A) 2.938.6 Joules
B) 1.138.8 Joules
C) 5.744.1 Joules
D) 7342.2 Joules
Correct Answer: C
Solution :
Here,\[{{P}_{1}}=10\,atm,\,{{P}_{2}}=1atm,T=300\,K,n=1,\]\[R=8.314\,J/K/mol\] As \[W=2.303\,nRT\log \frac{{{P}_{2}}}{{{P}_{4}}}\] \[=2.303\times 18.341\times 300\log \frac{1}{10}\] \[W=5744.4\,\text{Joule}\text{.}\] Hence, the correct option is (c).You need to login to perform this action.
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