/*! 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 16 During an isothermal compression... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

During an isothermal compression of an ideal gas, 410 J of heat must be removed from the gas to maintain constant temperature. How much work is done by the gas during the process?

Short Answer

Expert verified
410 J of work is done by the gas.

Step by step solution

01

Understand the Process

The process given is an isothermal compression of an ideal gas which means the temperature remains constant throughout the process. According to the first law of thermodynamics, the internal energy change \((\Delta U)\) in an isothermal process is zero since the temperature is constant for an ideal gas.
02

First Law of Thermodynamics for Isothermal Processes

For an isothermal process, the first law of thermodynamics is given by: \[ \Delta U = Q - W = 0 \]where \(Q\) is the heat added to the system, \(W\) is the work done by the system. Given \( \Delta U = 0\), we can say \[ Q = W \].
03

Substitute the Known Values

We are given that 410 J of heat is removed from the system, which means \( Q = -410 \text{ J} \). Since \( Q = W \), we have: \[ W = -410 \text{ J} \].
04

Interpret the Work Done

The negative sign indicates that the work is done on the gas (since it is being compressed), rather than by the gas. Thus, the magnitude of work done by the gas is 410 J.

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
An ideal gas is a theoretical concept that simplifies the study of real gases. It assumes that gas molecules:
  • Are in constant, random motion
  • Do not interact with each other (no forces of attraction or repulsion)
  • Occupy negligible volume compared to the total volume of the gas
These assumptions are crucial in simplifying calculations, especially when dealing with the laws of thermodynamics.

Ideal gases obey the Ideal Gas Law, represented 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
  • \( T \) is the temperature in Kelvin
This equation ties together the variables affecting gases under various conditions. While no real gases are ideally perfect, many behave enough like an ideal gas under certain conditions.
First Law of Thermodynamics
The First Law of Thermodynamics is essentially the principle of conservation of energy, applied to thermodynamic systems. It states that energy cannot be created or destroyed, only transformed from one form to another. Formally, it is expressed as:\[ \Delta U = Q - W \]Here:
  • \( \Delta U \) represents the change in the internal energy of a system
  • \( Q \) is the heat added to the system
  • \( W \) is the work done by the system
In an isothermal process involving an ideal gas, the first law simplifies significantly as the internal energy change, \( \Delta U \), is zero when temperature remains constant. The equation thus becomes:\[ Q = W \]This indicates that any heat added to the system is used entirely to do work. In our exercise, the heat removed results in work done on the gas, underlining energy conservation.
Work Done in Thermodynamics
In thermodynamics, work done by or on a system during a process is a core concept. It is often represented as:\[ W = \int P \cdot dV \]where:
  • \( W \) is the work done
  • \( P \) is the pressure
  • \( dV \) is the change in volume
During an isothermal compression of an ideal gas, as in our exercise, a specific amount of heat must be removed to maintain constant temperature. For an isothermal process within the boundaries of the Ideal Gas Law, we often see that the work done is equal in magnitude but opposite in direction to the heat flow.

In the given exercise, 410 J of heat are removed, indicating that the same amount of work, 410 J, is done on the gas. The negative sign in calculations reflects this direction, affirming that work is being done on the gas, not by the gas. Understanding this helps in comprehending energy flow in thermodynamic 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

During an adiabatic expansion the temperature of 0.450 mol of argon (Ar) drops from 66.0\(^\circ\)C to 10.0\(^\circ\)C. The argon may be treated as an ideal gas. (a) Draw a \(pV\)-diagram for this process. (b) How much work does the gas do? (c) What is the change in internal energy of the gas?

When water is boiled at a pressure of 2.00 atm, the heat of vaporization is 2.20 \(\times\) 10\(^6\) J/kg and the boiling point is 120\(^\circ\)C. At this pressure, 1.00 kg of water has a volume of 1.00 \(\times\) 10\(^{-3}\) m\(^3\), and 1.00 kg of steam has a volume of 0.824 m\(^3\). (a) Compute the work done when 1.00 kg of steam is formed at this temperature. (b) Compute the increase in internal energy of the water.

In another test, the valve of a 500-L cylinder full of the gas mixture at 2000 psi (gauge pressure) is opened wide so that the gas rushes out of the cylinder very rapidly. Why might some \(N_2O\) condense during this process? (a) This is an isochoric process in which the pressure decreases, so the temperature also decreases. (b) Because of the rapid expansion, heat is removed from the system, so the internal energy and temperature of the gas decrease. (c) This is an isobaric process, so as the volume increases, the temperature decreases proportionally. (d) With the rapid expansion, the expanding gas does work with no heat input, so the internal energy and temperature of the gas decrease.

Starting with 2.50 mol of N\(_2\) gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0\(^\circ\)C, a chemist first heats the gas at constant volume, adding 1.36 \(\times\) 10\(^4\) J of heat, then continues heating and allows the gas to expand at constant pressure to twice its original volume. Calculate (a) the final temperature of the gas; (b) the amount of work done by the gas; (c) the amount of heat added to the gas while it was expanding; (d) the change in internal energy of the gas for the whole process.

A large research balloon containing \(2.00 \times 10^{3} \mathrm{~m}^{3}\) of helium gas at 1.00 atm and a temperature of \(15.0^{\circ} \mathrm{C}\) rises rapidly from ground level to an altitude at which the atmospheric pressure is only 0.900 atm (Fig. \(\mathbf{P} 19.50\) ). Assume the helium behaves like an ideal gas and the balloon's ascent is too rapid to permit much heat exchange with the surrounding air. (a) Calculate the volume of the gas at the higher altitude. (b) Calculate the temperature of the gas at the higher altitude. (c) What is the change in internal energy of the helium as the balloon rises to the higher altitude?

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.