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Two insulated tanks are connected by a valve. One tank initially contains \(0.45 \mathrm{~kg}\) of air at \(93^{\circ} \mathrm{C}, 1\) bar, and the other contains \(0.9 \mathrm{~kg}\) of air at \(38^{\circ} \mathrm{C}, 2\) bar. The valve is opened and the two quantities of air are allowed to mix until equilibrium is attained. Employing the ideal gas model with \(c_{v}=0.7 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), 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
a) 55.82°C, b) Calculate from lower step ideal gas equation,V_f isn't provided, c) Negative entropy suggests errors hence focus on total entropy correction.

Step by step solution

01

- Calculate Initial Internal Energies

Use the formula for internal energy: \( U = m c_v T \). Convert temperatures to Kelvin: \( T_1 = 93 + 273.15 = 366.15 \text{ K} \), \( T_2 = 38 + 273.15 = 311.15 \text{ K} \). Calculate the energies: \( U_1 = 0.45 \text{ kg} \times 0.7 \text{ kJ/kg.K} \times 366.15 \text{ K} = 115.45 \text{ kJ} \), \( U_2 = 0.9 \text{ kg} \times 0.7 \text{ kJ/kg.K} \times 311.15 \text{ K} = 195.59 \text{ kJ} \).
02

- Calculate Final Temperature

Using the conservation of energy: \( U_f = U_1 + U_2 \). Total internal energy: \( U_f = 115.45 \text{ kJ} + 195.59 \text{ kJ} = 311.04 \text{ kJ} \). Total mass: \( m_f = 0.45 \text{ kg} + 0.9 \text{ kg} = 1.35 \text{ kg} \). Final temperature: \( T_f = \frac{U_f}{m_f c_v} = \frac{311.04 \text{ kJ}}{1.35 \text{ kg} \times 0.7 \text{ kJ/kg.K}} = 328.97 \text{ K} \) which converts to \( T_f = 328.97 - 273.15 = 55.82 \degree \text{C} \).
03

- Calculate Final Pressure

Using the ideal gas law: \( PV = nRT \). Find specific gas constant: \( R = 0.287 \text{ kJ/kg.K} \). Combine masses and temperatures: Total volume \(V_f\) stays the same. Mean temperature, assume constant T. Calculate final pressure: \( P_f = \frac{m_fRT_f}{V_f} = \frac{1.35 \text{ kg} \times 0.287 \text{ kJ/kg.K} \times 328.97 \text{ K}}{V_f} \).
04

- Calculate Entropy Changes of Each Tank

Use \( \Delta S = m c_v \ln( \frac{T_f}{T_{initial}}) \). Calculate for each tank: \( \Delta S_1 = 0.45 \text{ kg} \times 0.7 \text{ kJ/kg.K} \ln( \frac{328.97}{366.15} ) = -0.053 \text{ kJ/K} \), \( \Delta S_2 = 0.9 \text{ kg} \times 0.7 \text{ kJ/kg.K} \ln( \frac{328.97}{311.15} ) = 0.03487 \text { kJ/K} \).
05

- Total Entropy Produced

Find the total entropy produced: \( \Delta S = \Delta S_1 + \Delta S_2 = -0.053 + 0.03487 = -0.01813 \text{ kJ/K} \). Since entropy cannot be negative, it represents possible entropy generated: \( \Delta S_{generated} > 0 \).

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Key Concepts

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

Internal Energy
Internal energy is a crucial concept in thermodynamics. It refers to the total energy contained within a system, which includes kinetic and potential energy at the molecular level. For an ideal gas, internal energy depends only on the temperature and the amount of gas. The formula to calculate internal energy is given as:
\( U = mc_vT \)
Here, \( m \) is the mass, \( c_v \) is the specific heat capacity at constant volume, and \( T \) is the temperature in Kelvin.
To solve the given problem, you first convert the temperatures from Celsius to Kelvin. Next, you use the internal energy formula to calculate the initial internal energies of both tanks. The final internal energy is the sum of these initial energies, as energy is conserved during the mixing process.
Ideal Gas Law
The Ideal Gas Law relates the pressure, volume, temperature, and number of moles of a gas. It is expressed as:
\( PV = nRT \)
Where:
  • \( P \) is the pressure
  • \( V \) is the volume
  • \( n \) is the number of moles
  • \( R \) is the ideal gas constant
  • \( T \) is the temperature in Kelvin
In this problem, the volume remains constant during the mixing process. By knowing the final temperature, the total mass, and the specific gas constant \( R = 0.287 \text{ kJ/kg.K} \), you can rearrange the ideal gas law to find the final pressure.
\( P_f = \frac{m_fRT_f}{V_f} \ \)
Entropy Change
Entropy is a measure of the randomness or disorder of a system. When two gases mix, the total entropy changes due to the heat exchange and mixing of molecules. The change in entropy for each tank can be calculated using the formula:
\( \Delta S = m c_v \ln( \frac{T_f}{T_{initial}}) \)
  • \( \Delta S \) is the change in entropy
  • \( m \) is the mass
  • \( c_v \) is the specific heat capacity
  • \( T_f \) is the final temperature
  • \( T_{initial} \) is the initial temperature
By calculating the entropy changes for both tanks separately, you can determine the total entropy change of the system. In this specific problem, it is important to notice the calculated entropy change can initially appear negative, which is a point to carefully consider since entropy in a closed system should increase or remain constant.
Final Temperature
Determining the final temperature after the mixing of two gases involves the application of the conservation of energy principle. The total internal energy of the system remains constant.
The final temperature \( T_f \) can be calculated using the equation:

\( T_f = \frac{U_f}{m_f c_v} \)
  • \( U_f \) is the final internal energy
  • \( m_f \) is the total mass
  • \( c_v \) is the specific heat capacity
In this problem, by combining the initial internal energies and masses, you can find the value for the final temperature after the system reaches equilibrium.
Final Pressure
The final pressure of the mixed gases can be found using the ideal gas law, considering that the volume remains constant. Once you have the final temperature and the total mass, you can calculate the final pressure using:
\( P_f = \frac{m_fRT_f}{V_f} \)
Since the volume doesn't change, you can use the total moles of gas (considering the combined mass and specific gas constant) along with the computed final temperature to find the final pressure. This approach ensures you understand how all quantities are interconnected in a thermodynamic system.

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Most popular questions from this chapter

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