A) \[6.02\times {{10}^{23}}\]
B) \[1.2\times {{10}^{23}}\]
C) \[3.01\times {{10}^{23}}\]
D) \[2.01\times {{10}^{23}}\]
Correct Answer: A
Solution :
\[\frac{{{C}_{P}}}{{{C}_{V}}}=1.4\] so, given gas is diatomic 11.2L \[=3.01\times {{10}^{23}}\]molecules \[\therefore \] No. of atoms \[=3.01\times {{10}^{23}}\times 2\] \[=6.023\times {{10}^{23}}\]atomsYou need to login to perform this action.
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