Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A gas in a cylinder expands from a volume of 0.110 m\(^3\) to 0.320 m\(^3\). Heat flows into the gas just rapidly enough to keep the pressure constant at 1.65 \(\times\) 10\(^5\) Pa during the expansion. The total heat added is 1.15 \(\times\) 10\(^5\) J. (a) Find the work done by the gas. (b) Find the change in internal energy of the gas. (c) Does it matter whether the gas is ideal? Why or why not?

Short Answer

Expert verified
(a) Work done: 3.465 × 10⁴ J. (b) Change in internal energy: 8.035 × 10⁴ J. (c) It doesn't matter if the gas is ideal.

Step by step solution

01

Understanding the Problem

We are given the initial and final volumes of a gas in a cylinder, the constant pressure during its expansion, and the total heat added to the system. Our goals are to find the work done by the gas, the change in its internal energy, and consider whether the ideal gas model affects the outcome.
02

Calculate the Work Done by the Gas (a)

The work done by the gas during an isobaric (constant pressure) process is given by the equation: \[ W = P \Delta V \]where \( P \) is the pressure and \( \Delta V \) is the change in volume. Given:\( P = 1.65 \times 10^5 \) Pa, initial volume \( V_i = 0.110 \) m\(^3\), final volume \( V_f = 0.320 \) m\(^3\).Calculate \( \Delta V \):\[ \Delta V = V_f - V_i = 0.320 - 0.110 = 0.210 \text{ m}^3 \]Substitute into the equation for work:\[ W = 1.65 \times 10^5 \times 0.210 = 3.465 \times 10^4 \text{ J} \]
03

Calculate the Change in Internal Energy (b)

The change in internal energy \( \Delta U \) for a system can be determined using the first law of thermodynamics:\[ \Delta U = Q - W \]where \( Q \) is the heat added to the system and \( W \) is the work done by the system.Given that \( Q = 1.15 \times 10^5 \) J (heat added) and we previously found \( W = 3.465 \times 10^4 \) J:\[ \Delta U = 1.15 \times 10^5 - 3.465 \times 10^4 = 8.035 \times 10^4 \text{ J} \]
04

Consideration of Gas Type (c)

In this problem, it does not matter whether the gas is ideal when calculating work done and internal energy change because these calculations depend solely on measurable quantities such as pressure, volume, and heat, not on the type of gas. The ideal gas assumption typically affects state equations but not directly the conservation of energy relations used here.

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 Vaia!

Key Concepts

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

Isobaric Process
An isobaric process is a thermodynamic process that occurs at a constant pressure. This means that throughout the entire process, the pressure of the system does not change. In our exercise, the expansion of the gas happens isobarically at a steady pressure of 1.65 \( \times \) 10\(^5\) Pa.
Here are a few important points about isobaric processes:
  • The volume of the gas changes, while the pressure remains constant.
  • Work done by the gas in this process can be calculated using the formula \( W = P \Delta V \), where \( W \) is the work done, \( P \) is the constant pressure, and \( \Delta V \) is the change in volume.
  • This process commonly occurs in everyday contexts, such as the heating of a gas in a sealed container where external pressure is constant.
Isobaric processes are straightforward because they simplify the relationship between pressure, volume, and the work done on or by a gas. The simplification arises due to constant pressure, making calculations easier to manage directly compared to processes where pressure varies.
First Law of Thermodynamics
The first law of thermodynamics is a form of the law of conservation of energy, adapted for thermodynamic systems. It establishes a relationship between internal energy, heat added to the system, and work done by the system. The first law states that the change in a system's internal energy \( \Delta U \) is equal to the heat added to the system \( Q \) minus the work done by the system \( W \). This can be expressed with the formula:
\[ \Delta U = Q - W \]
Where:
  • \( \Delta U \) represents the change in internal energy.
  • \( Q \) is the total heat added to the system.
  • \( W \) is the work done by the system.
In our example, the heat added is 1.15 \( \times \) 10\(^5\) J,
and we calculated the work done by the gas as 3.465 \( \times \) 10\(^4\) J.
By utilizing the first law, we compute \( \Delta U \),
resulting in a significant shift in internal energy.
Understanding this principle is crucial because it enables us to analyze energy transfer in or out of any system, explaining how and why energy changes within the system.
Ideal Gas
In the context of thermodynamics, an 'ideal gas' is a hypothetical gas whose molecules occupy negligible space and have no interactions, except for elastic collisions. An ideal gas follows the ideal gas law, which is generally stated as \( PV = nRT \), where
  • \( P \) is the pressure,
  • \( V \) is the volume,
  • \( n \) is the number of moles of the gas,
  • \( R \) is the universal gas constant,
  • \( T \) is the absolute temperature.
The consideration of a gas being ideal does not affect the calculations for the change in internal energy or the work done in our original problem, because these calculations rely only on measurable parameters such as pressure, volume, and heat. It's when we need to understand how gases behave on a micro-level, considering relationships like pressure, volume, and temperature, that the ideal gas model becomes significant.
While real gases deviate from the ideal gas model under high pressure and low temperature, for many practical purposes, the ideal gas law provides helpful approximations, simplifying calculations and predictions about gas behavior. In our exercise, discussing whether the gas is ideal is more relevant to theoretical interests rather than affecting practical outcomes in these specific calculations.

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

In a cylinder, 1.20 mol of an ideal monatomic gas, initially at 3.60 \(\times\) 10\(^5\) Pa and 300 K, expands until its volume triples. Compute the work done by the gas if the expansion is (a) isothermal; (b) adiabatic; (c) isobaric. (d) Show each process in a \(pV\)-diagram. In which case is the absolute value of the work done by the gas greatest? Least? (e) In which case is the absolute value of the heat transfer greatest? Least? (f) In which case is the absolute value of the change in internal energy of the gas greatest? Least?

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 cylinder with a piston contains 0.150 mol of nitrogen at 1.80 \(\times\) 10\(^5\) Pa and 300 K. The nitrogen may be treated as an ideal gas. The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure. (a) Show the series of processes in a \(pV\)-diagram. (b) Compute the temperatures at the beginning and end of the adiabatic expansion. (c) Compute the minimum pressure.

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.

Nitrogen gas in an expandable container is cooled from 50.0\(^\circ\)C to 10.0\(^\circ\)C with the pressure held constant at 3.00 \(\times\) 10\(^5\) Pa. The total heat liberated by the gas is 2.50 \(\times\) 10\(^4\) J. Assume that the gas may be treated as ideal. Find (a) the number of moles of gas; (b) the change in internal energy of the gas; (c) 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?

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.

Sign-up for free