
KE of equal masses of methane, oxygen and sulphur dioxide under similar conditions will be in the ratio of:
(A) $ 1:1:1 $
(B) $ 1:2:3 $
(C) $ 4:2:1 $
(D) $ 1:2:4 $
Answer
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Hint: A theory released in chemistry with some ideal concepts for gaseous particles is the Kinetic theory of gases. It describes the kinetic energy of moving gaseous particles. At the same temperature the gases have the same average kinetic energy.
Complete step by step answer:
According to kinetic theory of gases, the gas particles have a constant motion with perfectly elastic collisions in them. The average kinetic energy of the gaseous particles is always proportional to the absolute temperature of the gas. At the same temperature the gases have the same average kinetic energy.
The formula related to kinetic energy is as follows:
$ K = \dfrac{{3R}}{{2{N_A}}}T $
Where K is kinetic energy, \[{N_A}\] is Avogadro’s number and T is temperature.
Now by converting formula we get;
$ K = \dfrac{3}{2}nRT $ [n is number of moles, R is gas constant and T is temperature]
so we can say that kinetic energy is directly proportional to number of moles ( $ K \propto n $ )
Therefore kinetic energy is inversely proportional to molecular mass ( $ K \propto \dfrac{1}{M} $ ) [M is molecular weight]
So we can say that $ {K_1}:{K_2}:{K_2} \equiv \dfrac{1}{{{M_1}}}:\dfrac{1}{{{M_2}}}:\dfrac{1}{{{M_3}}} $
And the molecular weight of methane, oxygen and sulphur dioxide are $ {\text{16,32 and 64}} $ respectively.
$ \Rightarrow {K_1}:{K_2}:{K_3} \equiv \dfrac{1}{{16}}:\dfrac{1}{{32}}:\dfrac{1}{{64}} $
$ \Rightarrow {K_1}:{K_2}:{K_3} \equiv 4:2:1 $
Hence option (C) is correct.
Note:
Postulates of the KTG (kinetic theory of gases) are as follows:
The gaseous particles are very small and occupy negligible volume.
These gas molecules have constant random motion in the provided volume.
The collisions of the particles with one another and the walls of the container are perfectly elastic, so there is no loss of kinetic energy.
The gas is considered ideal with these postulates and it also simplifies the numerical calculations.
Complete step by step answer:
According to kinetic theory of gases, the gas particles have a constant motion with perfectly elastic collisions in them. The average kinetic energy of the gaseous particles is always proportional to the absolute temperature of the gas. At the same temperature the gases have the same average kinetic energy.
The formula related to kinetic energy is as follows:
$ K = \dfrac{{3R}}{{2{N_A}}}T $
Where K is kinetic energy, \[{N_A}\] is Avogadro’s number and T is temperature.
Now by converting formula we get;
$ K = \dfrac{3}{2}nRT $ [n is number of moles, R is gas constant and T is temperature]
so we can say that kinetic energy is directly proportional to number of moles ( $ K \propto n $ )
Therefore kinetic energy is inversely proportional to molecular mass ( $ K \propto \dfrac{1}{M} $ ) [M is molecular weight]
So we can say that $ {K_1}:{K_2}:{K_2} \equiv \dfrac{1}{{{M_1}}}:\dfrac{1}{{{M_2}}}:\dfrac{1}{{{M_3}}} $
And the molecular weight of methane, oxygen and sulphur dioxide are $ {\text{16,32 and 64}} $ respectively.
$ \Rightarrow {K_1}:{K_2}:{K_3} \equiv \dfrac{1}{{16}}:\dfrac{1}{{32}}:\dfrac{1}{{64}} $
$ \Rightarrow {K_1}:{K_2}:{K_3} \equiv 4:2:1 $
Hence option (C) is correct.
Note:
Postulates of the KTG (kinetic theory of gases) are as follows:
The gaseous particles are very small and occupy negligible volume.
These gas molecules have constant random motion in the provided volume.
The collisions of the particles with one another and the walls of the container are perfectly elastic, so there is no loss of kinetic energy.
The gas is considered ideal with these postulates and it also simplifies the numerical calculations.
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