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Chapter 19: Problem 1

Two moles of an ideal gas are heated at constant pressure from \(T=27^{\circ} \mathrm{C}\) to \(T=107^{\circ} \mathrm{C}\) (a) Draw a \(p V\) -diagram for this process. (b) Calculate the work done by the gas.

Short Answer

Expert verified
The work done by the gas is 1326.24 Joules.

Step by step solution

01

Understand the Problem

This exercise deals with an ideal gas that is heated at constant pressure. We need to analyze the changes it undergoes and calculate the work done during these changes given the initial and final temperatures of the gas.
02

Set Up the Initial and Final States

Convert the given temperatures from Celsius to Kelvin for accurate calculations involving gases. The initial temperature \(T_i\) is \(27^{\circ}C = 300K\), and the final temperature \(T_f\) is \(107^{\circ}C = 380K\).
03

Plot the PV Diagram

In a \(pV\) diagram for a constant pressure process, the line is horizontal as pressure remains constant. Label the initial point (state 1) with \(V_1\) and the final point (state 2) with \(V_2\) on this horizontal line, where \(V_2 > V_1\) due to the increase in temperature.
04

Recall the Ideal Gas Law

The ideal gas law equation is \(pV = nRT\), where \(p\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin. Since pressure \(p\) is constant, the volume \(V\) is proportional to temperature \(T\).
05

Calculate Work Done by the Gas

The work done by the gas at constant pressure is given by \(W = p\Delta V\). However, we can also express this as \(W = nR(T_f - T_i)\) since \(\Delta V = \frac{nR(T_f - T_i)}{p}\). Substitute \(n = 2\ \mathrm{moles}, R = 8.314\ \mathrm{J/(mol\cdot K)}, T_f = 380\ \mathrm{K}, T_i = 300\ \mathrm{K}\) to calculate \(W = 2 \times 8.314 \times (380 - 300)\).
06

Perform the Calculation

Calculate the value: \[W = 2 \times 8.314 \times 80 = 1326.24\ \mathrm{J}\]. Thus, the work done by the gas is 1326.24 Joules.

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Key Concepts

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

Constant Pressure Process
When dealing with a constant pressure process, imagine you are inflating a balloon. The external pressure of the balloon remains the same as you add air. Similarly, in our exercise, the gas undergoes heating at a constant pressure. This is also known as an isobaric process. During this, although pressure remains unchanged, other variables like volume and temperature do change.

Some key points about a constant pressure process include:
  • Pressure remains the same throughout the process.
  • As the gas is heated, its volume will increase if it is not confined.
  • This is important in understanding how gases behave under varied conditions while pressure stays the same.
Since pressure is constant, according to Boyle's Law, any increase in temperature will lead to an increase in volume. Thus, the relationship between volume and temperature becomes direct.
Work Done by Gas
Work done by gas can be visualized as the energy transferred from the gas as it expands. In practical terms, think about pushing a piston in a car engine, where the gas pushes the piston out as it expands. The work done is essentially the amount of energy needed to move the piston鈥攖his is similar to what happens during the gas expansion.

For gases, specifically at constant pressure, the work done (W) by the gas can be calculated using:\[W = p \Delta V\]Or using the ideal gas law, reformed:\[W = nR(T_f - T_i)\]Where:
  • W is work done by the gas.
  • p is the constant pressure.
  • \Delta V is the change in volume.
  • n is the number of moles of the gas.
  • R is the universal gas constant (8.314 J/mol路K).
  • T_f and T_i are the final and initial temperatures in Kelvin, respectively.
This mathematical formulation helps in solving problems where thermal energy transfers result in mechanical work.
PV Diagram
The PV diagram, or pressure-volume diagram, is a graphical representation of the changes in pressure and volume in a system. It is a crucial tool in thermodynamics for visualizing how a gas undergoes different processes. In our specific exercise dealing with constant pressure, the PV diagram simplifies the visual understanding.

In a constant pressure process, the PV diagram will display a horizontal straight line. Here鈥檚 why this happens:
  • The horizontal axis represents volume (V), and the vertical axis represents pressure (P).
  • Since the pressure is constant, the line is horizontal instead of slanted or vertical.
  • The initial state of the gas starts at the left (lower volume), and as the gas heats, it expands, moving right (to a higher volume) along the line.
A PV diagram not only shows the initial and final states but also helps in calculating work done, as the area under the line is indicative of the work exerted by or on the system.

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Most popular questions from this chapter

Chinook. During certain seasons strong winds called chinooks blow from the west across the eastern slopes of the Rockies and downhill into Denver and nearby areas. Although the mountains are cool, the wind in Denver is very hot; within a few minutes after the chinook wind arrives, the temperature can climb 20 \(\mathrm{C}^{\circ}\) ("chinook" is a Native American word meaning "snow cater"). Similar winds occur in the Alps (called fochns) and in southern California (called Santa Anas). (a) Explain why the temperature of the chinook wind rises as it descends the slopes. Why is it important that the wind be fast moving? (b) Suppose a strong wind is blowing toward Denver (elevation 1630 \(\mathrm{m} )\) from Grays Peak \((80 \mathrm{km} \text { west of Denver, at an elevation of } 4350 \mathrm{m})\) , where the air pressure is \(5.60 \times 10^{4} \mathrm{Pa}\) and the air temperature is \(-15.0 ^{\circ} \mathrm{C}\) . The temperature and pressure in Denver before the wind arrives are \(20^{\circ} \mathrm{C}\) and \(8.12 \times 10^{4} \mathrm{Pa}\) . By how many Celsius degrees will the temperature in Denver rise when the chinook arrives?

On a warm summer day, a large mass of air (atmospheric pressure \(1.01 \times 10^{5} \mathrm{Pa}\) ) is heated by the ground to a temperature of \(26.0^{\circ} \mathrm{C}\) and then begins to rise through the cooler surrounding air. (This can be treated approximately as an adiabatic process; why? Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only \(0.850 \times 10^{5} \mathrm{Pa}\) . Assume that air is an ideal gas, with \(\gamma=1.40\) . (This rate of cool- ing for dry, rising air, corresponding to roughly \(1^{\circ} \mathrm{C}\) per 100 \(\mathrm{m}\) of altitude, is called the dry adiabatic lapse rate.)

Nitrogen gas in an expandable container is cooled from \(50.0^{\circ} \mathrm{C}\) to \(10.0^{\circ} \mathrm{C}\) with the pressure held constant at \(3.00 \times 10^{5} \mathrm{Pa}\) . The total heat liberated by the gas is \(2.50 \times 10^{4} \mathrm{J}\) . Assume that the gas may be treated as ideal. (a) Find the number of moles of gas. (b) Find the change in internal energy of the gas. (c) Find the work done by the gas. (d) How much heat would be liberated by the gas for the same temperature change if the volume were constant?

A monatomic ideal gas expands slowly to twice its original volume, doing 300 \(\mathrm{J}\) of work in the process. Find the heat added to the gas and the change in internal energy of the gas if the process is (a) isothermal; (b) adiabatic; (c) isobaric.

In a certain process, \(2.15 \times 10^{5} \mathrm{J}\) of heat is liberated by a system, and at the same time the system contracts under a constant external pressure of \(9.50 \times 10^{5} \mathrm{Pa}\) . The internal energy of the system is the same at the beginning and end of the process. Find the change in volume of the system. (The system is not an ideal gas.)
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