A) 80c
B) 88.9c
C) 66.7c
D) 100cc
Correct Answer: B
Solution :
88.9cc \[\frac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}\] \[{{\operatorname{P}}_{1}}=1atm,\,\,{{V}_{1}}=100\,\,cc\] \[{{\operatorname{T}}_{1}}=273K\] \[{{\operatorname{P}}_{2}}=1\frac{1}{2}=1.5\] \[{{\operatorname{V}}_{2}}=?\,\,\,\,\,,{{T}_{2}}=273+\frac{1}{3}\times 273\,\left( given \right)\] \[=\left( 273+91 \right)=364K\] \[\therefore \frac{1\times \left( 100 \right)}{1\times 273}=\frac{1.5\times {{V}_{2}}}{364}\] \[\therefore {{\operatorname{V}}_{2}}=88.9\operatorname{CC}\]
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