$\mathrm{Zn}+2\mathrm{HCl}⟶{\mathrm{ZnCl}}_2+{\mathrm{H}}_2\left(g\right)$                     …(1)

           $1\mathrm{mol}$                  $1\mathrm{mol}$   

   Since H₂ gas drives back the atmosphere in an open beaker, work is done by the gas which is negative.         

   $\therefore \mathrm{W}=-{\mathrm{P}}_{\text{ext }}\times \mathrm{\Delta V}$

 Since initial volume zero  so  $\mathrm{\Delta V}=$  Final volume 

         initial volume;$∆$ V = Final volume 

But           $\mathrm{\Delta V}=\frac{\mathrm{nRT}}{{\mathrm{P}}_{\text{ext }}}$

      $\begin{array}{l}\mathrm{W}=-{\mathrm{P}}_{\text{ext }}\times \frac{\mathrm{nRT}}{{\mathrm{P}}_{\text{ext }}}=-\mathrm{nRT}\\ \mathrm{n}=\text{ no. of mol of }\\ \mathrm{Zn}=\frac{\text{ wt. of }{\mathrm{Z}}_{\mathrm{n}}}{\text{ at. wt. of }\mathrm{Zn}}=\frac{6.5}{65}=\frac{1}{10}\end{array}$

From equation (1), 

          $1\text{ mol of }\mathrm{Zn}\equiv 1\text{ mol of }{\mathrm{H}}_2;\frac{1}{10}\text{ mol }$ of 

               $\begin{array}{l}\mathrm{Zn}\equiv \frac{1}{10}\mathrm{mol} \mathrm{of} {\mathrm{H}}_2\\ \mathrm{W}=-\mathrm{nRT};\\ \mathrm{W}=-\frac{1}{10}\mathrm{mol}\times 8.314{\mathrm{JK}}^{-}\mathrm{mol}\times 298\mathrm{K}=-247.76\mathrm{J}\\ \end{array}$