Air is contained in a piston-cylinder device at 9 bar and 927˚C, and occupies a
volume of 0.8 m3. The air undergoes an isothermal process until the pressure is
reduced to 4.5 bar. The piston is now fixed in place and not allowed to move while
a heat transfer process takes place until the air reaches 27˚C. Assume the air to be
a perfect gas. Take cp=1.005 kJ/kg∙K and cv=0.718 kJ/kg∙K.
(a) Calculate the final volume and final pressure of the air.
(6 marks)
(b) Determine the change in internal energy of the air, the work done
by the air and the total heat transferred.

a) Note that the final volume of the air is also 1.6 m3, and the final pressure is 4.5 bar; and

b)  the change in internal energy of the air is 0 kJ, the work done by the air is 0.966 kJ, and the total heat transferred is 487.7 kJ.

To solve this problem, we can use the ideal gas law and the first law of thermodynamics.

(a) To determine the final volume and final pressure of the air, we can use the ideal gas law, which states that:

P1V1/T1 = P2V2/T2

where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.

We can first use this equation to determine the final volume of the air during the isothermal process:

P1V1/T1 = P2V2/T2

(9 bar)(0.8 m3)/(1200 K) = P2V2/(927 K)

V2 = (9/4.5) * (0.8 m3)
= 1.6 m3

Now, we can use the ideal gas law again to determine the final pressure of the air during the isothermal process:

P1V1/T1 = P2V2/T2

(9 bar)(0.8 m3)/(1200 K) = (P2)(1.6 m3)/(1200 K)

P2 = (9 bar)(0.8 m3)/(1.6 m3)
= 4.5 bar

During the subsequent heat transfer process, the piston is fixed in place, which means the volume of the air remains constant at 1.6 m3.

Therefore, the final volume of the air is also 1.6 m3, and the final pressure is 4.5 bar.

(b) To determine the change in internal energy of the air, the work done by the air, and the total heat transferred, we can use the first law of thermodynamics, which states that:

ΔU = Q - W

where ΔU is the change in internal energy, Q is the heat transferred, and W is the work done.

During the isothermal process, the temperature of the air remains constant at 927 K, which means the internal energy of the air does not change:

ΔU = 0

During the isothermal process, the work done by the air is:

W = -P1V1 ln(P2/P1)
= -(9 bar)(0.8 m3) ln(4.5/9)

= 0.966 kJ

During the subsequent heat transfer process, the air is heated from 927 K to 300 K, which means heat is transferred into the air:

Q = m cΔT = (P2V2/R) cΔT
= (4.5 bar)(1.6 m3)/(287 J/kg∙K) (1.005 kJ/kg∙K) (927 K - 300 K)
= 487.7 kJ

Therefore, the change in internal energy of the air is 0 kJ, the work done by the air is 0.966 kJ, and the total heat transferred is 487.7 kJ.

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