Answer:
(a) The heat transferred is 2552.64 kJ    
(b) The entropy change of the mixture is 1066.0279 J/K
Explanation:
Here we have 
Molar mass of H₂ = 2.01588 g/mol
Molar mass of N₂ = 28.0134 g/mol
Number of moles of H₂ = 500/2.01588  = 248 moles
Number of moles of N₂ = 1200/28.0134 = 42.8 moles
P·V = n·R·T
V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³
Since the volume is doubled then 
V₂ = 2 × 7.25 = 14.51 m³
At constant pressure, the temperature is doubled, therefore
T₂ = 600 K 
If we assume constant specific heat at the average temperature, we have
Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂ 
  cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K
 cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K
Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 =  2552.64 kJ    
b)   
 
Where:
 and
 and  are the mole fractions of Hydrogen and nitrogen respectively.
 are the mole fractions of Hydrogen and nitrogen respectively.
Therefore,  = 248 /(248 + 42.8) = 0.83
 = 248 /(248 + 42.8) = 0.83
  = 42.8/(248 + 42.8) = 0.1472
 = 42.8/(248 + 42.8) = 0.1472
∴  =  1066.0279 J/K
 =  1066.0279 J/K