/*! 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 40 A monatomic ideal gas expands at... [FREE SOLUTION] | 91Ó°ÊÓ

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A monatomic ideal gas expands at constant pressure. (a) What percentage of the heat being supplied to the gas is used to increase the internal energy of the gas? (b) What percentage is used for doing the work of expansion?

Short Answer

Expert verified
60% is used to increase internal energy, and 40% is used for expansion work.

Step by step solution

01

Understand the Process

When a monatomic ideal gas expands at constant pressure, the heat added to the system, \( Q \), causes an increase in the internal energy, \( \Delta U \), and does work on the surroundings, \( W \). According to the first law of thermodynamics: \( Q = \Delta U + W \).
02

Equation for Internal Energy Change

For a monatomic ideal gas, the change in internal energy \( \Delta U \) is given by: \( \Delta U = \frac{3}{2}nR\Delta T \), where \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( \Delta T \) is the change in temperature.
03

Equation for Work Done

The work done by the gas when it expands at constant pressure \( P \) is given by: \( W = P\Delta V = nR\Delta T \).
04

Calculate the Total Heat Supplied

The total heat supplied, using the first law of thermodynamics, becomes: \( Q = \Delta U + W = \frac{3}{2}nR\Delta T + nR\Delta T = \frac{5}{2}nR\Delta T \).
05

Percentage for Internal Energy Increase

The percentage of heat used to increase the internal energy is: \( \frac{\Delta U}{Q} \times 100\% = \frac{\frac{3}{2}nR\Delta T}{\frac{5}{2}nR\Delta T} \times 100\% = 60\% \).
06

Percentage for Work of Expansion

The percentage of heat used to do work is: \( \frac{W}{Q} \times 100\% = \frac{nR\Delta T}{\frac{5}{2}nR\Delta T} \times 100\% = 40\% \).

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Key Concepts

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

Ideal Gas
An ideal gas is a theoretical model that helps us understand how gases behave under different conditions. In this model, gas particles are assumed to be perfectly elastic and of negligible size, meaning they don't interact with each other. This makes the math simpler and more manageable. The properties of an ideal gas are controlled by three variables: pressure (P), volume (V), and temperature (T). These variables are related through the Ideal Gas Law, expressed as \( PV = nRT \), where \( n \) is the number of moles of the gas and \( R \) is the ideal gas constant. The Ideal Gas Law helps to predict how a gas will respond to changes in pressure, volume, or temperature. This is essential in thermodynamic calculations, like in the original exercise with a monatomic ideal gas.
Constant Pressure Process
A constant pressure process, often referred to as isobaric, is where the pressure stays the same even though other properties of the gas may change. In these processes, the volume and temperature of an ideal gas will typically change to accommodate the constant pressure. This is important because, in an isobaric process, the work done by or on the system can be calculated simply. The work done by the gas can be given as \( W = P\Delta V \), where \( \Delta V \) is the change in volume.
This way, the constant pressure process allows one to simplify the calculations as it follows a straightforward formula. The original exercise involved a monatomic ideal gas expanding at a constant pressure, which required understanding the changes in internal energy and work done.
First Law of Thermodynamics
The first law of thermodynamics is essentially a version of the law of conservation of energy applied to thermodynamics. It states that the energy added to a system, \( Q \), is equal to the change in internal energy, \( \Delta U \), plus the work done by the system on its surroundings, \( W \). Mathematically, this is represented as \( Q = \Delta U + W \).
This fundamental concept serves as the backbone of the original exercise. It helps explain how energy distributes in the form of heat and work within a thermodynamic process. In the context of the exercise, it allows the determination of what fraction of heat contributes to the internal energy of the gas and what fraction is used for work during a constant pressure expansion. By understanding and applying the first law, students can calculate the proportions of energy conversion accurately.

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Most popular questions from this chapter

Two Carnot air conditioners, \(A\) and \(B,\) are removing heat from different rooms. The outside temperature is the same for both rooms, 309.0 \(\mathrm{K}\) . The room serviced by unit A is kept at a temperature of 294.0 \(\mathrm{K}\) , while the room serviced by unit \(\mathrm{B}\) is kept at 301.0 \(\mathrm{K}\) . The heat removed from either room is 4330 \(\mathrm{J}\) J. For both units, find the magnitude of the work required and the magnitude of the heat deposited outside.

See Multiple-Concept Example 10 to review the concepts that are important in this problem. The water in a deep underground well is used as the cold reservoir of a Carnot heat pump that maintains the temperature of a house at 301 \(\mathrm{K}\) . To deposit 14200 \(\mathrm{J}\) of heat in the house, the heat pump requires 800 \(\mathrm{J}\) of work. Determine the temperature of the well water.

A Carnot engine operates with an efficiency of 27.0\(\%\) when the temperature of its cold reservoir is 275 \(\mathrm{K}\) . Assuming that the temperature of the hot reservoir remains the same, what must be the temperature of the cold reservoir in order to increase the efficiency to 32.0\(\%\) ?

One-half mole of a monatomic ideal gas expands adiabatically and does 610 \(\mathrm{J}\) of work. By how many kelvins does its temperature change? Specify whether the change is an increase or a decrease.

Heat \(Q\) flows spontaneously from a reservoir at 394 \(\mathrm{K}\) into a reservoir at 298 \(\mathrm{K}\) . Because of the spontaneous flow, 2800 \(\mathrm{J}\) of energy is rendered unavailable for work when a Carnot engine operates between the reservoir at 298 \(\mathrm{K}\) and a reservoir at 248 \(\mathrm{K}\) . Find \(Q\) .

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