/*! 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 54 A cylinder contains \(0.250 \mat... [FREE SOLUTION] | 91Ó°ÊÓ

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A cylinder contains \(0.250 \mathrm{~mol}\) of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) gas at a temperature of \(27.0^{\circ} \mathrm{C}\). The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.00 atm on the gas. The cylinder is placed on a hot plate and \(920 \mathrm{~J}\) of heat flows into the gas, thereby raising its temperature to \(127.0^{\circ} \mathrm{C}\). Assume that the \(\mathrm{CO}_{2}\) may be treated as an ideal gas. (a) Draw a \(p V\) diagram of this process. (b) How much work is done by the gas in the process? (c) On what is this work done? (d) What is the change in internal energy of the gas? (e) How much work would have been done if the pressure had been 0.50 atm?

Short Answer

Expert verified
Work done by gas: calculate using \(W = p \Delta V\). On the piston. \(\Delta U = q - W\).

Step by step solution

01

Identify the Given Data

We are provided with the following information: initial number of moles \(n = 0.250\, \text{mol}\), initial temperature \(T_1 = 27.0\, ^{\circ}\text{C} = 300\, \text{K}\), final temperature \(T_2 = 127.0\, ^{\circ}\text{C} = 400\, \text{K}\), pressure \(p = 1.00\, \text{atm} = 101325\, \text{Pa}\), heat added \(q = 920\, \text{J}\).
02

Draw the pv Diagram

Since the pressure remains constant and the volume changes, the \(pV\) diagram is a horizontal line at \(1.00\, \text{atm}\) from the initial volume \(V_1\) to the final volume \(V_2\).
03

Calculate Work Done by the Gas

The work done \(W\) by the gas is given by \(W = p \Delta V\). Initially, calculate \(V_1\) and \(V_2\) using the ideal gas equation \(pV = nRT\). For \(V_1\):\[ V_1 = \frac{nRT_1}{p} = \frac{0.250 \times 8.314 \times 300}{101325} \] For \(V_2\):\[ V_2 = \frac{nRT_2}{p} = \frac{0.250 \times 8.314 \times 400}{101325} \] Then, \(\Delta V = V_2 - V_1\), and substitute in the work equation:\[ W = p \Delta V = 101325 \times (V_2 - V_1) \].
04

Determine the System the Work is Done On

The work done by the gas is performed on the piston of the cylinder, as the gas expands against the constant external pressure.
05

Calculate the Change in Internal Energy

Utilize the first law of thermodynamics: \(\Delta U = q - W\). Substitute the given heat \(q = 920\, \text{J}\) and the calculated work \(W\) from Step 3:\[ \Delta U = 920 - W \].
06

Calculate Work with Lower Pressure

If the pressure had been \(0.50\, \text{atm} = 50662.5\, \text{Pa}\), we calculate new work using: \[ W' = (0.50 \times 101325) \Delta V \]. Use the same \(\Delta V\) from Step 3.

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Key Concepts

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

Thermodynamics
Thermodynamics is the branch of physical science that deals with the relations between heat and other forms of energy, such as mechanical work. It is at the heart of many fundamental concepts in physics and chemistry.
In this exercise, we are particularly interested in the First Law of Thermodynamics, which states that energy cannot be created or destroyed, only transferred or changed in form. This is mathematically expressed as:\[\Delta U = q - W\]where:
  • \( \Delta U \) is the change in internal energy of the system.
  • \( q \) represents the heat added to the system.
  • \( W \) is the work done by the system.
The above equation helps us understand how energy changes within a given system - in this case, COâ‚‚ gas in a cylinder. Here, the system gains heat (making \( q \) positive) and does work as it expands.
Thermodynamics guides us in analyzing these changes to predict system behavior.
Work Done by Gas
When gas expands within a cylinder, it performs work. Understanding this concept involves knowing how gas under pressure can do mechanical work, particularly when the system involves a movable piston.
The work done by the gas can be calculated with:\[W = p \Delta V\]where:
  • \( W \) is the work done by the gas.
  • \( p \) is the pressure exerted by the gas.
  • \( \Delta V \) is the change in volume.
In this context, the formula shows that work is directly proportional to the pressure and the change in volume during expansion. When you apply heat to the gas, it expands, pushing the piston under constant pressure, and thereby doing work. Before calculating the work, the volumes before and after heating must be found using the ideal gas law.
This concept illustrates how gases can convert heat energy into mechanical work, showcasing one of the fascinating processes in physical science.
Change in Internal Energy
The internal energy in a system refers to the total energy contained within it due to both the kinetic and potential energies of its molecules.
To find the change in internal energy (\( \Delta U \)), we use the First Law of Thermodynamics,\[\Delta U = q - W\]where \( q \) is the heat input into the system and \( W \) is the work done. This formula essentially describes energy conservation inside the system, taking the input energy and subtracting the energy expended as work.
This concept is crucial because analyzing internal energy gives insights into how energy is stored and manipulated in physical systems, especially gases, when external conditions change. For instance, when heat is applied to our gas in the cylinder, we can calculate how much of that heat translates into internal energy versus how much is used for expansion.
Pressure-Volume Diagram
A pressure-volume (\( pV \)) diagram is a graphical representation of pressure changes with respect to volume. This diagram is fundamental in understanding the work done during thermodynamic processes.
In this exercise, the \( pV \) diagram is a straight line parallel to the volume axis since the process occurs at constant pressure (1.00 atm). These diagrams are especially helpful in visualizing different thermodynamic processes such as isochoric (constant volume), isobaric (constant pressure), isothermal (constant temperature), and adiabatic (no heat exchange) transforms.
Utilizing a \( pV \) diagram provides a clear visual representation of how gas expands within a cylinder, making it easier to see the relationship between pressure, volume, and work done by the gas over the course of the heating process. This visual tool is invaluable in thermodynamics to convey complex information in an immediate and intuitive format.

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Most popular questions from this chapter

The atmosphere of the planet Mars is \(95.3 \%\) carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) and about \(0.03 \%\) water vapor. The atmospheric pressure is only about \(600 \mathrm{~Pa}\), and the surface temperature varies from \(-30^{\circ} \mathrm{C}\) to \(-100^{\circ} \mathrm{C}\). The polar ice caps contain both \(\mathrm{CO}_{2}\) ice and water ice. Could there be liquid \(\mathrm{CO}_{2}\) on the surface of Mars? Could there be liquid water? Why or why not?

An automobile tire has a volume of \(0.0150 \mathrm{~m}^{3}\) on a cold day when the temperature of the air in the tire is \(5.0^{\circ} \mathrm{C}\) and atmospheric pressure is 1.02 atm. Under these conditions, the gauge pressure is measured to be 1.70 atm (about \(25 \mathrm{lb} /\) in. \(^{2}\) ). After the car is driven on the highway for \(30 \mathrm{~min}\), the temperature of the air in the tires has risen to \(45.0^{\circ} \mathrm{C}\) and the volume to \(0.0159 \mathrm{~m}^{3}\). What is the gauge pressure at that time?

If deep-sea divers rise to the surface too quickly, nitrogen bubbles in their blood can expand and prove fatal. This phenomenon is known as the bends. If a scuba diver rises quickly from a depth of \(25 \mathrm{~m}\) in Lake Michigan (which is freshwater), what will be the volume at the surface of an \(\mathrm{N}_{2}\) bubble that occupied \(1.0 \mathrm{~mm}^{3}\) in his blood at the lower depth? Does it seem that this difference is large enough to be a problem? (Assume that the pressure difference is due only to the changing water pressure, not to any temperature difference, an assumption that is reasonable, since we are warm-blooded creatures.)

A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains \(0.110 \mathrm{~m}^{3}\) of air at a pressure of 3.40 atm. The piston is slowly pulled out until the volume of the gas is increased to \(0.390 \mathrm{~m}^{3}\). If the temperature remains constant, what is the final value of the pressure?

Calculate the volume of \(1.00 \mathrm{~mol}\) of liquid water at a temperature of \(20^{\circ} \mathrm{C}\) (at which its density is \(998 \mathrm{~kg} / \mathrm{m}^{3}\) ), and compare this volume with the volume occupied by \(1.00 \mathrm{~mol}\) of steam at \(200^{\circ} \mathrm{C}\). Assume the steam is at atmospheric pressure and can be treated as an ideal gas.

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