Question 745: Thermodynamics & KTG

Question

' $N$ ' molecules of gas $A$, each having mass ' $m$ ' and ' 2 N ' molecules of gas B , each of mass ' 2 m ' are contained in the same vessel which is at constant temperature ' T '. The mean square velocity of $B$ is $V^2$ and mean square of x -component of A is $\omega^2$. The value of $\frac{\omega^2}{\mathrm{~V}^2}$ is

Accepted Answer

The problem involves calculating the ratio of the mean square of the x-component of velocity of gas A to the mean square velocity of gas B. Both gases are kept at constant temperature $ T $ in a vessel.
Key Equations
Mean Square Velocity for a Gas
:
$    	ext{Mean square velocity of a molecule} = \frac{3kT}{m}    $
Gas A Data
:
Let the mean square of the x-component of velocity of gas A be $\omega^2$.
Thus, the mean square velocity of a molecule of gas A is $ 3\omega^2 $.
Relation for Gas A
:
$    3\omega^2 = \frac{3kT}{m} \quad 	ext{(i)}    $
Gas B Data
:
Mean square velocity of a molecule of gas B is $ V^2 $.
Relation for Gas B
:
$    V^2 = \frac{3kT}{2m} \quad 	ext{(ii)}    $
Calculation
Using equations (i) and (ii):
$  \frac{3\omega^2}{V^2} = \frac{3kT}{m} 	imes \frac{2m}{3kT}  $
Simplifying the expression:
$  \frac{\omega^2}{V^2} = \frac{2}{3}  $
Thus, the ratio $\frac{\omega^2}{V^2}$ is $\frac{2}{3}$.

Answer

$3: 2$

Answer

$2: 3$

Answer

$1: 2$

Answer

$2: 1$

Thermodynamics & KTG

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