Question

Consider a 20 -L evacuated rigid bottle that is surrounded by the atmosphere at \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). A valve at the neck of the bottle is now opened and the atmospheric air is allowed to flow into the bottle. The air trapped in the bottle eventually reaches thermal equilibrium with the atmosphere as a result of heat transfer through the wall of the bottle. The valve remains open during the process so that the trapped air also reaches mechanical equilibrium with the atmosphere. Determine the net heat transfer through the wall of the bottle and the exergy destroyed during this filling process.

Short answer

Answer: The net heat transfer through the wall of the bottle is 0 kJ, and the exergy destroyed during the filling process is also 0 kJ.

Step-by-step solution

Determine the final conditions of the trapped air

We are given that the bottle finally reaches thermal and mechanical equilibrium with the atmosphere. Thus, the final pressure and temperature inside the bottle will be the same as atmospheric conditions: \(P_{final}=100 kPa\) and \(T_{final}=25^{\circ}C (298.15 K)\).

Calculate the amount of air trapped inside the bottle

To determine the amount of air trapped in the bottle, we can use the ideal gas law: \(PV = nRT\) In this case, we are given the volume \(V = 20 L\), the final pressure \(P_{final} = 100 kPa\), and the final temperature \(T_{final} = 298.15 K\). The gas constant for air is \(R = 0.287 kJ/(kg\cdot K)\). Solving for \(n\) gives: \(\displaystyle n = \frac{PV}{RT} = \frac{(100\,\text{kPa})(20\,\text{L})}{(0.287\,\text{kJ/(kg}\cdot\text{K})(298.15\,\text{K})} \approx 2.368\,\text{kg}\) So, the mass of air trapped inside the bottle is 2.368 kg.

Determine the energy balance equation and solve for heat transfer

As the volume of the bottle is constant, no work is done during the filling process (rigid bottle). Therefore, the energy balance can be written as: \(Q - W = \Delta U\) The change in internal energy of air can be determined using specific heat capacity at constant volume, \(C_v = 0.718\, kJ/(kg\cdot K)\), and the temperature change. Since the initial temperature of air was equal to the final temperature (due to thermodynamic equilibrium), the temperature change is zero: \(\Delta U = mc_v \Delta T = 0\) Therefore, from the energy balance, we can conclude that no heat is transferred during the process. So, \(Q = 0\).

Determine the exergy destroyed

The exergy balance equation is given by: \(\Delta E_{exergy} + Q - W + T_0 \Delta S_{gen} = E_{exergy, destroyed}\) Since \(Q=0\), \(W=0\), and \(\Delta E_{exergy} = 0\) (initial and final states are the same), the expression simplifies to: \(E_{exergy, destroyed} = T_0 \Delta S_{gen}\) To calculate \(\Delta S_{gen}\), the change in entropy, we can use the relationship for an ideal gas: \(\Delta S_{gen} = mc_p \ln{\frac{T_{final}}{T_{initial}}} - nR\ln{\frac{P_{final}}{P_{initial}}}\) Since \(T_{initial} = T_{final}\), \(\Delta T = 0\). Therefore, the first term becomes zero, and the expression simplifies to: \(\Delta S_{gen} = -nR\ln{\frac{P_{final}}{P_{initial}}}\) The initial pressure inside the empty bottle is \(P_{initial} = 0\,\text{kPa}\). However, when calculating entropy change, we should assume an isothermal process during the filling of the bottle occurred. Thus, \(\frac{P_{final}}{P_{initial}}\rightarrow\frac{P_{final}}{P_{final}}=1\). Therefore, \(\Delta S_{gen} = 0\). Finally, since \(\Delta S_{gen}=0\), we can conclude that the exergy destroyed is also zero: \(E_{exergy, destroyed} = 0\) So, the net heat transfer through the wall of the bottle is 0 kJ, and the exergy destroyed during the filling process is 0 kJ as well.