O2(g)→H2O(l) E°cell=1.23V (from given data) ∵ΔG°=−nFE°cell =−2×96500×1.23J∕mol Work derived using 70% efficiency and on consumption of 1.0×10−3‌mol of H2 ( g W=2×96500×1.23×0.7×1×10−3 =166.17J This work done = Change in internal energy of monoatomic gas. ⇒166.17=nCv,mΔT ⇒ΔT=