To calculate the change in entropy for the gas during this process, we need to use the formula:

ΔS = ∫(dq/T),

where ΔS is the change in entropy, dq is the differential amount of heat transferred, and T is the temperature.

In this case, since the process is irreversible, we cannot determine the exact path. However, we can assume an ideal gas, which allows us to use the ideal gas law:

PV = nRT,

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

Since both the initial and final states are given in terms of pressure and temperature, we can use the ideal gas law to determine the initial and final volumes and solve for ΔS.

Initial state:
P1 = 2 bar
T1 = 400°C + 273.15 = 673.15 K

Since we have two variables (P and T) and one equation (PV = nRT), we need another equation. Using the fact that P1V1 = P2V2 for an isothermal process:

P1V1 = P2V2

V1 = nRT1/P1

Final state:
P2 = 1 bar
T2 = 300°C + 273.15 = 573.15 K

Using the same equation:

V2 = nRT2/P2

Substituting these values into the ΔS equation:

ΔS = ∫(dq/T)

ΔS = ∫( dq1/T1 - dq2/T2 ),

where dq1 is the heat transferred at the initial state and dq2 is the heat transferred at the final state.

For an isobaric process, dq = CpdT, where Cp is the heat capacity at constant pressure.

ΔS = ∫( CpdT1/T1 - CpdT2/T2 ).

Assuming Cp is constant over the temperature range, we can take it out of the integral and integrate each term separately:

ΔS = Cp∫( dT1/T1 - dT2/T2 )

ΔS = Cp * ln(T1/T2)

ΔS = Cp * ln((673.15 K)/(573.15 K))

Using the specific heat capacity of an ideal gas at constant pressure (Cp = 29.1 J/molK, which is an approximation):

ΔS = 29.1 J/molK * ln((673.15 K)/(573.15 K)) ≈ 8.34 J/molK

Therefore, the change in entropy for the gas during this process is approximately 8.34 J/molK.

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