A) \[1kg{{H}_{2}}\]
B) \[1kgHe\]
C) \[\frac{1}{2}1kg{{H}_{2}}+\frac{1}{2}1kgHe\]
D) \[\frac{1}{4}1kg{{H}_{2}}+\frac{3}{4}1kgHe\]
Correct Answer: A
Solution :
[a] Total Kinetic Energy \[=U=\frac{f}{2}nRT\] In case of \[{{H}_{2}}\] degree of freedom is greatest and number of moles n is highest. So this is the case of maximum kinetic energy
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