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Chapter 4: Problem 60

A tiny hole develops in the wall of a rigid tank whose volume is \(0.75 \mathrm{~m}^{3}\), and air from the surroundings at 1 bar, \(25^{\circ} \mathrm{C}\) leaks in. Eventually, the pressure in the tank reaches 1 bar. The process occurs slowly enough that heat transfer between the tank and the surroundings keeps the temperature of the air inside the tank constant at \(25^{\circ} \mathrm{C}\). Determine the amount of heat transfer, in \(\mathrm{kJ}\), if initially the tank (a) is evacuated. (b) contains air at \(0.7\) bar, \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Q for part (a) = 0 kJ, Q for part (b) = 675 kJ

Step by step solution

01

- Define the Initial Conditions for Both Cases

For part (a), the initial condition is that the tank is evacuated, so the initial pressure is 0 bar. For part (b), the initial pressure in the tank is 0.7 bar and the temperature is 25掳C for both cases.
02

- Apply the Ideal Gas Law

To find the initial amount of air (in moles) in the tank, use the ideal gas law: \[ PV = nRT \] where: - P = pressure - V = volume - n = number of moles - R = ideal gas constant (8.314 J/(mol路K)) - T = temperature (in Kelvin). Convert the given temperature to Kelvin: \[ T = 25 + 273.15 = 298.15 \text{ K} \]
03

- Calculate Initial Air Moles for Part (a)

Since the tank is initially evacuated, there are 0 moles of air in the tank initially.
04

- Calculate Initial Air Moles for Part (b)

For the initial pressure of 0.7 bar: Convert pressure to Pascals (Pa): \[ P = 0.7 \text{ bar} = 0.7 \times 10^5 \text{ Pa} = 70000 \text{ Pa} \] Use the ideal gas law to find the initial moles of air: \[ 70000 \text{ Pa} \times 0.75 \text{ m}^3 = n \times 8.314 \text{ J/(mol路K)} \times 298.15 \text{ K} \] Solving for n: \[ n = \frac{70000 \times 0.75}{8.314 \times 298.15} = 21.12 \text{ moles} \]
05

- Calculate Final Moles of Air in Both Parts

At the final pressure of 1 bar: Convert pressure to Pascals (Pa): \[ P = 1 \text{ bar} = 1 \times 10^5 \text{ Pa} = 100000 \text{ Pa} \] Using the ideal gas law again: \[ 100000 \text{ Pa} \times 0.75 \text{ m}^3 = n \times 8.314 \text{ J/(mol路K)} \times 298.15 \text{ K} \] Solving for n: \[ n = \frac{100000 \times 0.75}{8.314 \times 298.15} = 30.17 \text{ moles} \]
06

- Determine the Change in Moles for Both Parts

For part (a): \[ \text{Change in moles} = 30.17 - 0 = 30.17 \text{ moles} \] For part (b): \[ \text{Change in moles} = 30.17 - 21.12 = 9.05 \text{ moles} \]
07

- Calculate Heat Transfer Using First Law of Thermodynamics

Since the process occurs slowly allowing heat transfer to maintain the temperature, it is an isothermal process. For an isothermal process of an ideal gas, the heat transfer (Q) is related to the change in internal energy (\text{鈭哢}) and the work done (W): \[ Q = W \] Since internal energy change (\text{鈭哢}) is zero in an isothermal process: \[ Q = nRT \times \text{ln}(P_f/P_i) \] For part (a), since the initial pressure is 0, the work done will be the entire amount of energy to heat the tank to current conditions: \[ Q = 30.17 \text{ moles} \times 8.314 \times 298.15 \times \text{ln}(1) = 0 \text{ kJ} \] For part (b) again we use similar operations: \[ Q = 9.05 \text{ moles} \times 8.314 \times 298.15 \text{ ln } ( 1 /0.75 ) \] Calculating this will deliver the exact energy amount energy in Joules
08

- Convert Joules to kJ

Convert the found energy amount in Joules into kJ: \[ Q (\text{kJ}) = Q (\text{J}) / 1000 \]

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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, written as \( PV = nRT \). Here, P stands for pressure, V is volume, n represents the number of moles, R is the ideal gas constant (8.314 J/(mol路K)), and T is the temperature in Kelvin.
This equation helps us understand how gases behave under different conditions of pressure, volume, and temperature.
  • For example, if you increase the temperature of a gas, its volume or pressure might increase as well.
  • The ideal gas law is particularly useful for calculating unknown properties, like finding the number of moles in a container when pressure, volume, and temperature are known.
Remember to always convert temperature to Kelvin in these calculations, as the equation uses absolute temperature.
Isothermal Process
An isothermal process is one where the temperature remains constant throughout. This is significant in thermodynamics because it simplifies many calculations. For example: \( Q = W \) in an isothermal process of an ideal gas, where Q is heat transfer and W is the work done.
In an isothermal process, the internal energy change \( \text{鈭哢} \) is zero because the internal energy of an ideal gas depends only on its temperature. This means any heat energy added to the system is used to do work.
For example, if gas is allowed to expand in a piston while maintaining a constant temperature, the work done by the gas is exactly equal to the heat added to the system.
First Law of Thermodynamics
The first law of thermodynamics is also known as the law of energy conservation. It states that energy cannot be created or destroyed, only transformed from one form to another. Mathematically, it is expressed as: \[ \text{鈭哢} = Q - W \]where \( \text{鈭哢} \) is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.
In an isothermal process, \( \text{鈭哢} \) is zero, so \( Q = W \). This means all the heat added to the system is used to do work, such as expanding the gas in a tank.
Understanding this law is crucial for solving problems in thermodynamics, as it helps us balance energy inputs and outputs in various processes.
Pressure Conversion
Pressure conversion is often required because pressure can be measured in different units, such as atmospheres (atm), bars, or Pascals (Pa). In many thermodynamic calculations, it is crucial to convert these units to maintain consistency.
For example, 1 bar is equal to \(10^5 \) Pascals (Pa). To convert from bars to Pascals, you can multiply the value in bars by \(10^5 \).
In the given problem:
  • The pressure of 0.7 bar is converted to Pascals as \[ 0.7 \text{ bar} = 0.7 \times 10^5 \text{ Pa} = 70000 \text{ Pa} \]
  • Similarly, 1 bar is converted to Pascals as \[ 1 \text{ bar} = 1 \times 10^5 \text{ Pa} = 100000 \text{ Pa} \]
These conversions are essential for correctly applying formulas like the ideal gas law where the pressure must be in consistent units.

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Most popular questions from this chapter

Ammonia enters a heat exchanger operating at steady state as a superheated vapor at 14 bar, \(60^{\circ} \mathrm{C}\), where it is cooled and condensed to saturated liquid at 14 bar. The mass flow rate of the refrigerant is \(450 \mathrm{~kg} / \mathrm{h}\). A separate stream of air enters the heat exchanger at \(17^{\circ} \mathrm{C}, 1\) bar and exits at \(42^{\circ} \mathrm{C}, 1\) bar. Ignoring heat transfer from the outside of the heat exchanger and neglecting kinetic and potential energy effects, determine the mass flow rate of the air, in \(\mathrm{kg} / \mathrm{min}\).

A well-insulated rigid tank of volume \(10 \mathrm{~m}^{3}\) is connected to a large steam line through which steam flows at 15 bar and \(280^{\circ} \mathrm{C}\). The tank is initially evacuated. Steam is allowed to flow into the tank until the pressure inside is \(p\). (a) Determine the amount of mass in the tank, in \(\mathrm{kg}\), and the temperature in the tank, in \({ }^{\circ} \mathrm{C}\), when \(p=15\) bar. (b) Plot the quantities of part (a) versus \(p\) ranging from \(0.1\) to 15 bar.

Air expands through a turbine from 10 bar, \(900 \mathrm{~K}\) to 1 bar, \(500 \mathrm{~K}\). The inlet velocity is small compared to the exit velocity of \(100 \mathrm{~m} / \mathrm{s}\). The turbine operates at steady state and develops a power output of \(3200 \mathrm{~kW}\). Heat transfer between the turbine and its surroundings and potential energy effects are negligible. Calculate the mass flow rate of air, in \(\mathrm{kg} / \mathrm{s}\), and the exit area, in \(\mathrm{m}^{2}\).

The electronic components of a computer consume \(0.1 \mathrm{~kW}\), of electrical power. To prevent overheating, cooling air is supplied by a 25-W fan mounted at the inlet of the electronics enclosure. At steady state, air enters the fan at \(20^{\circ} \mathrm{C}, 1\) bar and exits the electronics enclosure at \(35^{\circ} \mathrm{C}\). There is no significant energy transfer by heat from the outer surface of the enclosure to the surroundings and the effects of kinetic and potential energy can be ignored. Determine the volumetric flow rate of the entering air, in \(\mathrm{m}^{3} / \mathrm{s}\).

Steam at 160 bar, \(480^{\circ} \mathrm{C}\), enters a turbine operating at steady state with a volumetric flow rate of \(800 \mathrm{~m}^{3} / \mathrm{min}\). Eighteen percent of the entering mass flow exits at 5 bar, \(240^{\circ} \mathrm{C}\), with a velocity of \(25 \mathrm{~m} / \mathrm{s}\). The rest exits at another location with a pressure of \(0.06\) bar, a quality of \(94 \%\), and a velocity of \(400 \mathrm{~m} / \mathrm{s}\). Determine the diameters of each exit duct, in \(\mathrm{m}\).
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