Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 1140

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) $\mathrm{T}_{2}=\mathrm{T}_{1}\left(\mathrm{p}_{2} / \mathrm{p}_{1}\right)^{\gamma-1}$

Short Answer

Expert verified
(B) \(\mathrm{T}_{2}=\mathrm{T}_{1}\times\left\{\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}\right\}^{\frac{(\gamma-1)}{\gamma}}\)

Step by step solution

01

Recall the formula for adiabatic process

In an adiabatic process involving an ideal gas, the equation linking the initial and final states of the gas is given by \(\mathrm{T}_{1}\mathrm{P}_{1}^{(\gamma - 1) / \gamma} = \mathrm{T}_{2}\mathrm{P}_{2}^{(\gamma - 1) / \gamma}\) where \(\gamma\) is the ratio of specific heats, given as Cp/Cv.
02

Rearrange equation for final temperature

To find the equation for final temperature \(\mathrm{T}_{2}\) , you rearrange the adiabatic equation derived in Step 1 to give: \(\mathrm{T}_{2}=\mathrm{T}_{1}\times\left\{\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}\right\}^{\frac{(\gamma-1)}{ \gamma}}\)
03

Match the derived equation with the choices given

Your task is to check which equation in the given options matches with the rearranged equation obtained in Step 2 for the final temperature. Using the equation derived in Step 2, you see that the final temperature \(\mathrm{T}_{2} = \mathrm{T}_{1}\times\left\{\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}\right\}^{\frac{(\gamma-1)}{\gamma}}\)
04

Final Answer

Comparing your derived equation with the choice equations, the correct answer is (B) \(\mathrm{T}_{2}=\mathrm{T}_{1}\times\left\{\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}\right\}^{\frac{(\gamma-1)}{\gamma}}\). This matches with the rearranged equation obtained from Step 2. So, the final temperature of the gas after adiabatic compression is given by choice (B).

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

Instructions:Read the assertion and reason carefully to mask the correct option out of the options given below. (A) If both assertion and reason are true and the reason is the correct explanation of the assertion. (B) If both assertion and reason are true but reason is not be correct explanation of assertion. (C) If assertion is true but reason is false. (D) If the assertion and reason both are false. Assertion: The total translation kinetic energy of all the molecules of a given mass of an ideal gas is \(1.5\) times the product of its Pressure and its volume. Reason: The molecules of a gas collide with each other and velocities of the molecules change due to the collision (A) D (B) \(\mathrm{C}\) (C) A (D) B

A gas mixture consists of 2 mole of oxygen and 4 mole of argon at temperature \(\mathrm{T}\). Neglecting all vibrational modes, the total internal energy of the system is (A) \(11 \mathrm{RT}\) (B) \(9 \mathrm{RT}\) (C) \(15 \mathrm{RT}\) (D) \(4 \mathrm{RT}\)

Instructions:Read the assertion and reason carefully to mask the correct option out of the options given below. (A) If both assertion and reason are true and the reason is the correct explanation of the assertion. (B) If both assertion and reason are true but reason is not be correct explanation of assertion. (C) If assertion is true but reason is false. (D) If the assertion and reason both are false. Assertion: The carnot is useful in understanding the performance of heat engine Reason: The carnot cycle provides a way of determining the maximum possible efficiency achievable with reservoirs of given temperatures. (A) \(\mathrm{A}\) (B) B (C) \(\mathrm{C}\) (D) \(\mathrm{D}\)

Instructions:Read the assertion and reason carefully to mask the correct option out of the options given below. (A) If both assertion and reason are true and the reason is the correct explanation of the assertion. (B) If both assertion and reason are true but reason is not be correct explanation of assertion. (C) If assertion is true but reason is false. (D) If the assertion and reason both are false. Assertion: The melting point of ice decreases with increase of Pressure Reason: Ice contracts on melting. (A) \(\mathrm{C}\) (B) \(\mathrm{B}\) (C) \(\mathrm{A}\) (D) D

Starting with the same initial Conditions, an ideal gas expands from Volume \(\mathrm{V}_{1}\) to \(\mathrm{V}_{2}\) in three different ways. The Work done by the gas is \(\mathrm{W}_{1}\) if the process is purely isothermal, \(\mathrm{W}_{2}\) if purely isobaric and \(\mathrm{W}_{3}\) if purely adiabatic Then (A) \(\mathrm{W}_{2}>\mathrm{W}_{1}>\mathrm{W}_{3}\) (B) \(\mathrm{W}_{2}>\mathrm{W}_{3}>\mathrm{W}_{1}\) (C) \(\mathrm{W}_{1}>\mathrm{W}_{2}>\mathrm{W}_{3}\) (D) \(\mathrm{W}_{1}>\mathrm{W}_{3}>\mathrm{W}_{2}\)
See all solutions

Recommended explanations on English Textbooks

Summary Text

Read Explanation

English Grammar

Read Explanation

Multiple Choice Questions

Read Explanation

Lexis and Semantics

Read Explanation

International English

Read Explanation

Semiotics

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