A) \[1.4\,\,litre\]
B) \[1.8\,\,litre\]
C) \[0.4\,\,litre\]
D) \[0.8\,\,litre\]
Correct Answer: C
Solution :
Here: Initial volume\[({{V}_{1}})=0.2\,\,litre\] Initial temperature\[({{T}_{1}})={{0}^{o}}C={{273}^{o}}K\]; Final temperature\[({{T}_{2}})={{273}^{o}}K=546\,\,K\]. We know from the Charles law \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{T}_{1}}}{{{T}_{2}}}\] or \[\frac{0.2}{{{V}_{2}}}=\frac{273}{546}\] \[{{V}_{2}}=\frac{0.2\times 546}{273}=0.4\,\,litre\]
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