Question

# 56 g of iron reacts with dilute H2SO4 at 30 oC. Work done (in cal) in: (I) a closed vessel of fixed volume (II) an open vessel is: (given molar mass of Fe is 56 g/mol)

A
I II0606
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B
I II0 0
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C
I II600 600
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D
I II0303
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Solution

## The correct option is A I II0−606Fe(s)56 g=1 mol+H2SO4(aq)→FeSO4(aq)+H2(g)1 mol H2 Final volume of gas is due to 1 mole of H2 pV=nRT⇒V=nRTp (I) When the vessel is closed, △V=0, i.e. volume is constant. Thus, W=p△V=0 (II) For an open vessel, Δn (gaseous)=1 W=−PΔV=−ΔnRT Given, T=303 K R=2 cal/K.mol W=−1×2×303=−606 cal Theory : Expression for P-V or Mechanical Work: If the volume of the system is decreasing (compression), the mechanical work is being done on the system by the surrounding. If the volume of the system is increasing (expansion), the mechanical work is being done by the system on the surrounding. Work done on the system = Work done by the surrounding Work done on the system=Fsurr.dsurr Work done on the system=(Pext×A).(−dx) ∴ dW=−PextdV Where Fsurr= Force of surrounding dsurr= displacement of surrounding Pext= External pressure Case 1: Expansion dV>0 ∴ W<0 Case 2: Compression dV<0 ∴ W>0 Case 3: Isochoric Process dV=0 ∴ W=0 Case 4: Free Expansion Pext=0 ∴ W=0

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