The coefficient of performance (COP) for a Carnot refrigeration cycle is given by:
COP = QL / W
where QL is the amount of heat extracted from the cold reservoir and W is the work done on the system.
For an ideal Carnot cycle, the COP is given by:
COPideal = TL / (TH - TL)
where TH is the temperature of the hot reservoir and TL is the temperature of the cold reservoir.
Since we are not given the work done on the system, we cannot determine the ideal COP. However, we can still calculate the actual COP using the given temperatures and pressures.
First, we need to determine the temperatures at the two states of the cycle. Since the pressure ranges between 0.2 bar and 1 bar, we can assume that the air behaves as an ideal gas and 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 gas constant, and T is the temperature.
At state 1, the pressure is 1 bar and the volume is V1 = nRT1/P1, where T1 = 25°C + 273.15 = 298.15 K. Similarly, at state 2, the pressure is 0.2 bar and the volume is V2 = nRT2/P2, where T2 = -15°C + 273.15 = 258.15 K.
The amount of heat extracted from the cold reservoir is given by:
QL = nCp(T2 - T1)
where Cp is the specific heat capacity of air at constant pressure.
The work done on the system is given by:
W = nRT1ln(P2/P1)
Putting all these values together, we get:
COP = QL / W
COP = nCp(T2 - T1) / [nRT1ln(P2/P1)]
COP = Cp(T2 - T1) / [RT1ln(P2/P1)]
Substituting the values, we get:
COP = (1.005 kJ/kg K) (258.15 K - 298.15 K) / [(8.314 J/mol K) (298.15 K) ln(0.2/1)]
COP ≈ 2.5
Therefore, the coefficient of performance for the Carnot refrigeration cycle of one mole of air is approximately 2.5.