A) \[9.14\text{ }J\]
B) \[9.23J\]
C) \[9.43J\]
D) \[9.75J\]
Correct Answer: C
Solution :
\[\because \] \[\Delta G=2.303nRT+\log \frac{{{V}_{2}}}{{{V}_{1}}}\] Here \[\Delta G=\]maximum work obtained =? \[n=0.75\text{ }mole,\]\[T={{27}^{o}}C=27+273=300\text{ }K\] \[{{V}_{1}}=15\text{ }lit,\] \[{{V}_{2}}=25\text{ }lit,\]\[R=0.082\] \[\therefore \] \[\Delta G=2.303\times 0.75\times 0.082\times 300\log \frac{25}{15}\] \[=9.43J\]
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