A) 80
B) 257
C) 255
D) 256
Correct Answer: C
Solution :
[c] \[{{n}_{1}}=1\]mole, \[{{n}_{2}}=?\], \[{{P}_{1}}=1\]atm, \[{{P}_{2}}=4\]atm \[{{P}_{1}}{{V}_{1}}={{n}_{1}}RT\] \[{{P}_{2}}{{V}_{2}}={{n}_{2}}RT\] \[\frac{{{P}_{1}}{{V}_{1}}}{{{n}_{1}}}=\frac{{{P}_{2}}{{V}_{2}}}{{{n}_{2}}}\Rightarrow 1\times \frac{1}{4}\pi {{({{r}_{1}})}^{3}}=\frac{4}{{{n}_{2}}}\times \frac{4}{3}\pi {{({{r}_{2}})}^{3}}\] \[\Rightarrow {{({{r}_{1}})}^{3}}=\frac{4}{{{n}_{2}}}{{({{r}_{2}})}^{3}}\] (i) Now \[{{r}_{1}}=\frac{1}{2};{{r}_{2}}=?\] But, \[P\propto d\Rightarrow \frac{{{P}_{1}}}{{{d}_{1}}}=\frac{{{P}_{2}}}{{{d}_{2}}}\Rightarrow \frac{1}{4}=\frac{1}{{{d}_{2}}}\] \[{{d}_{2}}=4\]\[{{r}_{2}}=2\] (ii) \[\Rightarrow \]Form (ii) and (i), we have \[{{\left( \frac{1}{2} \right)}^{3}}=\frac{4}{{{n}_{2}}}{{(2)}^{3}}\] \[{{n}_{2}}=64\times 4=256\] \[\Rightarrow \]Number of moles added =256-1=255
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