/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 113 A rigid, insulated tank whose vo... [FREE SOLUTION] | 91Ó°ÊÓ

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A rigid, insulated tank whose volume is \(10 \mathrm{~L}\) is initially evacuated. A pinhole leak develops and air from the surroundings at 1 bar, \(25^{\circ} \mathrm{C}\) enters the tank until the pressure in the tank becomes 1 bar. Assuming the ideal gas model with \(k=1.4\) for the air, determine (a) the final temperature in the tank, in \({ }^{\circ} \mathrm{C}\), (b) the amount of air that leaks into the tank, in \(\mathrm{g}\), and (c) the amount entropy produced, in \(\mathrm{J} / \mathrm{K}\).

Short Answer

Expert verified
The final temperature can't be calculated exactly due to paradox; amount of air entering is approximately 0.1174 g.

Step by step solution

01

– Understanding the Setup

The system is a 10 L rigid and insulated tank. Initially, it is evacuated and then air from the surroundings at 1 bar and 25°C leaks in until the tank’s pressure reaches 1 bar. Given the ideal gas model with specific heat ratio (k) of 1.4, we need to find the final temperature, the mass of the air that enters, and the entropy produced.
02

– Apply Ideal Gas Law

Since the tank is rigid and of constant volume, we apply the ideal gas law. Initially, the tank is evacuated, so the initial volume is zero. As air enters the tank, P=1 bar, V=10 L. Use the ideal gas law to find the initial moles of air entering the tank:\[ PV = nRT \]We need to convert temperature from Celsius to Kelvin and volume from liters to cubic meters.
03

– Convert Units

Convert the surrounding air conditions:- Pressure (P) is 1 bar = 100 kPa- Temperature (T) is 25°C = 298.15 K- Volume (V) is 10 L = 0.01 m³
04

– Find the Initial Moles of Air

Using the ideal gas equation with the adjusted units:\[ n = \frac{PV}{RT} \ n = \frac{(100 \, \text{kPa})(0.01 \, \text{m}^3)}{(8.314 \, \text{J/mol·K})(298.15 \, \text{K})} = 0.00405 \, \text{mol} \]
05

– Calculate the Final Temperature

As the tank is insulated, it means there is no heat exchange (adiabatic process). Using the relationship for adiabatic processes,\[ \frac{T_final}{T_{initial}} = \left( \frac{P_final}{P_initial} \right)^{\frac{(1-1.4)}{1.4}} \ \frac{T_{final}}{298.15} = \left( \frac{1}{0} \right)^{\frac{(1-1.4)}{1.4}} \to Pressure initial P = 0 \ T_{final} = 0 \]Therefore, this equation serves as a paradox since the value for initial P = 0.
06

– Calculate the Mass of Air Entering

The molar mass of air, M, approximately equals 29 g/mol. Using the moles calculated,\[ m = nM \ m = 0.00405 \, \text{mol} \times 29 \, \text{g/mol} = 0.1174 \, \text{g} \]
07

– Calculate Entropy Produced

For entropy production, use the entropy change relationship in adiabatic processes. Since the tank is isolated,\[ S = nC_v \, ln \left( \frac{T_{final}}{T_{initial}} \right) + nR \, ln \left( \frac{V_{final}}{V_{initial}} \right) \ V_{initial} = 0 \to \text{no initial volume} = 0 need to correct since Process is irreversible.\]

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in thermodynamics that relates the pressure, volume, temperature, and quantity of an ideal gas. The law 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
The Ideal Gas Law is used to determine the amount of air entering the tank. By converting all units properly and applying the equation, we can calculate the number of moles of air that have entered the tank. In our exercise, the pressure P is 1 bar (or 100 kPa), the volume V is 10 liters (or 0.01 cubic meters), and the temperature T is 25°C (or 298.15 K). Substituting these values into the Ideal Gas Law gives:

\[ n = \frac{PV}{RT} = \frac{(100 \text{kPa}) (0.01 \text{m}^3)}{(8.314 \text{J/mol·K})(298.15 \text{K})} = 0.00405 \text{mol} \]

This calculation reveals that 0.00405 moles of air have leaked into the tank.
Adiabatic Process
An adiabatic process is one in which no heat is transferred into or out of the system. In the context of our exercise, the tank is insulated, hence it undergoes an adiabatic process. For such a process, the relationship between the temperature and pressure is given by:

\[ \frac{T_{final}}{T_{initial}} = \left( \frac{P_{final}}{P_{initial}} \right)^{\frac{(1-k)}{k}} \]

where k is the specific heat ratio (for air, k = 1.4). Initially, the tank is evacuated which means the initial pressure \( P_{initial} \) is 0, making the calculation paradoxical using this formula directly. Instead, understanding that an adiabatic process involves rapid compression or expansion without heat exchange, our final temperature can be inferred through the changes in internal energy that reflect such a process. Given the rigidity of the tank and no heat exchange, the system relies on the internal mechanism to reach thermal equilibrium, indirectly stabilizing at higher temperatures given the entrance of external air at ambient conditions.
Entropy Production
Entropy production is a key concept to understand the irreversibility of a process. In thermodynamics, entropy quantifies the amount of disorder within a system. For an adiabatic process where the tank is isolated, the calculation of entropy involves the following formula:

\[ \Delta S = nC_v \ln \left( \frac{T_{final}}{T_{initial}} \right) + nR \ln \left( \frac{V_{final}}{V_{initial}} \right) \]

Here, the tank initially has no volume because it is evacuated, so simplifying the irreversibility part of the process becomes critical. Simplistically, while the exact quantified value isn’t straightforward due to the undefined initial volume, conceptually, this entropic expression alerts to greater disorder due to molecular chaos upon intermolecular collision as air rushes in.

Irreversibilities in our scenario involve air entering the previously evacuated volume, creating significant entropy production. This results in analytic complexity as our zero initial volume means entropy directly quantifying getting tricky – reinforcing the practical impact of entropy as representing disorder's inevitable increase under such system constraints.

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Most popular questions from this chapter

One-tenth kmol of carbon monoxide \((\mathrm{CO})\) in a pistoncylinder assembly undergoes a process from \(p_{1}=150 \mathrm{kPa}\), \(T_{1}=300 \mathrm{~K}\) to \(p_{2}=500 \mathrm{kPa}, T_{2}=370 \mathrm{~K}\). For the process, \(\mathrm{W}=-300 \mathrm{~kJ}\). Employing the ideal gas model, determine (a) the heat transfer, in kJ. (b) the change in entropy, in \(\mathrm{kJ} / \mathrm{K}\). Show the process on a sketch of the \(T-s\) diagram.

Air enters the turbine of a jet engine at \(1190 \mathrm{~K}, 10.8\) bar and expands to \(5.2\) bar. The air then flows through a nozzle and exits at \(0.8\) bar. Operation is at steady state, and the flow is adiabatic. The nozzle operates with no internal irreversibilities, and the isentropic turbine efficiency is \(85 \%\). The air velocities at the turbine inlet and exit are negligible. Assuming the ideal gas model for the air, determine the velocity of the air exiting the nozzle, in \(\mathrm{m} / \mathrm{s}\).

One kg of air contained in a piston-cylinder assembly undergoes a process from an initial state where \(T_{1}=300 \mathrm{~K}\), \(v_{1}=0.8 \mathrm{~m}^{3} / \mathrm{kg}\) to a final state where \(T_{2}=420 \mathrm{~K}, v_{2}=\) \(0.2 \mathrm{~m}^{3} / \mathrm{kg}\). Can this process occur adiabatically? If yes, determine the work, in \(\mathrm{kJ}\), for an adiabatic process between these states. If no, determine the direction of the heat transfer. Assume the ideal gas model for air.

Steam enters a turbine operating at steady state at \(6 \mathrm{MPa}, 600^{\circ} \mathrm{C}\) with a mass flow rate of \(125 \mathrm{~kg} / \mathrm{min}\) and exits as saturated vapor at \(20 \mathrm{kPa}\), producing power at a rate of 2 MW. Kinetic and potential energy effects can be ignored. Determine (a) the rate of heat transfer, in \(\mathrm{kW}\), for a control volume including the turbine and its contents, and (b) the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), for an enlarged control volume that includes the turbine and enough of its surroundings that heat transfer occurs at the ambient temperature, \(27^{\circ} \mathrm{C}\).

By injecting liquid water into superheated steam, the desuperheater shown in Fig. P6.83 has a saturated vapor stream at its exit. Steady-state operating data are provided in the accompanying table. Stray heat transfer and all kinetic and potential energy effects are negligible. (a) Locate states 1,2 , and 3 on a sketch of the \(T-s\) diagram. (b) Determine the rate of entropy production within the desuperheater, in \(\mathrm{kW} / \mathrm{K}\). $$ \begin{array}{ccccccc} \text { State } & p(\mathrm{MPa}) & T\left({ }^{\circ} \mathrm{C}\right) & v \times 10^{3}\left(\mathrm{~m}^{3} / \mathrm{kg}\right) & u(\mathrm{k}) / \mathrm{kg}) & h(\mathrm{k}) / \mathrm{kg}) & s(\mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}) \\ \hline 1 & 2.7 & 40 & 1.0066 & 167.2 & 169.9 & 0.5714 \\ 2 & 2.7 & 300 & 91.01 & 2757.0 & 3002.8 & 6.6001 \\ 3 & 2.5 & \text { sat. vap. } & 79.98 & 2603.1 & 2803.1 & 6.2575 \end{array} $$

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