Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • An ideal gas is initially at temperature and volume <img alt="\mathrm{V}" src="https://entrancecorner

An ideal gas is initially at temperature \mathrm{T} and volume \mathrm{V}. It volume is increased by \Delta \mathrm{V} due to an increase in temperature \Delta \mathrm{T}, pressure remaining constant. The quantity \delta=\Delta \mathrm{V} / \mathrm{V} \Delta \mathrm{T} varies with temperature as :

Option: 1


Option: 2


Option: 3


Option: 4


Answers (1)

best_answer

For an ideal gas : \mathrm{PV}=\mathrm{nRT}

\mathrm{Since, \quad \mathrm{P}=\frac{n R T}{V} (given) = constant ( given)}
\mathrm{Therefore, \mathrm{P} \Delta \mathrm{N}=\mathrm{nR} \Delta \mathrm{T}}

\mathrm{\therefore \quad \frac{\Delta V}{D T}=\frac{n R}{P}=\frac{n R}{\frac{n R T}{V}}=\frac{V}{T}}


\mathrm{\therefore \quad \frac{\Delta V}{V \Delta T}=\frac{1}{T} \: or \: \delta=\frac{1}{T}}

Therefore, \delta is inversely proportional to temperature T. i.e., when T increases, \delta decreases and vice-versa.
Hence. \delta -T graph will be a rectangular hyperbola as shown in the above figure.

Posted by

Shailly goel

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: Hardwired to excel? Know BTech electrical engineering cut-offs at NITs
anam-khan

JEE Main 2025 chemistry needs conceptual understanding; 5 tips to score high
anam-khan

JEE Main 2025 exam city intimation slip for session 1 soon; paper pattern, exams from January 22

Latest Question