Question

A closed, rigid, insulated vessel having a volume of \(0.142 \mathrm{~m}^{3}\) contains oxygen \(\left(\mathrm{O}_{2}\right)\) initially at 100 bar, \(7^{\circ} \mathrm{C}\). The oxygen is stirred by a paddle wheel until the pressure becomes 150 bar. Determine the (a) final temperature, in \({ }^{\circ} \mathrm{C}\). (b) work, in \(\mathrm{kJ}\). (c) amount of exergy destroyed in the process, in \(\mathrm{kJ}\). Let \(T_{0}=7^{\circ} \mathrm{C}\).

Short answer

Final temperature: 146.85 °C. Work: 1781.94 kJ. Exergy destroyed: 1026.75 kJ.

Step-by-step solution

Identify Initial Conditions

The volume of the vessel is given as \(0.142 \, \text{m}^3\). The initial pressure \(P_1\) is 100 bar, and the initial temperature \(T_1\) is \(7^{\text{\tiny{o}}} \text{C}\), which is equal to 280 K (using \(T(K) = T(^{\text{o}} C) + 273.15\)).

Ideal Gas Law Application

Using the ideal gas law, calculate the initial number of moles of oxygen: \(P_1 V = n R T_1\), where \(P_1 = 100 \, \text{bar} \, (1 \, \text{bar} = 10^5 \, \text{Pa})\), \(V = 0.142 \, \text{m}^3\), \(R = 8.314 \, \text{J/(mol \, K)}\), \(T_1 = 280 \, \text{K}\). Let's find \(n\).

Calculate Initial Moles

Rearrange the ideal gas law: \(n = \frac{P_1 V}{R T_1}\). Substituting the known values: \(n = \frac{100 \times 10^5 \, \text{Pa} \times 0.142 \, \text{m}^3}{8.314 \, \text{J/(mol \, K)} \times 280 \, \text{K}} \approx 606.46 \, \text{mol}\).

Final Temperature Calculation

Since the vessel is rigid and insulated, it is an adiabatic process with no change in volume. Using the ideal gas law for the final state, where final pressure \(P_2 = 150 \, \text{bar}\) and volume \(V = 0.142 \, \text{m}^3\): \(P_2 V = n R T_2\). Rearrange to find \(T_2\): \(T_2 = \frac{P_2 V}{n R}\).

Calculate Final Temperature

Substitute the values into the equation: \(T_2 = \frac{150 \times 10^5 \, \text{Pa} \times 0.142 \, \text{m}^3}{606.46 \, \text{mol} \times 8.314 \, \text{J/(mol \, K)}} \approx 420 \, \text{K}\). Converting to Celsius: \(T_2 = 420 \, \text{K} - 273.15 \approx 146.85 \, ^{\text{o}} \text{C}\).

Work Done Calculation

The work done by the paddle wheel can be calculated using the change in internal energy: for an ideal gas, \(U = n C_V \Delta T\). Here, \(C_V\) (specific heat at constant volume for \(O_2\)) is 21 J/(mol K).

Find Work Done

Calculate the change in temperature: \(\Delta T = T_2 - T_1 \). Substitute the values: \(\Delta T = 420 \, \text{K} - 280 \, \text{K} = 140 \, \text{K}\). Then, \(W = n C_V \Delta T = 606.46 \, \text{mol} \times 21 \, \text{J/(mol \, K)} \times 140 \, \text{K} = 1781944.4 \, \text{J} = 1781.94 \, \text{kJ}\).

Calculate Exergy Destroyed

The exergy destroyed in the process can be found using \(T_0 = 280 \, \text{K}\). Exergy destroyed \( \Delta \Phi = T_0 \Delta S_{gen}\). First find \( \Delta S_{gen}\). For an ideal gas, \( \Delta S_{gen} = n C_V \ln\frac{T_2}{T_1} + nR \ln\frac{V_2}{V_1} \). In our case, volume does not change: \( \ln\frac{V_2}{V_1} = 0 \).

Simplify and Calculate \( \Delta S_{gen} \)

Simplify: \( \Delta S_{gen} = n C_V \ln\frac{T_2}{T_1} \). Substitute values: \( \Delta S_{gen} = 606.46 \times 21 \ln\frac{420}{280} = 606.46 \times 21 \times \ln\frac{3}{2} \approx 606.46 \times 21 \times 0.2877 \). Hence, \( \Delta S_{gen} = 3666.98 \text{J/K}\).

Calculate Exergy Destroyed

Finally, \( \Delta \Phi = T_0 \Delta S_{gen} = 280 \, \text{K} \times 3666.98 \, \text{J/K} = 1026754 \text{J} = 1026.75 \text{kJ}\).

Summary of Results

The final temperature is approximately 146.85 °C, the work done by the paddle wheel is 1781.94 kJ, and the amount of exergy destroyed is 1026.75 kJ.

Key concepts

Adiabatic Process

An adiabatic process is one where there is no heat exchange with the surroundings. This means that all changes in the system are due to work done on or by the system. In the case of the closed, rigid, insulated vessel, any change in the internal energy of the oxygen inside is due to work done by the paddle wheel. Since the system is insulated, the condition of no heat transfer holds true. An important property of adiabatic processes is that they occur without changing the entropy of the system. As a result, understanding how work and temperature changes relate in an adiabatic process is essential for solving problems like this one.

Ideal Gas Law

The ideal gas law, represented as `PV = nRT`, describes the relationship between the pressure (P), volume (V), temperature (T), and amount (n) of an ideal gas. For this exercise, you used the ideal gas law to determine the number of moles of oxygen and subsequently the final temperature after the work done by the paddle wheel. Given the initial conditions, rearranging the formula allowed you to find the initial quantity of moles: \[n = \frac{P_1 V}{R T_1}\] Later, to find the final temperature under the new pressure, you used: \[T_2 = \frac{P_2 V}{n R}\] These calculations are fundamental in transitioning between different states of the gas during the process.

Internal Energy

Internal energy, denoted as U, is the total energy stored within a system. For an ideal gas, internal energy is only a function of temperature and the amount of substance. The change in internal energy, ∆U, relates to the work done on the system in an adiabatic process. This can be calculated using the formula: \[\Delta U = n C_V \Delta T\] Here, \(C_V\) is the specific heat at a constant volume. By finding the change in temperature (∆T) and knowing the number of moles (n), you can determine the work performed on the gas. In this case: \[ W = n C_V \Delta T = 606.46 \,\text{mol} \times 21 \,\text{J/(mol \, K)} \times 140 \,\text{K} = 1781944.4 \,\text{J} = 1781.94 \,\text{kJ} \] This follows from the first law of thermodynamics in an adiabatic process where the change in internal energy equals the work done.

Entropy Generation

Entropy is a measure of disorder or randomness in a system. In any real process, there's often an increase in entropy, representing energy that becomes unavailable for doing useful work. The entropy generation \(\Delta S_{gen}\) is calculated by considering the change in temperature and any volume change. For this exercise, volume remains constant, simplifying the entropy calculation to: \( \Delta S_{gen} = n C_V \ln\frac{T_2}{T_1} \) After finding \(\Delta S_{gen}\), you can assess the extent of energy dispersion into less useful forms.

Exergy Destruction

Exergy is a measure of the useful work potential of a system. During real processes, some of this potential is destroyed due to irreversibilities, such as friction or unrestrained expansion. The exergy destruction \(\Delta \Phi\) in the process is given by: \[ \Delta \Phi = T_0 \Delta S_{gen} \] This involves calculating the entropy generated and multiplying it by the initial temperature of the system. In the problem: \(\ \Delta \Phi = 280 \, \text{K} \times 3666.98 \, \text{J/K} = 1026754 \text{J} \approx 1026.75 \text{kJ} \) This value indicates how much potential work is irrecoverably lost due to the inherent inefficiencies of the process.