/*! 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 42 A gas in a cylinder is held at a... [FREE SOLUTION] | 91Ó°ÊÓ

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A gas in a cylinder is held at a constant pressure of \(2.30 \times 10^{5} \mathrm{~Pa}\) and is cooled and compressed from \(1.70 \mathrm{~m}^{3}\) to \(1.20 \mathrm{~m}^{3}\). The internal energy of the gas decreases by \(1.40 \times 10^{5} \mathrm{~J}\). (a) Find the work done by the gas. (b) Find the amount of the heat that flowed into or out of the gas, and state the direction (inward or outward) of the flow.

Short Answer

Expert verified
(a) Work done by the gas: -1.15 x 10^5 J; (b) Heat flow: -2.55 x 10^5 J (outward).

Step by step solution

01

Calculate Work Done by the Gas

The work done \((W)\) by the gas at constant pressure can be calculated using the formula \[ W = P imes (V_f - V_i) \]where \(P\) is the constant pressure, \(V_i\) is the initial volume, and \(V_f\) is the final volume.Given:- \(P = 2.30 \times 10^5 \, \text{Pa}\)- \(V_i = 1.70 \, \text{m}^3\)- \(V_f = 1.20 \, \text{m}^3\)Substituting these values:\[ W = 2.30 \times 10^5 \, \text{Pa} \times (1.20 \, \text{m}^3 - 1.70 \, \text{m}^3) \]\[ W = 2.30 \times 10^5 \, \text{Pa} \times (-0.50 \, \text{m}^3) \]\[ W = -1.15 \times 10^5 \, \text{J} \]The negative sign indicates that the work is done on the gas.
02

Calculate Heat Flow into or out of the Gas

We use the first law of thermodynamics which states:\[ \Delta U = Q - W \]where \(\Delta U\) is the change in internal energy, \(Q\) is the heat exchanged, and \(W\) is the work done by the system.Given:- \(\Delta U = -1.40 \times 10^5 \, \text{J}\)- \(W = -1.15 \times 10^5 \, \text{J}\)Rearranging the formula to solve for \(Q\):\[ Q = \Delta U + W \]\[ Q = (-1.40 \times 10^5 \, \text{J}) + (-1.15 \times 10^5 \, \text{J}) \]\[ Q = -2.55 \times 10^5 \, \text{J} \]The negative sign indicates that the heat flows out of the gas.

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Key Concepts

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

First Law of Thermodynamics
The First Law of Thermodynamics is a fundamental concept in physics, forming the foundation of much of thermodynamic study. It states that energy cannot be created or destroyed, only transformed from one form to another. In mathematical terms, this can be expressed as:\[ \Delta U = Q - W \]Where:
  • \( \Delta U \) is the change in internal energy of the system
  • \( Q \) is the heat added to the system
  • \( W \) is the work done by the system

This equation implies that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In the context of our exercise, it helps connect the internal changes within the gas to the external influences, explaining how energy transfers occur within the gas system.
Work Done by Gas
Work done by a gas during a process is crucial to understanding how energy flows in thermodynamic systems. When a gas expands or contracts, it does work either on the surrounding environment or has work done onto it. When a gas at constant pressure changes volume, the work done \( W \) can be calculated as:\[ W = P \times (V_f - V_i) \]Here:
  • \( P \) is the constant pressure
  • \( V_f \) is the final volume
  • \( V_i \) is the initial volume

From the exercise, we see that the gas is compressed, indicated by the change from an initial volume of \( 1.70 \ \text{m}^3 \) to a final volume of \( 1.20 \ \text{m}^3 \). Since the work calculated is negative, \( -1.15 \times 10^5 \ \text{J} \), it tells us that work is being done on the gas, compressing it.
Heat Exchange
Heat exchange in thermodynamic processes is the transfer of thermal energy between the system and its surroundings. It can occur in two directions: heat can be added to the system or removed from it. The direction of heat flow is essential to understanding how a system's energy changes. If \( Q \) is positive, heat enters the system; if \( Q \) is negative, heat leaves.
In our problem, using the rearranged formula from the first law, we calculate that \( Q = -2.55 \times 10^5 \ \text{J} \). This negative sign indicates heat flows out of the gas. This outcome is consistent with the internal energy decreasing and work being performed on the gas.
Internal Energy Change
Internal energy change, \( \Delta U \), is the cumulative effect of energy transformations within a system. It encompasses the total kinetic and potential energy changes resulting from temperature, volume, and pressure variations.
In our exercise, the internal energy of the gas decreases by \( 1.40 \times 10^5 \ \text{J} \). Despite doing work on the gas (compressing it), the overall energy inside the system diminishes. This loss of energy correlates with heat being transferred from the system, as indicated by the negative internal energy change.
This concept shows how internal energy is a measure of the system's total energy status and its changes provide insights into how energy moves within the system in physical terms.

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Most popular questions from this chapter

(a) A deuteron, \({ }_{1}^{2} \mathrm{H}\), is the nucleus of a hydrogen isotope and consists of one proton and one neutron. The plasma of deuterons in a nuclear fusion reactor must be heated to about 300 million \(\mathrm{K}\). What is the rms speed of the deuterons? Is this a significant fraction of the speed of light \(\left(c=3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\right) ?\) (b) What would the temperature of the plasma be if the deuterons had an rms speed equal to \(0.10 c ?\)

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Oxygen \(\left(\mathrm{O}_{2}\right)\) has a molar mass of \(32.0 \mathrm{~g} / \mathrm{mol}\). (a) What is the root-mean-square speed of an oxygen molecule at a temperature of \(300 \mathrm{~K} ?\) (b) What is its average translational kinetic energy at that speed?

In a certain chemical process, a lab technician supplies \(254 \mathrm{~J}\) of heat to a system. At the same time, \(73 \mathrm{~J}\) of work are done on the system by its surroundings. What is the increase in the internal energy of the system?

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