/*! 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 178 Pressure of an ideal gas is incr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Pressure of an ideal gas is increased by keeping temperature constant. What is the effect on kinetic energy of molecules? [UP SEE 2006] (a) Increase (b) Decrease (c) No change (d) Can't be determined

Short Answer

Expert verified
No change in kinetic energy (Option c).

Step by step solution

01

Understand the Ideal Gas Law

The ideal gas law is given by the equation \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature. The problem states that the temperature is constant.
02

Relate Pressure and Volume

Given that temperature \( T \) is constant, we can use the relation \( PV = nRT \). If pressure \( P \) increases with a constant temperature, to maintain equality in the equation, the volume \( V \) must decrease.
03

Analyze the Effect on Kinetic Energy

The kinetic energy of molecules for an ideal gas is associated with its temperature and is given by \( KE = \frac{3}{2} k T \), where \( k \) is Boltzmann's constant. Since temperature \( T \) is constant, the kinetic energy, which depends on temperature, also remains constant.

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.

Pressure and Volume Relationship
When dealing with ideal gases, an important relationship to understand is how pressure and volume interact. This principle is highlighted by Boyle's Law, which is part of the overarching ideal gas law. The ideal gas law is expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles of gas, \( R \) is the universal gas constant, and \( T \) is the temperature. Under the conditions of constant temperature (isothermal conditions), Boyle's Law provides us with a simple insight: if you increase the pressure exerted by a gas, its volume decreases, and vice versa.
  • This is because, at a constant temperature, the energy of the gas doesn't change, so a higher pressure must be compensated by a decrease in volume.
  • This inverse relationship is crucial when performing calculations involving gases and predicting how a gas will behave when subjected to different pressures.
This relationship helps us understand changes in a gas’s behavior, such as when a bicycle tire is pumped with more air, thereby increasing the pressure and subsequently reducing the volume occupied by the gas molecules inside the tire.
Kinetic Theory of Gases
The kinetic theory of gases is a fundamental scientific theory that explains the behavior of gases. It is based on the idea that gas molecules are in a constant state of random motion. Here are some key points of the kinetic theory:
  • Gases consist of many small particles (atoms or molecules) that are in constant, random motion.
  • The pressure of a gas results from collisions between gas molecules and the walls of their container.
  • The average kinetic energy of the gas molecules is proportional to the temperature of the gas in Kelvin.
This theory helps to provide a molecular-level interpretation of the macroscopic properties such as pressure and temperature. According to this theory, the ideal gas behavior arises when gases are at a high temperature and low pressure, conditions under which the interactions between molecules are negligible.
Temperature and Kinetic Energy
In the context of ideal gases, temperature is a measure of the average kinetic energy of gas molecules. The relationship is simple but fundamental: higher temperatures mean higher average kinetic energy, and vice versa. The kinetic energy for an ideal gas is derived from the equation \( KE = \frac{3}{2} k T \), where \( k \) is Boltzmann's constant, and \( T \) is the absolute temperature.
  • In this equation, temperature plays a pivotal role because it directly influences the kinetic energy of the gas molecules.
  • If the temperature remains constant, as in the problem described, the kinetic energy of the gas molecules does not change.
  • This explains why, when pressure is increased through a decrease in volume, the kinetic energy of the gas does not change, as it depends solely on temperature.
Consequently, understanding this relationship helps to connect the dots between macroscopic observations (like an increase in pressure) and microscopic behavior (kinetic energy) in gases.

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

In an industrial process \(10 \mathrm{~kg}\) of water per hour is to be heated from \(20^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\). To do this steam at \(150^{\circ} \mathrm{C}\) is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at \(90^{\circ} \mathrm{C}\). How many kg of steam is required per hour? (Specific heat of steam \(=1\) calorie per \(\mathrm{g}^{\circ} \mathrm{C}\), Latent heat of vaporisation \(=540 \mathrm{cal} / \mathrm{g}\) ) (a) \(1 \mathrm{~g}\) (b) \(1 \mathrm{~kg}\) (c) \(10 \mathrm{~g}\) (d) \(10 \mathrm{~kg}\)

A wall has two layers \(A\) and \(B\), made of two different materials. The thermal conductivity of material \(A\) is twice that of \(B\). If the two layers have same thickness and under thermal equilibrium, the temperature difference across the wall is \(48^{\circ} \mathrm{C}\), the temperature difference across layer \(B\) is (a) \(40^{\circ} \mathrm{C}\) (b) \(32^{\circ} \mathrm{C}\) (c) \(16^{\circ} \mathrm{C}\) (d) \(24^{\circ} \mathrm{C}\)

The average energy and the rms speed of molecules in a sample of oxygen gas at \(400 \mathrm{~K}\) are \(7.21 \times 10^{-21} \mathrm{~J}\) and \(524 \mathrm{~ms}^{-1}\) respectively. The corresponding values at \(800 \mathrm{~K}\) are nearly (a) \(14.42 \times 10^{-21} 3,1048 \mathrm{~ms}^{-1}\) (b) \(10.18 \times 10^{-21} J, 741 \mathrm{~ms}^{-1}\) (c) \(7.21 \times 10^{-21} \mathrm{~J}, 1048 \mathrm{~ms}^{-1}\) (d) \(14.42 \times 10^{-21} \mathrm{~J}, 741 \mathrm{~ms}^{-1}\)

A child running at a temperature of \(101^{\circ} \mathrm{F}\) is given an antipyrin (i.e., medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to \(98^{\circ} \mathrm{F}\) in \(20 \mathrm{~min}\), what is the average rate of extra evaporation caused by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of child is \(30 \mathrm{~kg}\). The specific heat of the human body is approximately the same as that of water and latent heat of evaporation of water at that temperature is about \(580 \mathrm{cal} / \mathrm{g}\). (a) \(4.31 \mathrm{~g} / \mathrm{min}\) (b) \(4.31 \mathrm{~g} / \mathrm{s}\) (c) \(2.31 \mathrm{~g} / \mathrm{min}\) (d) \(2.31 \mathrm{~g} / \mathrm{s}\)

The average translatory energy and rms speed of molecules in a sample of oxygen gas at \(300 \mathrm{~K}\) are \(5.21 \times 10^{-21} \mathrm{~J}\) and \(484 \mathrm{~ms}^{-1}\) respectively. The sorresponding values at \(600 \mathrm{~K}\) are nearly (assuming deal gas behaviour) (a) \(\left.12.42 \times 10^{-21}\right\rfloor, 968 \mathrm{~ms}^{-1}\) (b) \(7.78 \times 10^{-21} \mathrm{~J}, 684 \mathrm{~ms}^{-1}\) (c) \(6.21 \times 10^{-21} \mathrm{~J}, 968 \mathrm{~ms}^{-1}\) (d) \(12.42 \times 10^{-21} \mathrm{~J}, 684 \mathrm{~ms}^{-1}\)

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.