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Chapter 12: Problem 27

An insulated tank having a total volume of \(0.6 \mathrm{~m}^{3}\) is divided into two compartments. Initially one compartment contains \(0.4 \mathrm{~m}^{3}\) of hydrogen \(\left(\mathrm{H}_{2}\right)\) at \(127^{\circ} \mathrm{C}, 2\) bar and the other contains nitrogen \(\left(\mathrm{N}_{2}\right)\) at \(27^{\circ} \mathrm{C}, 4\) bar. The gases are allowed to mix until an equilibrium state is attained. Assuming the ideal gas model with constant specific heats, determine (a) the final temperature, in \({ }^{\circ} \mathrm{C}\). (b) the final pressure, in bar. (c) the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{K}\).

Short Answer

Expert verified
The final temperature is approximately 48.73 掳C, the final pressure is 2.5018 bar, and the entropy produced is 0.338 kJ/K.

Step by step solution

01

Determine Initial Conditions

Calculate the initial moles of hydrogen (H2) and nitrogen (N2) using the ideal gas law: \[PV = nRT \rightarrow n = \frac{PV}{RT} \]For H2: \(P_1 = 2 \, \text{bar} = 200\, \text{kPa} \) (convert to kPa), \(V_1 = 0.4\, \text{m}^3 \), \(T_1 = 127^\text{C} = 400 \, \text{K} \)\[n_{H2} = \frac{(200 \, \text{kPa}) \cdot (0.4 \, \text{m}^3)}{(8.314 \, \text{J} / \text{mol} \cdot \text{K}) \cdot (400 \, \text{K})} \approx 24.07 \, \text{mol} \]For N2: \(P_2 = 4 \, \text{bar} = 400\, \text{kPa} \), \(V_2 = 0.2\, \text{m}^3 \), \(T_2 = 27 ^\text{C} = 300 \, \text{K} \)\[n_{N2} = \frac{(400 \, \text{kPa}) \cdot (0.2 \, \text{m}^3)}{(8.314 \, \text{J} / \text{mol} \cdot \text{K}) \cdot (300 \, \text{K})} \approx 32.12 \, \text{mol} \]
02

Calculate Final Temperature

The system is insulated, so the process is adiabatic. Use the conservation of energy principle. The final number of moles is the sum of both gases: \(n_{total} = n_{H2} + n_{N2} \approx 24.07 \, \text{mol} + 32.12 \, \text{mol} = 56.19 \, \text{mol} \)Assuming constant specific heats, set up the energy balance for the system: \[n_{H2}c_{v,H2}(T_f - T_{1,H2}) + n_{N2}c_{v,N2}(T_f - T_{1,N2}) = 0 \]With values: \(c_{v,H2} = 10.19 \, \text{J} / \text{mol} \cdot \text{K} \) and \(c_{v,N2} = 20.76 \, \text{J} / \text{mol} \cdot \text{K} \):\[24.07 \cdot 10.19 \cdot (T_f - 400) + 32.12 \cdot 20.76 \cdot (T_f - 300) = 0 \]Solve for \(T_f\):\[24.07 \cdot 10.19 \cdot T_f - 24.07 \cdot 10.19 \cdot 400 + 32.12 \cdot 20.76 \cdot T_f - 32.12 \cdot 20.76 \cdot 300 = 0 \]\[(244.87 + 666.3)T_f = 24487 + 19908 \]\[911.17T_f = 44395 \]\[T_f \approx 48.73 \degree \text{C} \]
03

Calculate Final Pressure

Use the ideal gas law for the final conditions: \[P_f V_f = n_{total} R T_f \rightarrow P_f = \frac{n_{total} R T_f}{V_f} \]Substitute \(n_{total} = 56.19 \, \text{mol} \), \(R = 8.314 \, \text{J} / \text{mol} \cdot \text{K} \), \(T_f = 321.88 \, \text{K} \), and \(V_f = 0.6 \, \text{m}^3\):\[P_f = \frac{(56.19 \, \text{mol}) (8.314 \, \text{J} / \text{mol} \cdot \text{K}) (321.88 \, \text{K})}{0.6 \, \text{m}^3} \approx 250.18 \, \text{kPa} \, (2.5018 \, \text{bar}) \]
04

Determine Entropy Production

For entropy change in each gas, use: \[\Delta S = n c_v \ln\frac{T_f}{T_i} + nR \ln\frac{V_f}{V_i} \]For H2: \[\Delta S_{H2} = 24.07 \times 10.19 \ln \frac{321.88}{400} + 24.07 \times 8.314 \ln \frac{0.6}{0.4} \]For N2: \[\Delta S_{N2} = 32.12 \times 20.76 \ln \frac{321.88}{300} + 32.12 \times 8.314 \ln \frac{0.6}{0.2} \]Sum both: \[\Delta S_{total} = \Delta S_{H2} + \Delta S_{N2} \]
05

Calculate Numerical Values for Entropy

Calculate each component: For H2 entropy: \[\Delta S_{H2} = 24.07 \times 10.19 \ln \frac{321.88}{400} + 24.07 \times 8.314 \ln \frac{0.6}{0.4} \approx -209.14 + 34.59 = -174.55 \, \text{J/K} \]For N2 entropy: \[\Delta S_{N2} = 32.12 \times 20.76 \ln \frac{321.88}{300} + 32.12 \times 8.314 \ln \frac{0.6}{0.2} \approx 425.18 + 87.52 = 512.7 \, \text{J/K} \]Therefore,\[\Delta S_{total} = -174.55 + 512.7 \approx 338.15 \, \text{J/K}\] (converting to \( \text{kJ} / \text{K} \)) \[\Delta S_{total} \approx 0.338 \, \text{kJ} / \text{K} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal gas law
In an ideal gas mixing problem, the Ideal Gas Law, represented by the equation \(PV = nRT\), is pivotal for calculating the state properties of gases. This law connects pressure (P), volume (V), the number of moles (n), the universal gas constant (R), and temperature (T). For example, in the given exercise, we use the Ideal Gas Law to find the initial moles of hydrogen and nitrogen in their respective compartments. By rearranging the equation to \(n = \frac{PV}{RT}\), and substituting the known values for pressure, volume, and temperature, we determined the initial number of moles for each gas.
Conservation of energy
When dealing with an adiabatic process in thermodynamics, the principle of Conservation of Energy is essential. In our exercise, the gases are allowed to mix in an insulated tank, meaning no heat is exchanged with the environment. This makes the system adiabatic, and energy conservation translates to internal energy balance. The total change in internal energy for the system is zero. Specifically, the total amount of energy before mixing should equal the total energy after mixing. We apply this principle by setting up an energy balance equation: \(n_{H2}c_{v,H2}(T_f - T_{1,H2}) + n_{N2}c_{v,N2} (T_f - T_{1,N2}) = 0\), where \(c_v\) represents the specific heat at constant volume. Solving this balance equation helps us find the final equilibrium temperature of the gases after mixing.
Specific heats
Specific heats are vital parameters when examining the thermal properties of gases. The specific heat at constant volume \(c_v\) is particularly relevant in the context of the given problem. It represents the amount of heat required to raise the temperature of one mole of gas by one degree Kelvin at constant volume. In our exercise, we use \(c_{v,H2} = 10.19 \mathrm{J}/\mathrm{mol} \cdot \mathrm{K}\) for hydrogen and \(c_{v,N2} = 20.76 \mathrm{J}/\mathrm{mol} \cdot \mathrm{K}\) for nitrogen. These values are crucial in calculating the final temperature by helping us establish the energy balance mentioned in the previous section.
Entropy change
Entropy is a measure of the disorder or randomness of a system, and entropy change indicates how much the disorder increases or decreases. For the given exercise, we calculate entropy production during the mixing of hydrogen and nitrogen. Using the formula \( \Delta S = n c_v \ln \frac{T_f}{T_i} + nR \ln \frac{V_f}{V_i} \), we can find the entropy change for each gas. For example, for hydrogen, the initial and final temperatures and volumes are substituted into the entropy formula to get \(\Delta S_{H2}\). Similarly, we calculate \(\Delta S_{N2}\) for nitrogen. Summing these values gives us the total entropy production \(\Delta S_{total}\), which is a key outcome showing the extent of irreversibility in the mixing process.
Adiabatic process
An adiabatic process is one where no heat is transferred into or out of the system. It is characterized by energy changes happening solely through work done on or by the system. In the provided exercise, the gases are mixed in an insulated tank, making the process adiabatic. The key characteristic of adiabatic mixing is that the final temperature and pressure are determined by internal energy changes due to the mixing, without heat exchange with the surrounding. By considering the specific heats and the initial state variables, the final state variables can be calculated. This principle allows us to find the final temperature and pressure in the tank post-mixing.

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Most popular questions from this chapter

Atmospheric air having dry-bulb and wet-bulb temperatures of 33 and \(29^{\circ} \mathrm{C}\), respectively, enters a wellinsulated chamber operating at steady state and mixes with air entering with dry-bulb and wet- bulb temperatures of 16 and \(12^{\circ} \mathrm{C}\), respectively. The volumetric flow rate of the lower temperature stream is twice that of the other stream. A single mixed stream exits. Determine for the exiting stream (a) the relative humidity. (b) the temperature, in \({ }^{\circ} \mathrm{C}\). Pressure is uniform throughout at \(1 \mathrm{~atm}\). Neglect kinetic and potential energy effects.

Air at \(30^{\circ} \mathrm{C}, 1.05\) bar, and \(80 \%\) relative humidity enters a dehumidifier operating at steady state. Moist air exits at \(15^{\circ} \mathrm{C}, 1\) bar, and \(95 \%\) relative humidity. Condensate exits in a separate stream at \(15^{\circ} \mathrm{C}\). A refrigerant flows through the cooling coil of the dehumidifier with an increase in its specific enthalpy of \(100 \mathrm{~kJ}\) per \(\mathrm{kg}\) of refrigerant flowing. Heat transfer between the humidifier and its surroundings and kinetic and potential energy effects can be ignored. Determine the refrigerant flow rate, in \(\mathrm{kg}\) per \(\mathrm{kg}\) of dry air.

Outside air at \(50^{\circ} \mathrm{F}, 1\) atm, and \(40 \%\) relative humidity enters an air-conditioning device operating at steady state. Liquid water is injected at \(45^{\circ} \mathrm{F}\) and a moist air stream exits with a volumetric flow rate of \(1000 \mathrm{ft}^{3} / \mathrm{min}\) at \(90^{\circ} \mathrm{F}, 1 \mathrm{~atm}\) and a relative humidity of \(40 \%\). Neglecting kinetic and potential energy effects, determine (a) the rate water is injected, in \(\mathrm{lb} / \mathrm{min}\). (b) the rate of heat transfer to the moist air, in Btu/h.

An equimolar mixture of helium (He) and carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) enters an insulated nozzle at \(260^{\circ} \mathrm{F}, 5 \mathrm{~atm}\), \(100 \mathrm{ft} / \mathrm{s}\) and expands isentropically to a velocity of \(1110 \mathrm{ft} / \mathrm{s}\). Determine the temperature, in \({ }^{\circ} \mathrm{F}\), and the pressure, in atm, at the nozzle exit. Neglect potential energy effects.

A mixture having a molar analysis of \(60 \% \mathrm{~N}_{2}\) and \(40 \%\) \(\mathrm{CO}_{2}\) enters an insulated compressor operating at steady state at \(1 \mathrm{bar}, 30^{\circ} \mathrm{C}\) with a mass flow rate of \(1 \mathrm{~kg} / \mathrm{s}\) and is compressed to 3 bar, \(147^{\circ} \mathrm{C}\). Neglecting kinetic and potential energy effects, determine (a) the power required, in \(\mathrm{kW}\). (b) the isentropic compressor efficiency. (c) the rate of exergy destruction, in \(\mathrm{kW}\), for \(T_{0}=300 \mathrm{~K}\).
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