Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 1146

Work done per mole in an isothermal ? change is (A) RT log \(_{e}\left(\mathrm{v}_{2} / \mathrm{v}_{1}\right)\) (B) \(\mathrm{RT} \log _{10}\left(\mathrm{v}_{2} / \mathrm{v}_{1}\right)\) (C) RT \(\log _{10}\left(\mathrm{v}_{1} / \mathrm{v}_{2}\right)\) (D) RT \(\log _{\mathrm{e}}\left(\mathrm{v}_{1} / \mathrm{v}_{2}\right)\)

Short Answer

Expert verified
The short answer is: \[W = RT \log _{e}\left(\frac{V_2}{V_1}\right)\]

Step by step solution

01

Recall the ideal gas law

The ideal gas law is given by: \[PV = nRT\] Where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature. Since we're given an isothermal process, the temperature (T) won't change.
02

Express work done in an isothermal process

The work done (W) in an isothermal process can be expressed as: \[W = -nRT \log_e \left(\frac{V_2}{V_1}\right)\]
03

Comparing the given options with the work formula

Comparing the formula with the given options, we can see that the option (A) has the same expression as the work formula.
04

Conclusion

Based on the comparison, we can conclude that the work done per mole in an isothermal process is: (A) RT log\(_e\left(\mathrm{v}_{2} / \mathrm{v}_{1}\right)\).

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

One mole of a monoatomic gas is heat at a constant pressure of 1 atmosphere from \(0 \mathrm{k}\) to \(100 \mathrm{k}\). If the gas constant $\mathrm{R}=8.32 \mathrm{~J} / \mathrm{mol} \mathrm{k}$ the change in internal energy of the gas is approximate ? (A) \(23 \mathrm{~J}\) (B) \(1.25 \times 10^{3} \mathrm{~J}\) (C) \(8.67 \times 10^{3} \mathrm{~J}\) (D) \(46 \mathrm{~J}\)

An engine is supposed to operate between two reservoirs at temperature \(727^{\circ} \mathrm{C}\) and \(227^{\circ} \mathrm{C}\). The maximum possible efficiency of such an engine is (A) \((3 / 4)\) (B) \((1 / 4)\) (C) \((1 / 2)\) (D) 1

A Carnot engine having a efficiency of \(\mathrm{n}=(1 / 10)\) as heat engine is used as a refrigerators. if the work done on the system is \(10 \mathrm{~J}\). What is the amount of energy absorbed from the reservoir at lowest temperature ! (A) \(1 \mathrm{~J}\) (B) \(90 \mathrm{~J}\) (C) \(99 \mathrm{~J}\) (D) \(100 \mathrm{~J}\)

Water of volume 2 liter in a container is heated with a coil of $1 \mathrm{kw}\( at \)27^{\circ} \mathrm{C}$. The lid of the container is open and energy dissipates at the rate of \(160(\mathrm{~J} / \mathrm{S}) .\) In how much time temperature will rise from \(27^{\circ} \mathrm{C}\) to $77^{\circ} \mathrm{C}\(. Specific heat of water is \)4.2\\{(\mathrm{KJ}) /(\mathrm{Kg})\\}$ (A) \(7 \mathrm{~min}\) (B) \(6 \min 2 \mathrm{~s}\) (C) \(14 \mathrm{~min}\) (D) \(8 \min 20 \mathrm{~S}\)

For which combination of working temperatures the efficiency of Carnot's engine is highest. (A) \(80 \mathrm{~K}, 60 \mathrm{~K}\) (B) \(100 \mathrm{~K}, 80 \mathrm{~K}\) (C) \(60 \mathrm{~K}, 40 \mathrm{~K}\) (D) \(40 \mathrm{~K}, 20 \mathrm{~K}\)
See all solutions

Recommended explanations on English Textbooks

Key Concepts in Language and Linguistics

Read Explanation

Phonetics

Read Explanation

Free Response Essay

Read Explanation

International English

Read Explanation

English Language Study

Read Explanation

Lexis and Semantics Summary

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