Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 48

A 25 -L sample of ideal gas with \(\gamma=1.67\) is at \(250 \mathrm{K}\) and \(50 \mathrm{kPa} .\) The gas is compressed isothermally to one-third of its original volume, then heated at constant volume until its state lies on the adiabatic curve that passes through its original state, and then allowed to expand adiabatically to that original state. Find the net work involved. Is net work done on or by the gas?

Short Answer

Expert verified
The net work done and whether the work is done on the system or by the system will be determined after performing the prescribed calculations. A positive net work indicates work done on the system, whereas a negative net work suggests work done by the system.

Step by step solution

01

Calculation of Initial and Final Pressure

The initial Pressure \(P_1\) and Volume \(V_1\) are given as \(50 kPa\) and \(25 L\) respectively. The process is an isothermal compression to one-third of its original volume, therefore, the final Volume \(V_2\) is \(V_1/3\). Use the ideal gas law (\(PV=nRT\)) for isothermal processes where temperature \(T\) and total number of moles \(n\) is constant. So \(P_1 V_1 = P_2 V_2\). Solving for the final pressure, \(P_2 = (P_1 V_1) / V_2 = 3 P_1 = 3 * 50 kPa = 150 kPa\).
02

Work Done in Isothermal Compression

The work done in an isothermal process is given by \(W_1= nRT \ln(P_1/P_2)\). However, the number of moles of gas \(n\) is not provided. We can substitute the ideal gas law into the equation to replace \(n\) and solve for work \(W_1\). The work done is on the system (since compression takes place), so it is negative.
03

Work Done in Heating at Constant Volume

In a constant volume process, no work is done because the volume does not change. Therefore, \(W_2 = 0\).
04

Calculation of Final Temperature

After heating at constant volume, the state lies on the adiabatic curve that passes through its original state. For an isentropic or adiabatic process, \(P_1 V_1^gamma = P_2 V_2^gamma\). Rearranging this equation we get \(V_1/V_2 = (P_2/P_1)^(1/gamma)\) and the initial and final volumes are the same according to the problem statement. Thus, \(T_1=(P_1/P_2)^(1-1/gamma)T_2\) or \(T_2 = T_1 (P_2/P_1)^{(gamma-1/gamma)}\).
05

Work Done in Adiabatic Expansion

The work done in an adiabatic process is given by \(W_3= Cv(T1-T2)\), where \(Cv= R/(gamma-1)\).
06

Calculation of Net Work Done

The net work done \(W_net\) on the system is the algebraic sum of the work done in all the three processes, i.e. \(W_net = W_1 + W_2 + W_3\).
07

Identify whether net work is done on or by the system

The sign of the net work done will indicate whether the work is done on the system or by the system. A positive \(W_net\) indicates work is done on the system, while negative \(W_net\) means work done by the system.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Why can't an irreversible process be described by a path in a \(p V\) diagram?

A \(40-\) W heat source is applied to a gas sample for 25 s, during which time the gas expands and does 750 J of work on its surroundings. By how much does the internal energy of the gas change?

A gas mixture contains monatomic argon and diatomic oxygen. An adiabatic expansion that doubles its volume results in the pressure dropping to one- third of its original value. What fraction of the molecules are argon?

Experimental studies show that the \(p V\) curve for a frog's lung can be approximated by \(p=10 v^{3}-67 v^{2}+220 v,\) with \(v\) in \(\mathrm{mL}\) and \(p\) in \(\mathrm{Pa}\). Find the work done when such a lung inflates from zero to \(4.5 \mathrm{mL}\) volume.

An 8.5 -kg rock at \(0^{\circ} \mathrm{C}\) is dropped into a well-insulated vat containing a mixture of ice and water at \(0^{\circ} \mathrm{C}\). When equilibrium is reached, there are \(6.3 \mathrm{g}\) less ice. From what height was the rock dropped?
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Magnetism

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Electric Charge Field and Potential

Read Explanation

Physical Quantities And Units

Read Explanation

Radiation

Read Explanation
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