A rigid, well-insulated cylinder is divided into two chambers bya fixed dividing wall. As the cylinder is well-insulated, heat transferoccurs only through the dividing wall: no heat is lost to thesurroundings. The first chamber contains 11 m3 of air at 800 kPa and0 °C. The second chamber contains 1 m3 of refrigerant R134a at 800kPa and 60 °C. The refrigerant is allowed to cool until the pressure ofthe R134a chamber reaches 600 kPa.a) Assuming air is an ideal gas, calculate the initial mass of air inthe chamber.b) Calculate the initial mass of R134a in the chamber.c) Considering the whole cylinder as the system, use the first lawof thermodynamics to derive a relationship between the internalenergy of the air and the internal energy of the refrigerant.d) Determine the temperature of the air chamber when the R134areaches 600 kPa.e) Has the system reached thermal equilibrium? Give yourreasoningHeat R134a1 m3P = 800 kPaT = 60 CAir11 m3P = 800 kPaT = 0 C

a) To calculate the initial mass of air in the chamber, we can 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.

First, we need to convert the temperature from Celsius to Kelvin. T = 0°C = 273.15 K.

Then, we can rearrange the equation to solve for n (the number of moles): n = PV/RT.

Substituting the given values (P = 800 kPa = 800,000 Pa, V = 11 m^3, R = 287 J/(kgK) for air, and T = 273.15 K), we get n = (800,000 Pa * 11 m^3) / (287 J/(kgK) * 273.15 K) = 105.16 kg.

So, the initial mass of air in the chamber is 105.16 kg.

b) To calculate the initial mass of R134a in the chamber, we can use the specific volume (v) of R134a at the given state (P = 800 kPa, T = 60°C). The specific volume is the reciprocal of the density (ρ), so we can write the equation as m = V/ρ.

From the R134a tables, we find that the specific volume at P = 800 kPa and T = 60°C is approximately 0.001 m^3/kg.

Substituting the given values (V = 1 m^3 and v = 0.001 m^3/kg), we get m = 1 m^3 / 0.001 m^3/kg = 1000 kg.

So, the initial mass of R134a in the chamber is 1000 kg.

c) The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In this case, there is no work done (since the volume is constant), and the heat transfer occurs only between the air and the refrigerant. Therefore, the change in internal energy of the air (ΔU_air) is equal to the negative of the change in internal energy of the refrigerant (ΔU_R134a), or ΔU_air = -ΔU_R134a.

d) To determine the temperature of the air chamber when the R134a reaches 600 kPa, we need to use the first law of thermodynamics again. The change in internal energy of the air is equal to the mass of the air times the specific heat at constant volume (cv) times the change in temperature (ΔT), or ΔU_air = m_air * cv_air * ΔT_air.

Similarly, the change in internal energy of the R134a is equal to the mass of the R134a times the change in specific internal energy (u) from the initial state to the final state (P = 600 kPa), or ΔU_R134a = m_R134a * Δu_R134a.

Setting these two equations equal to each other (since ΔU_air = -ΔU_R134a), we can solve for ΔT_air.

This calculation requires the specific heat of air and the specific internal energy of R134a, which can be found in thermodynamic tables.

e) The system has reached thermal equilibrium when the temperatures of the air and the R134a are equal. This can be determined by comparing the final temperature of the air (calculated in part d) with the final temperature of the R134a (which can be found in the R134a tables at P = 600 kPa). If the temperatures are equal, then the system is in thermal equilibrium. If not, then the system has not yet reached thermal equilibrium.