A) \[6.02\times {{10}^{23}}\]
B) \[1.2\times {{10}^{24}}\]
C) \[3.01\times {{10}^{23}}\]
D) \[2.01\times {{10}^{23}}\]
Correct Answer: A
Solution :
Since \[\frac{{{C}_{P}}}{{{C}_{V}}}=1.4\], the gas should be diatomic. If volume is 11.2 lt then, no. of moles = \[\frac{1}{2}\] \ no. of molecules = \[\frac{1}{2}\]´ Avagadro?s No. no. of atoms = 2 ´ no. of molecules 2 ´ \[\frac{1}{2}\]´ Avagadro?s No. \[=6.0223\times {{10}^{23}}\]
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