of both gases = V  (initial and final), as we ASSUME they are in fixed containers.number of moles of each gas = n  (assume)Initial temperature of both gases = TQuantity of heat given to each gas = QMono-atomic gas      Ratio of specific heats = γ₁ = 5/3     Molar Heat capacity at constant volume = = 3/2 R     Pressure  initial = P₁    ,    final = P₁',         Final temperature = T₁'          P₁ V = n R T ,       P₁'  V = n R T₁'      =>    P₁' / P₁ = T₁' / TDiatomic gas       ratio of specific heats = γ₂ = 7/5     Molar Heat capacity at constant volume = = 5/2 R      pressure  initial  = P₂  ,    final pressure = P₂'  ,       final temperature = T₂'         P₂ V = n R T   ,       P₂'  V = n R T₂'        =>      P₂' / P₂ = T₂' / T              ==>    P₂  =  P₁          and           P₂' / P₁'  =   T₂' / T₁'=================There is no change in volumes.  Hence the WORK done W by each gas is zero.                   Q = ΔU + W     Q = ΔU = change in internal energy = As molar heat capacity for the diatomic gas is more than that of monoatomic gas, the temperature INCREASE for the diatomic gas will be less.   Thus, the mono-atomic gas will have higher temperature.