases are:
-No energy is gained or lost when molecules collide as the collisions are assumed to be completely elastic.
-The individual particles in a gas take up a negligible amount of space with respect to the container they occupy.
-The molecules of gas are in constant linear motion.
The degrees of freedom for a molecule or atom can be found by the formula given below:
$\text{f = 3N}$
Where,
N stands for the number of atoms present in one gas molecule.
${{O}_{2}}$ has 6 degrees of freedom as it is diatomic and $S{{O}_{2}}$ has 9 degrees of freedom as it is triatomic.
However ${{O}_{2}}$ has 1 and $S{{O}_{2}}$has 2 vibrational degrees of freedom which are not taken into consideration as its effect is seen only at high temperatures. Hence the degrees of freedom are 5 and 7 respectively.
Let us calculate the kinetic energies of ${{O}_{2}}$ and $S{{O}_{2}}$ by substituting the values of f calculated above.
$\text{K}\text{.E = }\dfrac{\text{f}}{2}{{\text{K}}_{b}}\text{ x T}$
\[\dfrac{\text{K}\text{.}{{\text{E}}_{{{O}_{2}}}}}{\text{K}\text{.}{{\text{E}}_{S{{O}_{2}}}}}\text{ = }\dfrac{\dfrac{{{\text{f}}_{{{O}_{2}}}}}{2}{{\text{K}}_{b}}\text{ x T}}{\dfrac{{{\text{f}}_{S{{O}_{{}}}}}}{2}{{\text{K}}_{b}}\text{ x T}}\]
\[\dfrac{\text{K}\text{.}{{\text{E}}_{{{O}_{2}}}}}{\text{K}\text{.}{{\text{E}}_{S{{O}_{2}}}}}\text{ = }\dfrac{\text{5}{{\text{K}}_{b}}\text{ }}{\text{7}{{\text{K}}_{b}}}\]
So, the correct answer is “Option B”.
Note: The kinetic theory of gases is defined only for ideal gases. The equations of kinetic theory of gases do not hold good for real gases. The assumptions made to apply the above equations are:
- The size of individual particles is very small compared to the size occupied by the gas.
- There is no interaction between the particles.
- The attractive and repulsive forces are non-existent.
- The average kinetic theory of the gas is proportional to the temperature.
Formula:$\text{K}\text{.E = }\dfrac{\text{f}}{2}{{\text{K}}_{b}}\text{ x T}$
Where,
K.E stands for kinetic energy of the molecule,
f stands for degrees of freedom,
T stands for temperature of the container,
${{\text{K}}_{b}}$ stands for the Boltzmann constant
Complete step by step answer:
The kinetic theory of gases is a simple model of the behavior of gases, based on which many thermodynamic concepts have been established.
The model describes a gas as a large number of identical microscopic particles (for atoms or molecules), all of which are in constant, Brownian motion. Their size is assumed to be much smaller than the average distance between the colliding particles.
The particles undergo random elastic collisions and with the walls of the container.
The three main conclusions of the kinetic theory of gases are:
-No energy is gained or lost when molecules collide as the collisions are assumed to be completely elastic.
-The individual particles in a gas take up a negligible amount of space with respect to the container they occupy.
-The molecules of gas are in constant linear motion.
The degrees of freedom for a molecule or atom can be found by the formula given below:
$\text{f = 3N}$
Where,
N stands for the number of atoms present in one gas molecule.
${{O}_{2}}$ has 6 degrees of freedom as it is diatomic and $S{{O}_{2}}$ has 9 degrees of freedom as it is triatomic.
However ${{O}_{2}}$ has 1 and $S{{O}_{2}}$has 2 vibrational degrees of freedom which are not taken into consideration as its effect is seen only at high temperatures. Hence the degrees of freedom are 5 and 7 respectively.
Let us calculate the kinetic energies of ${{O}_{2}}$ and $S{{O}_{2}}$ by substituting the values of f calculated above.
$\text{K}\text{.E = }\dfrac{\text{f}}{2}{{\text{K}}_{b}}\text{ x T}$
\[\dfrac{\text{K}\text{.}{{\text{E}}_{{{O}_{2}}}}}{\text{K}\text{.}{{\text{E}}_{S{{O}_{2}}}}}\text{ = }\dfrac{\dfrac{{{\text{f}}_{{{O}_{2}}}}}{2}{{\text{K}}_{b}}\text{ x T}}{\dfrac{{{\text{f}}_{S{{O}_{{}}}}}}{2}{{\text{K}}_{b}}\text{ x T}}\]
\[\dfrac{\text{K}\text{.}{{\text{E}}_{{{O}_{2}}}}}{\text{K}\text{.}{{\text{E}}_{S{{O}_{2}}}}}\text{ = }\dfrac{\text{5}{{\text{K}}_{b}}\text{ }}{\text{7}{{\text{K}}_{b}}}\]
So, the correct answer is “Option B”.
Note: The kinetic theory of gases is defined only for ideal gases. The equations of kinetic theory of gases do not hold good for real gases. The assumptions made to apply the above equations are:
- The size of individual particles is very small compared to the size occupied by the gas.
- There is no interaction between the particles.
- The attractive and repulsive forces are non-existent.
- The average kinetic theory of the gas is proportional to the temperature.
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