A) 1.0 mol of\[{{O}_{2}}\]at 560 K
B) 0.50 mol of Neat 500 K
C) 0.20 mol of\[C{{O}_{2}}\]at 440 K
D) 2.0 mol of He at 140 K
Correct Answer: D
Solution :
Average molecular speed\[=\sqrt{\frac{8RT}{\pi m}}\]is dependent on T and M. Thus, greater the value of\[\sqrt{\frac{T}{M}},\]greater the molecular speed. Thus, \[{{v}_{av}}\propto \sqrt{\frac{T}{M}}\] \[{{O}_{2}}=\sqrt{\frac{560}{32}}=4.18\] \[Ne=\sqrt{\frac{500}{10}}=5.0\] \[C{{O}_{2}}=\sqrt{\frac{440}{44}}=3.16\] \[He=\sqrt{\frac{140}{4}}-5.91\]
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