A) \[{{34}^{o}}C\]
B) \[{{17}^{o}}C\]
C) \[{{136}^{o}}C\]
D) \[{{887}^{o}}C\]
Correct Answer: D
Solution :
Given\[{{V}_{1}}=V,\,\,{{V}_{2}}=\frac{1}{8}V\] \[{{T}_{1}}={{17}^{o}}C=290\,\,K,\,\,r=1.667\](for monoatomic gases)\[{{T}_{2}}=?\] \[\frac{{{T}_{2}}}{{{T}_{1}}}={{\left[ \frac{{{V}_{1}}}{{{V}_{2}}} \right]}^{r-1}}\] or \[\frac{{{T}_{2}}}{290}={{\left( \frac{V}{1/8\,\,V} \right)}^{1.667-1}}\] or \[\frac{{{T}_{2}}}{290}={{(8)}^{0.667}}\] \[\therefore \] \[{{T}_{2}}=1160\,\,K=1160-273={{887}^{o}}C\]
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