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Volume 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 = T
Quantity of heat given to each gas = Q

Mono-atomic gas
     Ratio of specific heats = γ₁ = 5/3
     Molar Heat capacity at constant volume = C_v = 3/2 R
     Pressure  initial = P₁    ,    final = P₁',         Final temperature = T₁'

          P₁ V = n R T ,       P₁'  V = n R T₁'      =>    P₁' / P₁ = T₁' / T

Diatomic gas 
     ratio of specific heats = γ₂ = 7/5
     Molar Heat capacity at constant volume = C_v = 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 = n\ C_v\ \Delta T

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.

2 5 2
is there a shorter method ?
the last 6 lines after the ========== line are enough.
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