A) \[12.13J{{K}^{-1}}mo{{l}^{-1}}\]
B) \[15.17J{{K}^{-1}}mo{{l}^{-1}}\]
C) \[17.15\text{ }J{{K}^{-1}}mo{{l}^{-1}}\]
D) \[19.15J{{K}^{-1}}mo{{l}^{-1}}\]
Correct Answer: D
Solution :
\[\Delta S=2.303\text{ }nR\text{ }log\frac{{{V}_{2}}}{{{V}_{1}}}\] Here, \[n=1,\text{ }R=8.314\text{ }J,\text{ }T=25+273=298\text{ }K\] Suppose, \[{{V}_{1}}=V,{{V}_{2}}=10V\] \[\therefore \] \[\Delta S=2.303\times 1\times 8.34\text{ }log\frac{10V}{V}\] \[=9.15JK\,mo{{l}^{-1}}\]
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