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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 II0606
Fe(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=nRTV=nRTp

(I) When the vessel is closed,
V=0, i.e. volume is constant.
Thus, W=pV=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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