/*! 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 5 If \(1500 \mathrm{lb}_{w}\) of a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 5

If \(1500 \mathrm{lb}_{w}\) of air is pumped into a previously empty \(900 \mathrm{ft}^{3}\) storage tank and the air temperature in the tank is uniformly \(70^{\circ} \mathrm{F}\), what is the air pressure in the tank in atmospheres?

Short Answer

Expert verified
The air pressure in the tank is approximately 1.59 atm.

Step by step solution

01

Identify Given Values

We have been given the mass of air as 1500 lb_w (where lb_w represents weight in pounds), the volume of the tank as 900 ft鲁, and the uniform temperature of air as 70掳F.
02

Convert Temperature to Kelvin

First, we need to convert the temperature from Fahrenheit to Kelvin. Using the formula \[ T(K) = \frac{5}{9}(T(掳F) - 32) + 273.15 \]we have \[ T(K) = \frac{5}{9}(70 - 32) + 273.15 = 294.26 \] Kelvin.
03

Calculate the Mass in Slugs

Since 1 lb_w = 1/32.2 slugs (where slugs are the unit of mass), the mass of air is \[ \text{mass (slugs)} = \frac{1500}{32.2} \approx 46.58 \text{ slugs} \]
04

Use the Ideal Gas Law for Pressure

The Ideal Gas Law is given by: \[ PV = nRT \]where:- \( P \) is the pressure in atm,- \( V \) is the volume in ft鲁,- \( n \) is the number of moles,- \( R \) is the ideal gas constant,- \( T \) is the temperature in Kelvin.We note that for 1 slug of air, 1 mole of air is about 1 slug = 29.27 lb (using the molar mass of air of 29.27 lb/lbm), meaning the number of moles \( n \) can be approximated by the mass in slugs divided by the molar mass of air:\[ n \approx \frac{46.58 \text{ slugs}}{29.27} \approx 1.59 \text{ moles} \]
05

Determine Pressure Constant

The ideal gas constant \( R \) in lb ft/(slug R) is 1716. To convert this constant to atmospheres, use the conversion \( 1 \text{ atm} = 2116.22 \text{ lb/ft}^2 \). Therefore, we have:\[ R = \frac{0.0821 \times 298.15 \times 0.0375}{900} \]We utilize the relation:\[ P = \frac{nRT}{V} \approx \frac{1.59 \times 1716 \times 294.26}{900} \]So the pressure \( P \approx 1.59 \text{ atm} \).
06

Solving for Pressure

Simplifying the expression, the air pressure in the tank in atm can be calculated as:\[ P \approx 1.59 \text{ atm} \].

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.

Thermodynamics and Ideal Gas Law
Thermodynamics is a branch of physics that helps us understand how heat and energy interact with each other. One of its critical applications is the Ideal Gas Law, which connects how gases behave under different conditions. This law can be expressed using the formula \( PV = nRT \), which connects four essential variables:
  • \( P \) - Pressure: It indicates how much force the gas exerts on the walls of its container. Usually measured in atmospheres (atm).
  • \( V \) - Volume: This represents the space the gas occupies, commonly measured in cubic feet (ft鲁).
  • \( n \) - Moles: The quantity of gas, which depends on the amount of substance present. Calculated in moles.
  • \( R \) - Gas Constant: This is a constant that makes sure the equation balances out through changes in units or conditions.
  • \( T \) - Temperature: Expressed in Kelvin to standardize temperature measurements.
These variables are key to calculating the behavior of gases, such as the pressure in a storage tank when we know the volume, temperature, and quantity of gas inside. Remember, the Ideal Gas Law is an approximation that works well for many gases under normal conditions.
Gas Pressure Calculation
Calculating gas pressure is crucial in various applications like understanding weather changes or designing engines. In this particular exercise, we calculate the pressure using the Ideal Gas Law. The procedure involves:
  • Determine Moles: Use the mass of the gas in slugs. Convert the given weight from pounds of air into slugs. Compute the number of moles using this mass, by dividing each mass in slugs by the molar mass of air.
  • Apply the Formula: Insert the known values into the Ideal Gas Law. Ensure that the calculations are consistent with the units, such as using Kelvin for temperature and slugs for mass to keep the results in line with the constant \( R \).
  • Convert to Atmospheres: Finally, solve for the pressure \( P \). It鈥檚 important to consider the conversion of pressure into the standard units of atmospheres to make results more comprehensible.
By setting up the Ideal Gas Law equation properly and understanding each component, we can accurately solve for pressure, providing insight into the conditions within the tank.
Temperature Conversion in Thermodynamics
A key step in solving thermodynamic problems is converting temperature into Kelvin, which is the standard unit in physics. This conversion ensures consistency across various scientific equations.

Steps for Conversion

To convert Fahrenheit to Kelvin, you use the formula:
  • First, subtract 32 from the Fahrenheit temperature.
  • Multiply the result by \( \frac{5}{9} \) to convert to Celsius.
  • Add 273.15 to convert from Celsius to Kelvin.
In our example, converting 70掳F to Kelvin is done by:
  • 70 - 32 = 38
  • 38 脳 \( \frac{5}{9} \) = 21.11掳C
  • 21.11 + 273.15 = 294.26 K
Using Kelvin allows us to eliminate negative temperature values, which aren鈥檛 feasible in physics calculations, thereby simplifying and stabilizing the equation solving process. Conversion is straightforward when you follow these step-by-step operations, ensuring you can consistently utilize thermodynamic equations like the Ideal Gas Law.

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

At a point in the test section of a supersonic wind tunnel, the air pressure and temperature are \(0.5 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\) and \(240 \mathrm{~K}\), respectively. Calculate the specific volume.

In a gas turbine jet engine, the pressure of the incoming air is increased by flowing through a compressor, the air then enters a combustor that looks vaguely like a long can (sometimes called the combustion can). Fuel is injected in to the combustor and burns with the air, and then the burned fuel- air mixture exits the combustor at a higher temperature than the air coming into the combustor. (Gas turbine jet engines are discussed in Ch. 9.) The pressure of the flow through the combustor remains relatively constant; that is, the combustion process is at constant pressure. Consider the case where the gas pressure and temperature entering the combustor are \(4 \times 10^{6} \mathrm{~N} / \mathrm{m}^{2}\) and \(900 \mathrm{~K}\), respectively, and the gas temperature exiting the combustor is \(1500 \mathrm{~K}\). Calculate the gas density at \((a)\) the inlet to the combustor and \((b)\) the exit of the combustor. Assume that the specific gas constant for the fuel-air mixture is the same as that for pure air.

The X-15 (see Fig. 5.92) was a rocketpowered research airplane designed to probe the mysteries of hypersonic flight. In 2014 , the X-15 still holds the records for the fastest and highest flying piloted airplane (the Space Shuttle and Spaceship One, in this context, are space ships, not airplanes). On August 22, 1963, pilot Joseph Walker set the unofficial world altitude record of 354,200 feet. On October 3 , 1967 , pilot William J. Knight set the world speed record of \(4520 \mathrm{mph}\) (Mach \(6.7\) ) (a) Convert Walker's altitude record to meters and kilometers. (b) Convert Knight's speed record to meters per second.

The velocity of the Space Shuttle (Fig. 2.24) at the instant of burnout of the rocket booster is \(26,000 \mathrm{ft} / \mathrm{sec}\). What is this velocity in \(\mathrm{km} / \mathrm{sec}\) ?

The X-15 is air-launched from under the wing of a B-52 mother ship. Immediately after launch, the pilot starts the XLR-99 rocket engine, which provides 57,000 Ib of thrust. For the first moments, the X-15 accelerates in horizontal flight. The gross weight of the airplane at the start is \(34,000 \mathrm{lb}\). Calculate the initial acceleration of the airplane.
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Translational Dynamics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Torque and Rotational Motion

Read Explanation

Famous Physicists

Read Explanation

Physical Quantities And Units

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.