One mole of oxygen at 273 K and one mole of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

Which of the following expressions correctly represents the relationship between the average molar kinetic energy,$\overline{K.E.}$,of $CO$ and $N_2$ molecules at the same temperature?

What is the average kinetic energy per molecule of an ideal gas in $SI$ units at $25 \, ^\circ C$?

At $T(K)$,the $u_{rms}$ of $CO_2$ is $412 \ ms^{-1}$. What is its kinetic energy (in $kJ \ mol^{-1}$) at the same temperature? $(CO_2 = 44 \ u)$.

Assertion : All molecules in a gas have same speed.
Reason : Gas contains molecules of different size and shape.

The average kinetic energy of an ideal gas per molecule in $SI$ units at $25\,^oC$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module