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Physics Question on Atoms

An electron tube was sealed-off during manufacture at a pressure of 1.2×10−7mm1.2\times 10^{-7}mm of mercury at 27∘C27^{\circ}C . Its volume is 100 cm3100\,cm^{3} . The number of molecules that remain in the tube is

A

2×10162\times {{10}^{16}}

B

3×10153\times {{10}^{15}}

C

3.86×10113.86\times {{10}^{11}}

D

5×10115\times {{10}^{11}}

Answer

3.86×10113.86\times {{10}^{11}}

Explanation

Solution

pV=nRTpV=nRT ⇒(1.2×10−10×13.6×103×9.8)×10−4\Rightarrow (1.2\times 10^{-10}\times 13.6\times 10^{3}\times 9.8)\times 10^{-4} =n×8.31×(273+27)=n\times 8.31\times (273+27) ∴\therefore Number of moles n=6.42×10−13n=6.42\times 10^{-13} Hence, number of molecules =6.42×10−13×6.02×1023=3.86×1011=6.42\times 10^{-13}\times 6.02\times 10^{23}=3.86\times 10^{11}