
If the total number of \[{H_2}\] molecules is double of the \[{O_2}\]molecules then the ratio of total kinetic energies of \[{H_2}\] to that of \[{O_2}\] at 300 K is :-
(A) 1:1
(B) 1:2
(C) 2:1
(D) 1:3
Answer
232.8k+ views
Hint: Here, we need to find the formula for kinetic energy of gases. Apply the given condition in two separate cases for two hydrogen molecules and one oxygen molecule. Find the ratio between the two.
Complete Step By Step Solution
We are given that the total number of \[{H_2}\] is double that of the number of \[{O_2}\]molecules. Which means that\[{n_{{H_2}}} = 2 \times {n_{{O_2}}}\]. We need to obtain its energy relation by using the kinetic theory of gases. According to the equi-partition theorem of gases, in thermal equilibrium, the kinetic energy of the particle is a product of average energy created by movement of a gas particle in each of its degrees of freedom. The average energy per particle is given as \[{k_b}T/2\], where \[{k_b}\]is called as Boltzmann’s constant and T is the temperature of the gas particle.
Now according to kinetic theory of gases and equipartition theorem
\[E = \dfrac{3}{2}{k_b}T\] (Since a gas particle have 3 degrees of freedom)
Now, the energy [produced by 2 molecules of hydrogen is given as
\[{E_{{H_2}}} = 2n \times E\]
Now the energy produced by one molecule of oxygen is given as :
\[{E_{{O_2}}} = E\]
Finding the ratio by dividing the above terms, we get,
\[\dfrac{{{E_{{H_2}}}}}{{{E_{{O_2}}}}} = \dfrac{2}{1}\](Common term E is cancelled out)
Therefore the ratios of the energy produced at 300K is \[2:1\]
Hence, Option(c) is the right answer for the given question.
Note Kinetic theory of gases states that gases molecules do not have any force of attraction between them and volume of the gas molecules are significantly smaller than the space occupied by gases. In order to explain this theory, five postulates were constructed which are
1. The molecules of gas are very far apart.
2. Gas molecules move in constant random motion
3. Molecules can collide with each other when they’re within a specified boundary.
4. Due to collision, no molecules lose kinetic energy and are said to be perfectly elastic.
5. Molecules have no attractive or repulsive force but have intermolecular force.
Complete Step By Step Solution
We are given that the total number of \[{H_2}\] is double that of the number of \[{O_2}\]molecules. Which means that\[{n_{{H_2}}} = 2 \times {n_{{O_2}}}\]. We need to obtain its energy relation by using the kinetic theory of gases. According to the equi-partition theorem of gases, in thermal equilibrium, the kinetic energy of the particle is a product of average energy created by movement of a gas particle in each of its degrees of freedom. The average energy per particle is given as \[{k_b}T/2\], where \[{k_b}\]is called as Boltzmann’s constant and T is the temperature of the gas particle.
Now according to kinetic theory of gases and equipartition theorem
\[E = \dfrac{3}{2}{k_b}T\] (Since a gas particle have 3 degrees of freedom)
Now, the energy [produced by 2 molecules of hydrogen is given as
\[{E_{{H_2}}} = 2n \times E\]
Now the energy produced by one molecule of oxygen is given as :
\[{E_{{O_2}}} = E\]
Finding the ratio by dividing the above terms, we get,
\[\dfrac{{{E_{{H_2}}}}}{{{E_{{O_2}}}}} = \dfrac{2}{1}\](Common term E is cancelled out)
Therefore the ratios of the energy produced at 300K is \[2:1\]
Hence, Option(c) is the right answer for the given question.
Note Kinetic theory of gases states that gases molecules do not have any force of attraction between them and volume of the gas molecules are significantly smaller than the space occupied by gases. In order to explain this theory, five postulates were constructed which are
1. The molecules of gas are very far apart.
2. Gas molecules move in constant random motion
3. Molecules can collide with each other when they’re within a specified boundary.
4. Due to collision, no molecules lose kinetic energy and are said to be perfectly elastic.
5. Molecules have no attractive or repulsive force but have intermolecular force.
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