/*! 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 110 A tank whose volume is \(0.01 \m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A tank whose volume is \(0.01 \mathrm{~m}^{3}\) is initially evacuated. A pinhole develops in the wall, and air from the surroundings at \(21^{\circ} \mathrm{C}, 1\) bar enters until the pressure in the tank is 1 bar. If the final temperature of the air in the tank is \(21^{\circ} \mathrm{C}\), determine (a) the final mass in the tank, in \(g\), and (b) the heat transfer between the tank contents and the surroundings, in kJ.

Short Answer

Expert verified
Final mass of air: around 0.41 gHeat transfer: 0 kJ (isothermal) as no internal energy change.

Step by step solution

01

Understand the Given Data

Identify and write down all the given data. Volume of the tank: \(V = 0.01 \text{ m}^3\)Initial pressure of the tank: \(P_1 = 0 \text{ bar}\) (evacuated)Final pressure of the tank: \(P_2 = 1 \text{ bar}\)Initial and final temperature: \(T = 21^\text{C} = 294.15 \text{ K}\)We need to determine:(a) The final mass of air in the tank in grams (b) The heat transferred in kJ.
02

Apply the Ideal Gas Law

Use the Ideal Gas Law to find the final mass in the tank.The Ideal Gas Law is given by:\[ PV = nRT \]where: - \( R \) is the gas constant (\( R = 0.0821 \text{ bar}\text{·}\text{L}\text{·}\text{K}^{-1}\text{·}\text{mol}^{-1}\) for convenience).Rearranging the equation for the number of moles (\( n \)):\[ n = \frac{PV}{RT} \]
03

Calculate the Moles of Air in the Tank

Substitute the known values into the Ideal Gas Law to find the moles of air in the tank:\[ n = \frac{(1 \text{ bar})(0.01 \text{ m}^3)}{(0.0821 \text{ bar}\text{·}\text{L}\text{·}\text{K}^{-1}\text{·}\text{mol}^{-1})(294.15 \text{ K})} \]Convert \( 0.01 \text{ m}^3 \) to \( 10 \text{ L} \):\[ n = \frac{(1 \text{ bar})(10 \text{ L})}{(0.0821 \text{ bar}\text{·}\text{L}\text{·}\text{K}^{-1}\text{·}\text{mol}^{-1})(294.15 \text{ K})} \]Calculate the value of \( n \).
04

Convert Moles to Mass

We know that the molar mass of air is approximately \( 28.97 \text{ g/mol} \). Use this to convert moles to grams:\[ m = n \times \text{molar mass} \]
05

Calculate the Heat Transfer

Considering the tank initially evacuated and assuming air behaves ideally, we use the First Law of Thermodynamics for a closed system where the change in internal energy (\( \text{dU} \)) equals the heat added (\( Q \)). Since the process is isothermal, we have:\[ \text{dU} = 0 = Q - W \]Then considering the work done on the gas, calculate \( W \), which is the work done by the gas as it enters the tank.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 chemistry and physics. It relates the pressure, volume, temperature, and number of moles of an ideal gas. Written as \[[ PV = nRT ]\], the law allows us to predict one state variable if the others are known. Here:
- \( P \) is the pressure in the tank
- \( V \) is the volume of the tank
- \( n \) is the number of moles of gas
- \( R \) is the ideal gas constant
- \( T \) is the temperature in Kelvin
For our problem, the final pressure, volume, and temperature are given. We can rearrange the law to solve for the number of moles, \( n \): \[ n = \frac{PV}{RT} \]. By substituting the known values, we calculate the number of moles of gas added to the tank through the pinhole.
Thermodynamics
Thermodynamics is the scientific study of energy, heat, and work. It includes various laws that describe how energy moves and changes in a system. In this problem, we are dealing with a fundamental part of thermodynamics: the behavior of gases under different conditions.
Key concepts:
- Energy can be transferred as heat or work
- Energy is conserved< br>The Ideal Gas Law is one of the tools used in thermodynamics to model the behavior of gases. Understanding how gases expand, compress, and transfer energy is crucial in many scientific and engineering applications. For this exercise, knowing thermodynamics helps us determine how much work was done by the gas when it entered the tank.
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 transferred or converted from one form to another. Written as \[ \Delta U = Q - W ]\], where:
- \( \Delta U \) is the change in internal energy
- \( Q \) is the heat added to the system
- \( W \) is the work done by the system
This law is particularly useful in analyzing closed systems where mass does not enter or leave.
In this problem, the air enters the tank, and because the process is isothermal, \( \Delta U = 0 \). Therefore, we have \( Q = W \), implying that the heat added to the tank is entirely converted into work done by the gas as it enters. Given the process conditions (isothermal), calculation steps are simplified since internal energy change is zero.
Isothermal Process
An isothermal process is when a system's temperature remains constant during gas expansion or compression. For ideal gases, this means that any heat added to the system will convert entirely into doing work.
Key features:
- Temperature (\( T \)) is constant
- Internal energy (\( \Delta U \)) does not change
- Ideal gases follow Part of thermodynamic transformations
In our problem, the temperature inside the tank remains at 21°C (294.15 K). Because it's isothermal, we do not have to worry about changes in internal energy. The heat added to the gas as it enters the tank is used entirely to do work. Thus, applying the First Law of Thermodynamics simplifies our calculations concerning the heat transfer.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Refrigerant \(134 \mathrm{a}\) enters an air conditioner compressor at 4 bar, \(20^{\circ} \mathrm{C}\), and is compressed at steady state to 12 bar, \(80^{\circ} \mathrm{C}\). The volumetric flow rate of the refrigerant entering is \(4 \mathrm{~m}^{3} /\) \(\min\). The work input to the compressor is \(60 \mathrm{~kJ}\) per \(\mathrm{kg}\) of refrigerant flowing. Neglecting kinetic and potential energy effects, determine the heat transfer rate, in \(\mathrm{kW}\).

A \(380-\mathrm{L}\) tank contains steam, initially at \(400^{\circ} \mathrm{C}, 3\) bar. A valve is opened, and steam flows out of the tank at a constant mass flow rate of \(0.005 \mathrm{~kg} / \mathrm{s}\). During steam removal, a heater maintains the temperature within the tank constant. Determine the time, in s, at which \(75 \%\) of the initial mass remains in the tank; also determine the specific volume, in \(\mathrm{m}^{3} / \mathrm{kg}\), and pressure, in bar, in the tank at that time.

Figure P4.8 shows a mixing tank initially containing \(2000 \mathrm{lb}\) of liquid water. The tank is fitted with two inlet pipes, one delivering hot water at a mass flow rate of \(0.8 \mathrm{lb} / \mathrm{s}\) and the other delivering cold water at a mass flow rate of \(1.2 \mathrm{lb} / \mathrm{s}\). Water exits through a single exit pipe at a mass flow rate of \(2.5 \mathrm{lb} / \mathrm{s}\). Determine the amount of water, in lb, in the tank after one hour.

Air, modeled as an ideal gas, is compressed at steady state from \(1 \mathrm{bar}, 300 \mathrm{~K}\), to \(5 \mathrm{bar}, 500 \mathrm{~K}\), with \(150 \mathrm{~kW}\) of power input. Heat transfer occurs at a rate of \(20 \mathrm{~kW}\) from the air to cooling water circulating in a water jacket enclosing the compressor. Neglecting kinetic and potential energy effects, determine the mass flow rate of the air, in \(\mathrm{kg} / \mathrm{s}\).

Air modeled as an ideal gas enters a combustion chamber at \(20 \mathrm{lbf} / \mathrm{in}^{2}\) and \(70^{\circ} \mathrm{F}\) through a rectangular duct, \(5 \mathrm{ft}\) by \(4 \mathrm{ft}\). If the mass flow rate of the air is \(830,000 \mathrm{lb} / \mathrm{h}\), determine the velocity, in \(\mathrm{ft} / \mathrm{s}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.