A) 190 L atm
B) 170.17 L atm
C) 192.25 L atm
D) 185.23 L atm
Correct Answer: B
Solution :
\[\operatorname{W}=\,\,2.303\,nRT\,log{{\,}_{10}}\,\frac{{{V}_{2}}}{{{V}_{1}}}\] \[=\,\,\,2.303\times 3\times 0.0821\times 300\,lo{{g}_{10}}\frac{10}{1}\] \[=\,\,\,170.17\,L\,atm\]You need to login to perform this action.
You will be redirected in
3 sec