Thermodynamics....try this if you can solve this?

Assume ideal gas

The number of moles in each compartment can be found from ideal gas law

p·V = N·R·T => N = (p·V)/(R·T)

First find the equilibrium temperature.

The internal energy of an ideal gas is solely a function of temperature:

ΔU = N·Cv·ΔT

Since there is no exchange of heat and work with the surrounding the sum of the internal energies of the compartments stay constant.

ΔU₁ + ΔU₂ = 0

<=>

N₁·Cv·ΔT₁ + N₂·Cv·ΔT₂ = 0

<=>

(p₁·V)/(R·T₁)·Cv·(T' - T₁) + (p₂·V)/(R·T₂)·Cv·(T'-T₂) = 0

(assume constant Cv)

=>

T' = (p₁ + p₂) / (p₁/T₁ + p₂/T₂)

= (137.8kPa + 413.4kPa) / (137.8kPa/300K + 413.4kPa/450K)

= 400K = 127°C

The change of entropy of an ideal gas is

ΔS = N·Cv·ln(T'/T) + N·R·ln(V'/V)

since the volume of the compartments does not change the second term is equal to zero.

Air is mainly consisting of the diatomic gases nitrogen and oxygen. The specific heat capacity at constant volume of diatomic ideal gases is:

Cv = (5/2)·R

The total change of the entropy inside the vessel is the sum of entropy changes in the compartments:

ΔS = ΔS₁ + ΔS₂

= N₁·Cv·ln(T'/T₁) + N₂·Cv·ln(T'/T₂)

= (p₁·V)/(R·T₁)·(5/2)·R · ln(T'/T₁) + (p₂·V)/(R·T₂)·(5/2)·R · ln(T'/T₂)

= (5/2)· V · [ (p₁/T₁)·ln(T'/T₁) + (p₂/T₂)·ln(T'/T₂) ]

= (5/2)· 0.029m³ · [ (137.8kPa/300K)·ln(400/300) + (413.4kPa/450K)·ln(400/450) ]

= 2.874×10-3 kJ/K