A) 80 cc
B) 88.9 cc
C) 66.7 cc
D) 100 cc
Correct Answer: B
Solution :
\[{{V}_{2}}=\frac{{{P}_{ & 1}}{{V}_{1}}{{T}_{2}}}{{{P}_{2}}{{T}_{1}}}\]Þ \[{{P}_{1}}=P\] ; \[{{T}_{1}}={{273}^{o}}K\] \[{{P}_{2}}=\frac{3}{2}P\]; \[{{T}_{2}}={{T}_{1}}+\frac{{{T}_{1}}}{3}=\frac{4}{3}\times {{273}^{o}}K\] \[{{V}_{2}}=\frac{2P}{3P}\times \frac{4}{3}\times \frac{273}{273}\times 100cc=\frac{800}{9}cc=88.888cc\] = 88.9 cc
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