SATHEE: Kinetic Theory - Result Question 20

Kinetic Theory - Result Question 20

20. $N$ molecules each of mass $m$ of a gas $A$ and $2 N$ molecules each of mass $2 m$ of gas $B$ are contained in the same vessel which is maintained at temperature $T$ . The mean square velocity of molecules of $B$ type is $v^{2}$ and the mean square rectangular component of the velocity of $A$ type is denoted by $ω^{2}$ . Then $ω^{2} / v^{2}$

(a) 2

(b) 1

(c) $1 / 3$

(d) $2 / 3$

[1991]

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Answer:

Correct Answer: 20. (d)

Solution:

(d) The mean square velocity of A type molecules $= ω^{2} + ω^{2} + ω^{2} = 3 ω^{2}$

Therefore, $\frac{1}{2} m (3 ω^{2}) = \frac{1}{2} (2 m) v^{2}$

This gives $ω^{2} / v^{2} = 2 / 3$

Mean kinetic energy of the two types of molecules should be equal.

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Kinetic Theory - Result Question 20

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