A) -312J
B) +123J
C) -213J
D) + 231 J
Correct Answer: C
Solution :
\[W=-\int_{{{V}_{1}}}^{{{V}_{2}}}{PdV=-P\,\,({{V}_{2}}-{{V}_{1}})}\] \[W=-1\,\,(20-10)=-10\,\,d{{m}^{3}}\,\,atm\] \[=-10\,\,d{{m}^{3}}\times \frac{8.314\,J{{K}^{-1}}mo{{l}^{-1}}}{0.821\,\,d{{m}^{3}}\,{{K}^{-1}}mo{{l}^{-1}}}\] = - 1013 J From 1st law of thermodynamic \[\Delta U=q+W\] \[=800\,\,J+(-1013\,J)=-213\,J\]You need to login to perform this action.
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