A) 0.1 R
B) 2.303 R
C) 10.0 R
D) 100.0 R
Correct Answer: B
Solution :
\[{{q}_{rev.}}=-{{W}_{rev.}}=2.303\,nRT\log \frac{{{V}_{2}}}{{{V}_{1}}}\] \[\Delta S=\frac{{{q}_{rev.}}}{T}=2.303\,nR\,\log \frac{{{V}_{2}}}{{{V}_{1}}}\] \[n=1\]and\[\frac{{{V}_{2}}}{{{V}_{1}}}=10\] \[\Delta S=2.303\times 1\times R\times \log 10\] \[=2.303\text{ }R\]
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