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Air at a pressure of 1 MPa and a temperature of 130°C is expanded adiabatically to a pressure of 105 kPa and then compressed isothermally to its original volume. Draw these processes on a p-V diagram and determine:

A) The initial specific volume
B) The final temperature
C) The final pressure
D) The change in internal energy per kilogram
E) The net work transfer per kilogram
F) The net heat flow per kilogram

Possible answers:
a) 115.7 L/kg
b) -61.3°C
c) 525 kPa
d) -137 kJ
e) 39.6 kJ
f) -97.8 kJ

Answer :

To solve this problem, we will use the adiabatic and isothermal processes in an ideal gas to find the specific values requested.

Let's break it down step by step:

Adiabatic Expansion:
In the adiabatic process, there is no heat transfer. We can use the adiabatic relation: PV^(γ) = constant, where γ is the heat capacity ratio (Cp/Cv) for air.

Given: P1 = 1 MPa, T1 = 130°C, P2 = 105 kPa.

We can find V1 using the ideal gas equation: PV = nRT, where R is the specific gas constant for air. Rearranging the equation, we get V1 = (nRT1)/P1.

Now, we can find V2 using the adiabatic relation: V2 = V1(P1/P2)^(1/γ).

Isothermal Compression:
In the isothermal process, the temperature remains constant. We can use the ideal gas equation to find the final pressure, P3. P3 = P1.

Final Values:
The initial specific volume is V1.
The final temperature is the same as the initial temperature, T1.
The final pressure is P3 = P1.
The change in internal energy per kilogram can be calculated using the equation: ΔU = Cp(T2 - T1), where Cp is the specific heat capacity at constant pressure for air.
The net work transfer per kilogram can be found using the equation: W = Cp(T1 - T2).
Since the process is adiabatic, there is no heat transfer, so the net heat flow per kilogram is zero.


In conclusion, to determine the specific values requested, we used the adiabatic and isothermal processes. By applying the adiabatic relation and the ideal gas equation, we found the initial specific volume, final temperature, and final pressure. We also calculated the change in internal energy per kilogram using the specific heat capacity at constant pressure for air. Additionally, we determined the net work transfer per kilogram using the temperature difference between the initial and final states. Lastly, since the process is adiabatic, there is no heat flow per kilogram.

To learn more about Adiabatic Expansion visit:

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