/*! 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 76 During a compression at a consta... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

During a compression at a constant pressure of \(250 \mathrm{~Pa},\) the volume of an ideal gas decreases from \(0.80 \mathrm{~m}^{3}\) to \(0.20 \mathrm{~m}^{3}\). The initial temperature is \(360 \mathrm{~K},\) and the gas loses \(210 \mathrm{~J}\) as heat. What are (a) the change in the internal energy of the gas and (b) the final temperature of the gas?

Short Answer

Expert verified
(a) \(-360 \text{ J}\). (b) \(90 \text{ K}\).

Step by step solution

01

Identify the Given Information

We are given the following variables:- Initial pressure, \(P = 250 \text{ Pa}\)- Initial volume, \(V_i = 0.80 \text{ m}^3\)- Final volume, \(V_f = 0.20 \text{ m}^3\)- Initial temperature, \(T_i = 360 \text{ K}\)- Heat lost, \(Q = -210 \text{ J}\) (since it is lost)We need to find:(a) Change in internal energy, \(\Delta U\)(b) Final temperature, \(T_f\) of the gas.
02

Use the First Law of Thermodynamics

The first law of thermodynamics is given by:\[ \Delta U = Q + W \]where \(Q = -210 \text{ J}\) and \(W\) (the work done by the gas) can be calculated as \(W = P \Delta V\).First, calculate \(\Delta V = V_f - V_i = 0.20 \text{ m}^3 - 0.80 \text{ m}^3 = -0.60 \text{ m}^3\).
03

Calculate Work Done by the Gas

Work done by the gas is given by:\[ W = P \Delta V = 250 \text{ Pa} \times (-0.60 \text{ m}^3) = -150 \text{ J} \] The work is negative since the volume is decreasing (compression).
04

Calculate Change in Internal Energy

Using the first law of thermodynamics:\[ \Delta U = Q + W = -210 \text{ J} - 150 \text{ J} = -360 \text{ J} \] Thus, the change in the internal energy of the gas is \(-360 \text{ J}\).
05

Apply Ideal Gas Law to Find Final Temperature

Using the ideal gas law, \(PV = nRT\), we can rearrange for temperature:\[ T = \frac{PV}{nR} \]Since \(P\) and \(nR\) are constants, the initial and final states give:\[ \frac{T_i}{V_i} = \frac{T_f}{V_f} \]Thus,\[ T_f = T_i \times \frac{V_f}{V_i} = 360 \text{ K} \times \frac{0.20 \text{ m}^3}{0.80 \text{ m}^3} = 90 \text{ K} \] Therefore, the final temperature is \(90 \text{ K}\).
06

Conclusion

The change in internal energy of the gas is \(-360 \text{ J}\), and the final temperature is \(90 \text{ K}\).

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 91Ó°ÊÓ!

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 principle that relates the change in internal energy of a system to heat exchange and work done. Simply put, it states that energy can neither be created nor destroyed, only transformed. This law can be expressed with the equation: \( \Delta U = Q + W \).
This equation breaks down the components involved when energy changes occur within a system:
  • \( \Delta U \) represents the change in internal energy.
  • \( Q \) is the heat added to the system. It's positive if heat is absorbed and negative if it's lost.
  • \( W \) is the work done on or by the system. Work done by the system is negative in compression, while it's positive when expansion occurs.
In the context of the given exercise, heat loss is stated as \(-210 \text{ J}\) and the work done during compression is \(-150 \text{ J}\). This ultimately leads to the change in internal energy being \(-360 \text{ J}\), indicating how the system's energy has decreased due to these exchanges. Understanding this helps students grasp how energy conversion works and is a vital principle in many thermodynamic processes.
Internal Energy
Internal energy refers to the total energy contained within a substance due to molecular movement and molecular interactions. It is a key concept in thermodynamics, encapsulating both kinetic and potential energy within molecules.
In an ideal gas, molecules move freely without interactions that hold or repulse each other, meaning that their internal energy purely depends on the temperature. Consequently, any changes to the internal energy for an ideal gas are directly linked to temperature changes.
Consider this scenario where heat is lost and work is performed in compression. The internal energy decreases as a result of heat loss \((-210 \text{ J})\) and work done \((-150 \text{ J})\), leading to a total change of \( -360 \text{ J} \). While this may seem simple, it showcases that energy transitions affect molecular motion, thus influencing internal energy.
Ideal Gas Law
The ideal gas law is a key equation that links the four properties of gases: pressure, volume, temperature, and the number of moles. It is presented as \( PV = nRT \), where:
  • \(P\) is the pressure of the gas.
  • \(V\) is the volume.
  • \(n\) is the number of moles.
  • \(R\) is the ideal gas constant.
  • \(T\) is the temperature in Kelvin.
In practice, this law is engaged for calculations when you know three of the four variables, and you aim to solve for the fourth. In the exercise, gas compresses from an initial volume to a smaller one, while pressure remains constant. Hence, using the proportionality \( \frac{T_i}{V_i} = \frac{T_f}{V_f} \) helped find the final temperature, showing the effect on temperature as volume shrinks. From here, one can see how interconnected these properties are in defining the state of an ideal gas.

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

Suppose \(1.80 \mathrm{~mol}\) of an ideal gas is taken from a volume of \(3.00 \mathrm{~m}^{3}\) to a volume of \(1.50 \mathrm{~m}^{3}\) via an isothermal compression at \(30^{\circ} \mathrm{C}\). (a) How much energy is transferred as heat during the compression, and (b) is the transfer to or from the gas?

Ten particles are moving with the following speeds: four at \(200 \mathrm{~m} / \mathrm{s},\) two at \(500 \mathrm{~m} / \mathrm{s},\) and four at \(600 \mathrm{~m} / \mathrm{s} .\) Calculate their (a) average and (b) rms speeds. (c) Is \(v_{\text {rms }}>v_{\text {avg }} ?\)

Oxygen \(\left(\mathrm{O}_{2}\right)\) gas at \(273 \mathrm{~K}\) and \(1.0 \mathrm{~atm}\) is confined to a cubical container \(10 \mathrm{~cm}\) on a side. Calculate \(\Delta U_{g} / K_{\text {avg }},\) where \(\Delta U_{g}\) is the change in the gravitational potential energy of an oxygen molecule falling the height of the box and \(K_{\text {avg }}\) is the molecule's average translational kinetic energy.

The volume of an ideal gas is adiabatically reduced from \(200 \mathrm{~L}\) to \(74.3 \mathrm{~L}\). The initial pressure and temperature are \(1.00 \mathrm{~atm}\) and \(300 \mathrm{~K}\). The final pressure is 4.00 atm. (a) Is the gas monatomic, diatomic, or polyatomic? (b) What is the final temperature? (c) How many moles are in the gas?

In a certain particle accelerator, protons travel around a circular path of diameter \(23.0 \mathrm{~m}\) in an evacuated chamber, whose residual gas is at \(295 \mathrm{~K}\) and \(1.00 \times 10^{-6}\) torr pressure. (a) Calculate the number of gas molecules per cubic centimeter at this pressure. (b) What is the mean free path of the gas molecules if the molecular diameter is \(2.00 \times 10^{-8} \mathrm{~cm} ?\)

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.