Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains 0.110 m\(^3\) of air at a pressure of 0.355 atm. The piston is slowly pulled out until the volume of the gas is increased to 0.390 m\(^3\). If the temperature remains constant, what is the final value of the pressure?

Short Answer

Expert verified
The final pressure is 0.100 atm.

Step by step solution

01

Identify the Known Variables

First, note down the given information from the problem. The initial volume of the gas is \( V_1 = 0.110 \, \text{m}^3 \), and the initial pressure is \( P_1 = 0.355 \, \text{atm} \). The final volume of the gas once the piston is pulled out is \( V_2 = 0.390 \, \text{m}^3 \). The temperature is constant, so we are assuming an isothermal process.
02

Use Boyle’s Law

Since the temperature is constant, we use Boyle’s Law, which states that \( P_1 \times V_1 = P_2 \times V_2 \). This can be rearranged to find the final pressure: \( P_2 = \frac{P_1 \times V_1}{V_2} \).
03

Substitute the Known Values

Substitute the known values into the equation from Step 2: \( P_2 = \frac{0.355 \, \text{atm} \times 0.110 \, \text{m}^3}{0.390 \, \text{m}^3} \).
04

Calculate the Final Pressure

Perform the calculation: \( P_2 = \frac{0.03905 \, \text{atm} \cdot \text{m}^3}{0.390 \, \text{m}^3} = 0.1001282 \, \text{atm} \). Rounding the answer, the final pressure is \( 0.100 \, \text{atm} \).

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 Vaia!

Key Concepts

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

Isothermal Process
An isothermal process is a fundamental concept in thermodynamics where the temperature of the system remains constant. This is crucial because even though the pressure and volume might change, the energy related to temperature remains stable. To ensure this, heat must be exchanged with the environment. In our exercise scenario, the air inside the cylindrical tank undergoes an expansion by pulling the piston out. Since the temperature stays the same, energy that might have been gained or lost due to volume change is compensated by the exchange of heat with the surroundings. This keeps the internal energy of the gas constant, which is a key characteristic of an isothermal process.
Pressure-Volume Relationship
The pressure-volume relationship, described by Boyle's Law, is pivotal in understanding gas behaviors. Boyle's Law states that for a fixed amount of gas at constant temperature, the pressure of the gas is inversely proportional to its volume.In mathematical terms, it's expressed as \( P \times V = \, \text{constant} \), or \( P_1 \times V_1 = P_2 \times V_2 \). This equation illustrates that when you increase the volume (like pulling a piston out), the pressure must decrease, provided the temperature does not change.In our given problem, we had an increase in volume from 0.110 m³ to 0.390 m³. As a result, Boyle's Law helps us calculate and understand the resultant decrease in pressure from 0.355 atm to approximately 0.100 atm.
Ideal Gas Law
The ideal gas law is an umbrella equation in thermodynamics that brings together pressure, volume, and temperature for an ideal gas. It is represented as \( PV = nRT \), where:
  • \( P \) is the pressure of the gas,
  • \( V \) is the volume of the gas,
  • \( n \) is the amount of substance (in moles),
  • \( R \) is the ideal gas constant,
  • \( T \) is the temperature in Kelvin.
In the context of an isothermal process, changes in pressure and volume do not alter the product of \( n \), \( R \), and \( T \), as temperature remains constant. Although our problem specifically centers on Boyle's Law, it is a component of the broader ideal gas law, which helps us in partially understanding varied gas behaviors under different conditions. Understanding how these equations interconnect provides clarity into gas dynamics within closed systems.

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 20.0-L tank contains \(4.86 \times 10{^-}{^4}\) kg of helium at 18.0\(^\circ\)C. The molar mass of helium is 4.00 g/mol. (a) How many moles of helium are in the tank? (b) What is the pressure in the tank, in pascals and in atmospheres?

An empty cylindrical canister 1.50 m long and 90.0 cm in diameter is to be filled with pure oxygen at 22.0\(^\circ\)C to store in a space station. To hold as much gas as possible, the absolute pressure of the oxygen will be 21.0 atm. The molar mass of oxygen is 32.0 g/mol. (a) How many moles of oxygen does this canister hold? (b) For someone lifting this canister, by how many kilograms does this gas increase the mass to be lifted?

How many moles are in a 1.00-kg bottle of water? How many molecules? The molar mass of water is 18.0 g/mol.

A vertical cylindrical tank contains 1.80 mol of an ideal gas under a pressure of 0.300 atm at 20.0\(^\circ\)C. The round part of the tank has a radius of 10.0 cm, and the gas is supporting a piston that can move up and down in the cylinder without friction. There is a vacuum above the piston. (a) What is the mass of this piston? (b) How tall is the column of gas that is supporting the piston?

You have two identical containers, one containing gas \(A\) and the other gas \(B\). The masses of these molecules are \(m$$_A\) = 3.34 \(\times\) 10\({^-}{^2}{^7}\) kg and \(m$$_B\) = 5.34 \(\times\) 10\({^-}{^2}{^6}\) kg. Both gases are under the same pressure and are at 10.0\(^\circ\)C. (a) Which molecules (\(A\) or \(B\)) have greater translational kinetic energy per molecule and rms speeds? (b) Now you want to raise the temperature of only one of these containers so that both gases will have the same rms speed. For which gas should you raise the temperature? (c) At what temperature will you accomplish your goal? (d) Once you have accomplished your goal, which molecules (\(A\) or \(B\)) now have greater average translational kinetic energy per molecule?

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.

Sign-up for free