A) 6.43 J
B) 12.87 J
C) 2.92 J
D) 8.57 J
Correct Answer: B
Solution :
\[{{W}_{\max }},{{\,}_{iso}}=-2.303\,\text{nRT}\log \frac{{{V}_{2}}}{{{V}_{1}}}\] \[=-2.303-1-0.0821-310-\log \frac{25}{15}\] \[=-58.61-0.22=-12.87\,J\] Here -ve sign shows the work of expansion, i.e., work done by the system. Hence, the correct option is (b).
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