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(B) The average kinetic energy $(KE)$ of one mole of an ideal gas is given by the formula: $KE = \frac{3}{2} RT$.
Since $R$ is a constant,$KE \propto

Vedclass · Accepted Answer

Kinetic energy of $O_{2} < $ kinetic energy of $SO_{2}$

Vedclass · Answer

(B) The average kinetic energy $(KE)$ of one mole of an ideal gas is given by the formula: $KE = \frac{3}{2} RT$.
Since $R$ is a constant,$KE \propto T$.
For oxygen $(O_{2})$: $T_{O_{2}} = 273 \ K$.
For sulphur dioxide $(SO_{2})$: $T_{SO_{2}} = 546 \ K$.
Comparing the kinetic energies: $\frac{KE_{O_{2}}}{KE_{SO_{2}}} = \frac{T_{O_{2}}}{T_{SO_{2}}} = \frac{273}{546} = \frac{1}{2}$.
Therefore,$KE_{SO_{2}} = 2 	imes KE_{O_{2}}$,which implies $KE_{SO_{2}} > KE_{O_{2}}$ or $KE_{O_{2}} < KE_{SO_{2}}$.

One mole of oxygen at $273 \ K$ and one mole of sulphur dioxide at $546 \ K$ are taken in two separate containers,then:

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