A) 0, ? 965.84 cal
B) ? 965.84 cal, + 965.84 cal
C) + 865.58 cal, ? 865.58 cal
D) ? 865.58 cal, ? 865.58 cal
Correct Answer: A
Solution :
\[W=2.303\,nRT\,\log \frac{{{P}_{2}}}{{{P}_{1}}}\] \[=2.303\times 1\times 2\times 300\,\log \frac{10}{2}=965.84\] at constant temperature, \[\Delta E=0.\] \[\Delta E=q+w\]; \[q=-w=-965.84\,cal\].You need to login to perform this action.
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