Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 1149

An adiabatic Bulk modulus of an ideal gas at Pressure \(\mathrm{P}\) is (A) \(\gamma \mathrm{P}\) (B) \((\mathrm{p} / \gamma)\) (C) \(\mathrm{P}\) (D) \((\mathrm{p} / 2)\)

Short Answer

Expert verified
The short answer is: The adiabatic Bulk modulus of an ideal gas at pressure \(P\) is (A) \(\gamma P\), where \(\gamma\) is the adiabatic index (heat capacity ratio).

Step by step solution

01

Write down the adiabatic bulk modulus formula

The formula for the adiabatic bulk modulus of an ideal gas (\(B_{\text{adiabatic}}\)) in terms of pressure (\(P\)) and adiabatic index (\(\gamma\)) is: \[B_{\text{adiabatic}} = \gamma P\]
02

Compare the formula with the given options

We will now compare the formula \(B_{\text{adiabatic}} = \gamma P\) with the given options to determine which one is correct: (A) \(\gamma P\) (B) \(\dfrac{P}{\gamma}\) (C) \(P\) (D) \(\dfrac{P}{2}\)
03

Select the correct option

By comparing the formula \(B_{\text{adiabatic}} = \gamma P\) with the given options, we see that the correct option is (A) \(\gamma P\).

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

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) \(100 \mathrm{~J}\)

A diatomic gas initially at \(18^{\circ} \mathrm{C}\) is Compressed adiabatically to one eighth of its original volume. The temperature after Compression will be (A) \(10^{\circ} \mathrm{C}\) (B) \(668 \mathrm{~K}\) (C) \(887^{\circ} \mathrm{C}\) (D) \(144^{\circ} \mathrm{C}\)

In an isothermal reversible expansion, if the volume of \(96 \mathrm{~J}\) of oxygen at \(27^{\circ} \mathrm{C}\) is increased from 70 liter to 140 liter, then the work done by the gas will be (A) \(300 \mathrm{R} \log _{\mathrm{e}}^{(2)}\) (B) \(81 \mathrm{R} \log _{\mathrm{e}}^{(2)}\) (C) \(2.3 \times 900 \mathrm{R} \log _{10} 2\) (D) \(100 \mathrm{R} \log _{10}^{(2)}\)

One mole of an ideal gas $\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right)=\gamma$ at absolute temperature \(\mathrm{T}_{1}\) is adiabatically compressed from an initial pressure \(\mathrm{P}_{1}\) to a final pressure \(\mathrm{P}_{2}\) The resulting temperature \(\mathrm{T}_{2}\) of the gas is given by. (A) $\mathrm{T}_{2}=\mathrm{T}_{1}\left\\{\mathrm{p}_{2} / \mathrm{p}_{1}\right\\}^{\\{\gamma /(\gamma-1)\\}}$ (B) $\mathrm{T}_{2}=\mathrm{T}_{1}\left\\{\mathrm{p}_{2} / \mathrm{p}_{1}\right\\}^{\\{(\gamma-1) / \gamma\\}}$ (C) $\mathrm{T}_{2}=\mathrm{T}_{1}\left\\{\mathrm{p}_{2} / \mathrm{p}_{1}\right\\}^{\gamma}$ (D) $\mathrm{T}_{2}=\mathrm{T}_{1}\left(\mathrm{p}_{2} / \mathrm{p}_{1}\right)^{\gamma-1}$

An ideal gas heat engine is operating between \(227^{\circ} \mathrm{C}\) and \(127^{\circ} \mathrm{C}\). It absorbs \(10^{4} \mathrm{~J}\) Of heat at the higher temperature. The amount of heat Converted into. work is \(\ldots \ldots\) J. (A)2000 (B) 4000 (C) 5600 (D) 8000
See all solutions

Recommended explanations on English Textbooks

Syntax

Read Explanation

Rhetorical Analysis Essay

Read Explanation

Essay Plan

Read Explanation

Essay Prompts

Read Explanation

Lexis and Semantics

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