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 = 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.