Consider the following two half-cell reactions along with the standard reduction potential given : $$ \begin{array}{ll} \mathrm{CO}_2+6 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow \mathrm{CH}_3 \mathrm{OH}+\mathrm{H}_2 \mathrm{O} & \mathrm{E}_{\mathrm{red}}^{\mathrm{o}}=0.02 \mathrm{~V} \\ \frac{1}{2} \mathrm{O}_2+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2 \mathrm{O} & \mathrm{E}_{\mathrm{red}}^{\mathrm{o}}=1.23 \mathrm{~V} \end{array} $$ A fuel cell was set up using the above two reactions such that the cell operates under the standard condition of 1 bar pressure and 298 K temperature. The fuel cell works with $80 \%$ efficiency. If the work derived from the cell using 1 mol of $\mathrm{CH}_3 \mathrm{OH}$ is used to compress an ideal gas isothermally against a constant pressure of 1 kPa , then the change in the volume of the gas, $$ \Delta \mathrm{V}=\ldots \mathrm{m}^3 \text {. (nearest integer) } $$ Given: $\mathrm{F}=96500 \mathrm{Cmol}^{-1}$