/*! 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 128 Nitrogen gas is contained in a r... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 128

Nitrogen gas is contained in a rigid \(1-\mathrm{m}\) tank, initially at 10 bar, \(300 \mathrm{~K}\). Heat transfer to the contents of the tank occurs until the temperature has increased to \(400 \mathrm{~K}\). During the process, a pressure-relief valve allows nitrogen to escape, maintaining constant pressure in the tank. Neglecting kinetic and potential energy effects, and using the ideal gas model with constant specific heats evaluated at \(350 \mathrm{~K}\), determine the mass of nitrogen that escapes, in \(\mathrm{kg}\), and the amount of energy transfer by heat, in \(\mathrm{kJ}\).

Short Answer

Expert verified
79.24 kg of nitrogen escapes, and 17674.73 kJ of energy is transferred by heat.

Step by step solution

01

Determine Initial Mass of Nitrogen

Use the ideal gas law to find the initial mass of nitrogen. The ideal gas law is given by \[ PV = nRT \] where \( P = 10 \text{ bar} = 10^6 \text{ Pa} \) \( V = 1 \text{ m}^3 \) \( R = 8314 \text{ J/(kmol路K)} / 28 \text{ kg/kmol} = 296.2 \text{ J/(kg路K)} \) and \( T = 300 \text{ K} \) . First, find the number of moles, \( n \) , using \[ n = \dfrac{PV}{RT} \] Then convert moles to mass, \( m \) , by using \( m = nM \) where \( M \) is the molar mass of nitrogen. Simplified: \[ m_1 = \dfrac{PV}{RT} \times M = \dfrac{(10^6 \text{ Pa})(1 \text{ m}^3)}{(296.2 \text{ J/(kg路K)})(300 \text{ K})} \times 28 \text{ kg/kmol} = 316.96 \text{ kg} \]
02

Determine Final Mass of Nitrogen

Since the pressure-relief valve maintains constant pressure, use the final temperature to determine the final mass of nitrogen in the tank using the formula \[ m_2 = \dfrac{PV}{RT} \times M = \dfrac{(10^6 \text{ Pa})(1 \text{ m}^3)}{(296.2 \text{ J/(kg路K)})(400 \text{ K})} \times 28 \text{ kg/kmol} = 237.72 \text{ kg} \]
03

Calculate Mass of Nitrogen that Escapes

Calculate the difference in mass before and after heating to get the mass of nitrogen that escapes: \[ m_{escaped} = m_1 - m_2 = 316.96 \text{ kg} - 237.72 \text{ kg} = 79.24 \text{ kg} \]
04

Determine Change in Internal Energy

Using the specific heat at constant volume, \( c_v \), calculate the change in internal energy. For nitrogen, \( c_v = 0.744 \text{ kJ/(kg路K)} \): \[ \Delta U = m_2 c_v \Delta T = (237.72 \text{ kg})(0.744 \text{ kJ/(kg路K)})(400 \text{ K} - 300 \text{ K}) = 17674.73 \text{ kJ} \]
05

Determine Energy Transfer by Heat

Since energy transfer by heat can be calculated as the change in internal energy plus work done (\[ Q = \Delta U + W \]), with constant pressure, the work done is \( W = P \Delta V = 0 \text { (since volume is constant)} \) , thus: \[ Q = \Delta U = 17674.73 \text{ kJ} \]

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 91影视!

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 describing the behavior of an ideal gas. It is represented as


PV = nRT

In this equation:
  • P is the pressure of the gas, measured in Pascals (Pa)
  • V is the volume of the gas, measured in cubic meters (m鲁)
  • n is the number of moles of the gas
  • R is the universal gas constant, equal to 8.314 J/(mol路K)
  • T is the temperature of the gas, measured in Kelvin (K)

The ideal gas law establishes a relationship between these variables, allowing us to calculate one if the others are known. For example, in the given problem, we used this law to determine the initial mass of nitrogen contained within a rigid tank.
Specific Heats
Specific heats are properties of substances that determine how much energy is needed to raise the temperature of a given amount of that substance. Two specific heats are important in thermodynamics:
  • c_v: The specific heat at constant volume, which indicates how much energy is needed to raise the temperature at constant volume.
  • c_p: The specific heat at constant pressure, which indicates how much energy is needed to raise the temperature at constant pressure.

For nitrogen, we used a specific heat, c_v, of 0.744 kJ/(kg路K), to calculate the change in internal energy during the process. This value changes depending on the temperature, but we assumed it to be constant at 350 K in this problem.
Energy Transfer by Heat
Energy transfer by heat is the process of energy moving from one system to another due to a temperature difference. In thermodynamics, it is symbolized by Q. In our problem, we needed to determine the amount of heat transferred to the nitrogen gas as it was heated from 300 K to 400 K.

The formula for energy transfer by heat, when the volume is constant and no work is done, is:

Q = 螖U

Here, 螖U is the change in internal energy. Therefore, by calculating the change in internal energy of the nitrogen gas, we directly obtained the energy transfer by heat.
Internal Energy Change
The internal energy of a gas is related to the kinetic energy of its molecules. When the temperature of a gas increases, its internal energy increases.

The change in internal energy, 螖U, for our nitrogen gas was calculated using the specific heat at constant volume:

螖U = m_2 * c_v * 螖T

Where:
  • m_2 is the final mass of nitrogen in the tank
  • c_v is the specific heat at constant volume
  • 螖T is the change in temperature

In our case, we found that the internal energy change provided a direct measure of the heat transferred to the nitrogen.
Pressure-Relief Valve
A pressure-relief valve is a safety device used in many systems to control or limit the pressure in a vessel that can build up during a process. In our problem:
  • The pressure-relief valve allowed nitrogen to escape the tank, maintaining a constant pressure of 10 bar throughout the heating process.
  • This ensured the pressure did not increase beyond the safety limit as the temperature rose from 300 K to 400 K.
  • By using the pressure-relief valve, we could simplify our calculations since the pressure remained constant even as some gas escaped.

Therefore, understanding the operation and implications of the pressure-relief valve was crucial for solving the problem.

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

A well-insulated rigid tank of volume \(15 \mathrm{~m}^{3}\) is connected to a large steam line through which steam flows at \(1 \mathrm{MPa}\) and \(320^{\circ} \mathrm{C}\). 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 \(p=500 \mathrm{kPa}\). (b) Plot the quantities of part (a) versus \(p\) ranging from \(0 \mathrm{kPa}\) to \(500 \mathrm{kPa}\).

Liquid propane enters an initially empty cylindrical storage tank at a mass flow rate of \(10 \mathrm{~kg} / \mathrm{s}\). Flow continues until the tank is filled with propane at \(20^{\circ} \mathrm{C}, 9\) bar. The tank is \(25 \mathrm{~m}\) long and has a \(4-\mathrm{m}\) diameter. Determine the time, in minutes, to fill the tank.

A rigid tank having a volume of \(0.1 \mathrm{~m}^{3}\) initially contains water as a two-phase liquid-vapor mixture at 1 bar and a quality of \(1 \%\). The water is heated in two stages: Stage 1: Constant-volume heating until the pressure is 20 bar. Stage 2: Continued heating while saturated water vapor is slowly withdrawn from the tank at a constant pressure of 20 bar. Heating ceases when all the water remaining in the tank is saturated vapor at 20 bar. For the water, evaluate the heat transfer, in kJ, for each stage of heating. Ignore kinetic and potential energy effects.

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 \(\mathrm{kW}\).

Propane vapor enters a valve at \(1.0 \mathrm{MPa}, 60^{\circ} \mathrm{C}\), and leaves at \(0.3 \mathrm{MPa}\). If the propane undergoes a throttling process, what is the temperature of the propane leaving the valve, in \({ }^{\circ} \mathrm{C}\) ?
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Physical Quantities And Units

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Physics of Motion

Read Explanation

Electrostatics

Read Explanation

Space Physics

Read Explanation
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.