MCQ
$2\,\,mole$ of zinc is dissolved in $HCl$ at $25\,^oC.$ The work done in open vessel is
- A$-2.477\,\,kJ$
- ✓$-4.955\,\,kJ$
- C$0.0489\,\,kJ$
- D$+2.47\,\,kJ$
✓
Answer
Correct option: B.
$-4.955\,\,kJ$
b
Given $n =2 ; T =25^{\circ} C =298 K ; R =8.314 J / mol . K$
Given $n =2 ; T =25^{\circ} C =298 K ; R =8.314 J / mol . K$
$W$ is the work done.
$=- P \Delta V =- nRT W =-2 8.314 \times 298$
$=-4.955 kJ$
Hence,option B is correct.
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