Question

Consider a \(50-\mathrm{L}\) evacuated rigid bottle that is surrounded by the atmosphere at \(95 \mathrm{kPa}\) and \(27^{\circ} \mathrm{C}\). 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 entropy generation during this filling process.

Short answer

The net heat transfer during the process is 11.82 kJ, and the entropy generation is 0.132 kJ/K.

Step-by-step solution

Calculate the initial and final states of the air in the bottle

Initially, the bottle is evacuated, which means that the initial mass and internal energy of the air inside the bottle are both zero. The final state is when the air in the bottle reaches thermal and mechanical equilibrium with the surrounding atmosphere. In this state, the final pressure and temperature inside the bottle are the same as the atmospheric pressure and temperature, \(95\ \mathrm{kPa}\) and \(27^\circ\mathrm{C}\), respectively.

Determine the mass of air that flows into the bottle

To find the mass of the air that flows into the bottle, we can use the ideal gas law: $$PV = mRT$$ Rearranging the equation to solve for m, we get: $$m = \frac{PV}{RT}$$ Plug in the given values for the pressure (\(P = 95\ \mathrm{kPa}\)), volume (\(V = 50\ \mathrm{L}\)), temperature (\(T = 27^\circ\mathrm{C} = 300\ \mathrm{K}\)), and the specific gas constant for air (\(R = 0.287\ \mathrm{kPa\cdot L/(mol\cdot K)}\)): $$m = \frac{95\ \mathrm{kPa} \cdot 50\ \mathrm{L}}{0.287\ \mathrm{kPa\cdot L/(mol\cdot K)} \cdot 300\ \mathrm{K}} = 55.2\ \mathrm{g}$$

Apply the first law of thermodynamics to calculate the heat transfer

Applying the first law of thermodynamics for a closed system, we get: $$Q - W = \Delta U$$ Since the bottle is rigid, no work is done during the process (\(W = 0\)). Also, the initial internal energy is zero. Thus, the net heat transfer (\(Q\)) is equal to the final internal energy (\(U_f\)): $$Q = U_f = m u_f$$

Use the specific internal energy table for the final state

From the air property tables, we can find the specific internal energy for the final state at the given temperature and pressure: $$u_f = 214.16\ \mathrm{kJ/kg}$$

Calculate the net heat transfer

Now we can calculate the net heat transfer: $$Q = m u_f$$ $$Q = (0.0552\ \mathrm{kg}) (214.16\ \mathrm{kJ/kg}) = 11.82\ \mathrm{kJ}$$

Apply the second law of thermodynamics to calculate the entropy generation

Applying the second law of thermodynamics for a closed system, we get: $$\Delta S = S_f - S_i = m(s_f - s_i) + \frac{Q}{T_0}$$ Since the initial entropy is zero, the entropy change is simplfied as follows: $$\Delta S = m s_f + \frac{Q}{T_0}$$

Use the specific entropy table for the final state

From the air property tables, we can find the specific entropy for the final state at the given temperature and pressure: $$s_f = 0.9595\ \mathrm{kJ/(kg\cdot K)}$$

Calculate the entropy generation

Now we can calculate the entropy generation: $$\Delta S = m s_f + \frac{Q}{T_0}$$ $$\Delta S = (0.0552\ \mathrm{kg})(0.9595\ \mathrm{kJ/(kg\cdot K)}) + \frac{11.82\ \mathrm{kJ}}{300\ \mathrm{K}} = 0.132\ \mathrm{kJ/K}$$ The net heat transfer through the wall of the bottle during the filling process is \(11.82\ \mathrm{kJ}\), and the entropy generation during the process is \(0.132\ \mathrm{kJ/K}\).