The surface of copper gets tarnished by the formation of copper...

The surface of copper gets tarnished by the formation of copper oxide. ${N}_{2}$ gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the ${N}_{2}$ gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below: $2Cu\left(s\right) + {H}_{2}O\left(g\right) \to C{u}_{2}O ( s ) + {H}_{2} ( g )$ ${P}_{{H}_{2}}$ Is the minimum partial pressure of ${H}_{2}$ (in bar) needed to prevent the oxidation at 1250 K. The value of $\ln \left(p{H}_{2}\right)$ is ____. (Given: total pressure = 1 bar, R (universal gas constant) $= 8J{K}^{- 1}mo{l}^{- 1}, \ln \left(10\right) = 2.3 . Cu ( s ) andC{u}_{2}O ( s )$ are mutually immiscible. At $1250K: 2Cu\left(s\right) + \dfrac{1}{2}{O}_{2}\left(g\right) \to C{u}_{2}O ( s ) ; ∆ {G}^{⊝} = - 78,000Jmo{l}^{- 1}$ ${H}_{2}\left(g\right) + \dfrac{1}{2}{O}_{2}\left(g\right) \to {H}_{2}O ( g ) ; {∆}^{⊝} = - 1,78,000Jmo{l}^{- 1}; G$ is the Gibbs energy)