Given the problem involving a 5-kg mass of air in a piston-cylinder undergoing an adiabatic expansion, we can analyze it step by step. Let's break down the required parts:

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Part (a) Determine the entropy change of the air

Known dataGiven the problem involving a 5-kg mass of air in a piston-cylinder undergoing an adiabatic expansion, we can analyze it step by step. Let's break down the required parts:

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Part (a) Determine the entropy change of the air

Known

• Mass of air: $m = 5 \, \text{kg}$
• Initial temperature: $T_1 = 427^\circ C = 700 \, K$
• Initial pressure: $P_1 = 600 \, \text{kPa}$
• Final pressure: $P_2 = 100 \, \text{kPa}$
• Work output: $W = 600 \, \text{kJ}$
• Specific heat ratio for air $\gamma = 1.4$ (for diatomic gases like air)
• The air is assumed to have constant specific heats evaluated at 300 K.

Since the process is adiabatic, there is no heat transfer ($Q = 0$).

Adiabatic Process Entropy Change:

In an adiabatic process, entropy change ($\Delta S$) can be directly determined by the formula:

$\Delta S = m c_v \ln \left(\frac{T_2}{T_1}\right) + m R \ln \left(\frac{V_2}{V_1}\right)$

But we can simplify things because for adiabatic processes in ideal gases, the relation between pressures and temperatures is given by:

$\frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma - 1}{\gamma}}$

Using this equation, we can determine the final temperature $T_2$.

$T_2 = T_1 \left( \frac{P_2}{P_1} \right)^{\frac{\gamma - 1}{\gamma}}$

Substituting values:

$T_2 = 700 \left( \frac{100}{600} \right)^{\frac{1.4 - 1}{1.4}} = 700 \times (0.1667)^{0.2857}$

Calculating $T_2$:

$T_2 \approx 700 \times 0.5848 = 409.4 \, \text{K}$

Thus, $T_2 \approx 409.4 \, K$.

Entropy Change:

For an adiabatic reversible process, the change in entropy $\Delta S$ is zero. However, for an irreversible adiabatic process, we need to calculate the entropy change. Since heat transfer is zero, the entropy change is often calculated using:

$\Delta S = m c_v \ln \left( \frac{T_2}{T_1} \right)$

Where:

• $c_v$ is the specific heat at constant volume. For air at constant specific heat, $c_v \approx 0.718 \, \text{kJ/kg} \cdot \text{K}$.

Substitute values:

$\Delta S = 5 \times 0.718 \times \ln \left( \frac{409.4}{700} \right)$

Calculating:

$\Delta S = 5 \times 0.718 \times \ln(0.5848)$ $\Delta S \approx 5 \times 0.718 \times (-0.5371)$ $\Delta S \approx -1.929 \, \text{kJ/kg} \cdot \text{K}$

Thus, the entropy change is approximately:

$\Delta S \approx -1.929 \, \text{kJ/kg} \cdot \text{K}$

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Part (b) Is the process realistic?

Since the process is adiabatic, there is no heat transfer to or from the system. However, the entropy change is negative, which suggests that the system's disorder decreases. In real-world systems, a decrease in entropy without any external heat input indicates that the process is likely irreversible, as real processes typically involve some form of friction, turbulence, or non-ideal gas behavior.

Second Law of Thermodynamics:

The second law states that for any isolated system, the total entropy can only stay the same (reversible process) or increase (irreversible process). A decrease in entropy in an adiabatic process can only happen if the system is not isolated (for example