A) \[2\,d{{m}^{3}}\]
B) \[3d{{m}^{3}}\]
C) \[4\,d{{m}^{3}}\]
D) \[8\,d{{m}^{3}}\]
Correct Answer: C
Solution :
\[\frac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}\]; \[\frac{{{P}_{1}}\times 4}{{{T}_{1}}}=\frac{2{{P}_{1}}\times {{V}_{2}}}{2{{T}_{1}}}\] \[8=2\times {{V}_{2}}\] so \[{{V}_{2}}=4d{{m}^{3}}\]
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