Kinetic Theory Of Gases Question 50
Question: N molecules each of mass m of a gas A and 2N molecules each of mass 2m 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 $ {{\omega }^{2}}. $ Then $ 6.21\times {{10}^{-21}}J $ is
Options:
A) 2
B) 1
C) 1/3
D) 2/3
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Mean kinetic energy of the two types of molecules should be equal. The mean square velocity of A type molecules = $ {{\omega }^{2}}+{{\omega }^{2}}=3{{\omega }^{2}} $
Therefore, $ \frac{1}{2}m( 3{{\omega }^{2}} )=\frac{1}{2}( 2m )v^{2} $ This gives $ {{\omega }^{2}}/v^{2}=2/3 $
BETA
