Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard
  1. Home
  2. Engineering
  3. I have a doubt in this
# Engineering

‘n’ moles of an ideal gas undergoes a process A→B as shown in the figure.  The maximum temperature of the gas during the process will be :

Answers (1)
S Sayak

P as a function of V for the given process would be as follows

\\\frac{P-P_{0}}{V-V_{0}}=\frac{2P_{0}-P_{0}}{V_{0}-2V_{0}}\\ P=-\frac{P_{0}}{V_{0}}V+2P_{0}

Temperature is given as

\\T=\frac{PV}{nR}\\ T=\frac{P_{0}}{nR}\left (- \frac{V^{2}}{V_{0}}+2V\right )

To find the point where temperature would be maximum we see where the slope is 0

\\\frac{dT}{dV}=0\\ \frac{d}{dV}\left ( \frac{P_{0}}{nR}\left (- \frac{V^{2}}{V_{0}}+2V\right ) \right )=0\\ \frac{d}{dV}\left (-\frac{V^{2}}{V_{0}}+2V \right )=0 \\-\frac{2V}{V_{0}}+2=0\\ V=V_{0}

Maximum temperature is therefore at point (V0, 2P0)

T_{max}=\frac{2P_{0}V_{0}}{nR}

Similar Questions

Latest Asked Questions

Most Viewed Questions

Preparation Products

Knockout JEE Main (Six Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 9999/- ₹ 8499/-
Buy Now
Knockout JEE Main (Nine Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 13999/- ₹ 12499/-
Buy Now
Test Series JEE Main 2024

Chapter/Subject/Full Mock Tests for JEE Main, Personalized Performance Report, Weakness Sheet, Complete Answer Key,.

₹ 7999/- ₹ 4999/-
Buy Now
JEE Main Rank Booster 2023

Booster and Kadha Video Lectures, Unlimited Full Mock Test, Adaptive Time Table, Faculty Support.

₹ 9999/- ₹ 6999/-
Buy Now
JEE Foundation + Knockout JEE Main 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports,.

₹ 3999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
Start Now
 
Exams
Articles
Questions