/*! 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 3 CALC Two moles of an ideal gas a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

CALC Two moles of an ideal gas are compressed in a cylinder at a constant temperature of \(65.0^{\circ} \mathrm{C}\) until the original pressure has tripled. (a) Sketch a pV-diagram for this process. (b) Calculate the amount of work done.

Short Answer

Expert verified
The work done on the gas is approximately -6187 J.

Step by step solution

01

Understand the Problem

We have two moles of an ideal gas that undergoes an isothermal (constant temperature) compression where the initial pressure triples. We need to sketch the process on a pV diagram and calculate the work done during compression.
02

Sketch the pV-Diagram

On a pV-diagram, a constant temperature (isothermal) process follows a hyperbolic curve. Initially, the gas is at some pressure \(P_1\) and volume \(V_1\). After compression, the pressure becomes \(3P_1\) and the volume \(V_2\). The curve should show that as pressure increases, volume decreases.
03

Recall the Equation for Work Done

For an isothermal process involving an ideal gas, the work done \(W\) is given by the formula: \[ W = nRT \ln\left(\frac{V_f}{V_i}\right) \] where \(n\) is the number of moles, \(R\) is the ideal gas constant, \(T\) is temperature in Kelvin, \(V_f\) is final volume, and \(V_i\) is initial volume.
04

Convert Temperature to Kelvin

Convert the temperature from Celsius to Kelvin using the formula: \[ T(K) = T(^{\circ}C) + 273.15 \] So, \(T = 65 + 273.15 = 338.15 \text{ K}\).
05

Use Ideal Gas Law to Relate Volumes and Pressures

Using the ideal gas law \( PV = nRT \), we know \(P_1V_1 = P_3V_2\) because temperature is constant. With \(P_3 = 3P_1\), it follows that \(V_2 = \frac{V_1}{3}\).
06

Calculate Work Done Using the Formula

Substitute the values into the work formula: \[ W = nRT \ln\left(\frac{V_2}{V_1}\right) = 2 \times 8.314 \times 338.15 \times \ln\left(\frac{1}{3}\right) \] Calculate this to find the work done.
07

Compute the Final Result

Calculating the above equation: \[ W = 2 \times 8.314 \times 338.15 \times (-1.0986) \approx -6187.21 \text{ J} \] The negative sign indicates work is done on the gas.

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.

Isothermal Process
An isothermal process is a thermodynamic process during which the temperature of the system remains constant. This is a key concept under ideal gas laws and can be applied to gases under particular conditions. When a gas undergoes an isothermal process, its internal energy remains unchanged as there is no temperature variation. This implies that any heat added to the system is used to do work.In the case of an ideal gas, an isothermal change ensures that the product of pressure and volume remains constant. Hence, the relationship can be expressed using Boyle's Law, which states:\[ PV = ext{constant} \]This process is usually depicted graphically on a pV-diagram as a hyperbolic curve, which shows an inverse relationship between pressure and volume at a constant temperature. Understanding an isothermal process helps in analyzing the behavior of gases in thermal equilibrium and predicting the corresponding changes in physical properties.
pV-Diagram
A pV-diagram, or pressure-volume diagram, is a graphical representation of the changes in pressure and volume of a gas within a thermodynamic system. It is a powerful tool in visualizing processes undergone by gases, such as compression, expansion, heating, or cooling. In an isothermal process, the pV-curve is characterized by a hyperbola. Let's break down why this happens:
  • Pressure (p) and volume (V) have an inverse relationship; as one increases, the other decreases.
  • The area under the curve on a pV-diagram represents the work done during the process.
  • For isothermal conditions, the curve moves smoothly from a higher volume and lower pressure to lower volume and higher pressure during compression.
Visualizing these processes on a pV-diagram can simplify the understanding of work done and energy transformations within a gas. It essentially offers a window into the gas's thermodynamic behavior and helps identify the efficiency of processes.
Work Done in Thermodynamics
In thermodynamics, work done is a critical concept that helps in understanding energy transfer within processes. In the context of an ideal gas undergoing processes like expansion or compression, calculating the work done reveals important insights about the system's energy interactions.For isothermal processes, the work done on or by the gas is calculated using:\[ W = nRT \ln\left(\frac{V_f}{V_i}\right) \]Where:
  • \(n\) is the number of moles of gas.
  • \(R\) is the ideal gas constant (8.314 J/mol·K).
  • \(T\) is the temperature in Kelvin.
  • \(V_f\) and \(V_i\) represent the final and initial volumes respectively.
This formula effectively takes into account the constant temperature condition by incorporating it into the natural logarithm of the volume ratio. When interpreting the result:
  • A positive work value denotes work done by the gas (expansion).
  • A negative value indicates work done on the gas (compression).
Harnessing this formula makes it possible to accurately quantify the thermal dynamics of gases, facilitating a deeper understanding of how heat, work, and energy interplay within a system.

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

CALC A cylinder with a frictionless, movable piston like that shown in Fig. 19.5 contains a quantity of helium gas. Initially the gas is at a pressure of \(1.00 \times 10^{5}\) Pa, has a temperature of 300 \(\mathrm{K}\) , and occupies a volume of 1.50 \(\mathrm{L}\) . The gas then undergoes two processes. In the first, the gas is heated and the piston is allowed to move to keep the temperature equal to 300 \(\mathrm{K}\) . This continues until the pressure reaches \(2.50 \times 10^{4}\) Pa. In the second process, the gas is compressed at constant pressure until it returns to its original volume of 1.50 L. Assume that the gas may be treated as ideal. (a) In a \(p V\) -diagram, show both processes. (b) Find th volume of the gas at the end of the first process, and find the pressure and temperature at the end of the second process. (c) Fin the total work done by the gas during both processes. (d) Final would you have to do to the gas to return it to its original pres sure and temperature?

Three moles of an ideal monatomic gas expands at a constant pressure of 2.50 atm; the volume of the gas changes from \(3.20 \times 10^{-2} \mathrm{m}^{3}\) to \(4.50 \times 10^{-2} \mathrm{m}^{3} .\) (a) Calculate the initial and final temperatures of the gas. (b) Calculate the amount of work the gas does in expanding. (c) Calculate the amount of heat added to the gas. (d) Calculate the change in internal energy of the gas.

Five moles of an ideal monatomic gas with an initial temperature of \(127^{\circ} \mathrm{C}\) expand and, in the process, absorb 1200 \(\mathrm{J}\) of heat and do 2100 \(\mathrm{J}\) of work. What is the final temperature of the gas?

High-Altitude Research. A large research balloon containing \(2.00 \times 10^{3} \mathrm{m}^{3}\) of helium gas at 1.00 atm and a temperature of \(15.0^{\circ} \mathrm{C}\) rises rapidly from ground level to an altitude at which the atmospheric pressure is only 0.900 atm (Fig. Pl9.58. Assume the helium behaves like an ideal gas and the balloon's ascent is too rapid to permit much heat exchange with the surrounding air. (a) Calculate the volume of the gas at the higher altitude. (b) Calculate the temperature of the gas at the higher altitude. (c) What is the change in internal energy of the helium as the balloon rises to the higher altitude?

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 \(3.00 \times 10^{3}\) Pa. The total heat liberated by the gas is \(2.50 \times 10^{4}\) . 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?

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.