Question

$56\text{g}$ of iron reacts with dilute $H_{2}S{O}_4$ at $30{}^o\text{C}$. Work done (in cal) in:
$\left(I\right)$ a closed vessel of fixed volume
$\left(II\right)$ an open vessel
is:
(given molar mass of $Fe$ is $56\text{g/mol}$)

• A. $\begin{array}{cc}I& II\\ 0& -606\end{array}$
• B. $\begin{array}{cc}I& II\\ 0& 0\end{array}$
• C. $\begin{array}{cc}I& II\\ -600& 600\end{array}$
• D. $\begin{array}{cc}I& II\\ 0& -303\end{array}$

Answer 1

The correct option is A $\begin{array}{cc}I& II\\ 0& -606\end{array}$

$\underset{56g=1\text{mol}}{Fe\left(s\right)}+H_{2}S{O}_4\left(aq\right)\to FeS{O}_4\left(aq\right)+\underset{1\text{mol}{H}_2}{H_2\left(g\right)}$

Final volume of gas is due to 1 mole of $H_2$
$pV=nRT\Rightarrow V=\frac{nRT}{p}$

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

(II) For an open vessel,
$\Delta n\text{(gaseous)}=1$
$W=-P\Delta V=-\Delta nRT$
Given, $T=303K$
$R=2\text{cal/K.mol}$
$W=-1\times 2\times 303=-606\text{cal}$

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
$\text{Work done on the system}=F_{surr}.d_{surr}$
$\text{Work done on the system}=(P_{ext}\times A).(-dx)$
$\therefore dW=-P_{ext}dV$
Where $F_{surr}=$ Force of surrounding
$d_{surr}=$ displacement of surrounding
$P_{ext}=$ External pressure
Case 1: Expansion
$dV>0$
$\therefore W<0$
Case 2: Compression
$dV<0$
$\therefore W>0$
Case 3: Isochoric Process
$dV=0$
$\therefore W=0$
Case 4: Free Expansion
$P_{ext}=0$
$\therefore W=0$