A) \[1:1\]
B) \[2:3\]
C) \[1:2\]
D) \[1:4\]
Correct Answer: A
Solution :
Key Idea: Use the following formula \[KE=\frac{3}{2}kT=\frac{3}{2}n\frac{R}{{{N}_{0}}}T\] Given number of Hz molecules\[=2n\] No. of \[{{O}_{2}}\] molecules\[=n\] \[\therefore \] Average\[KE\]of\[{{O}_{2}}=\frac{3}{2}\frac{nRT}{{{N}_{0}}\times n}\] \[\therefore \] Average\[KE\]of\[{{H}_{2}}=\frac{3}{2}\times \frac{2nRT}{{{N}_{0}}\times n}\] \[\therefore \]\[KE\]of per molecule of \[{{H}_{2}}=KE\]per molecule of \[{{O}_{2}}\] \[\therefore \] \[1:1\]
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