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 :
[a] \[W=2.303\,nRT\,\log \frac{{{P}_{2}}}{{{P}_{1}}}\] |
\[=2.303\times 1\times 2\times 300\,\log \frac{10}{2}=965.84\] |
at constant temperature, \[\Delta E=0.\] |
\[\Delta E=q+w\]; \[q=-w=-965.84\,cal\]. |
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