A) 4938.8 J
B) 4138.8 J
C) 5744.1 J
D) 6257.2 J
Correct Answer: C
Solution :
Given that \[{{P}_{1}}=10\,atm\], \[{{P}_{2}}=1\,atm\], \[T=300K\], \[n=1\] \[R=8.314\,J/K/\,mol\] Now, by using \[W=2.303\,nRT\,\,{{\log }_{10}}\frac{{{P}_{2}}}{{{P}_{1}}}\] \[=2.303\times 1\times 8.314\times 300\,\,{{\log }_{10}}\frac{1}{10}\] \[W=5744.1\,Joule\]
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