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 expands isothermally from a volume and <img alt="\mathrm{V}_{2}" src="https:

An ideal gas expands isothermally from a volume \mathrm{V}_{1} and \mathrm{V}_{2} and then compressed to original volume \mathrm{V}_{1} adiabatically. Initial pressure is \mathrm{P}_{1} and final pressure is \mathrm{P}_{3}.The total work done is \mathrm{W}Then:

Option: 1

\mathrm{P}_{3}>\mathrm{P}_{1}, \mathrm{~W}>0


Option: 2

\mathrm{P}_{3}<\mathrm{P}_{1}, \mathrm{~W}<0


Option: 3

\mathrm{P}_{3}>\mathrm{P}_{1}, \mathrm{~W}<0


Option: 4

\mathrm{P_{3}=P_{1}, W=0}


Answers (1)

best_answer

Slope of adiabatic process at a given state (\mathrm{P}, \mathrm{V}, \mathrm{T}) is more than the slope of isothermal process. The corresponding \mathrm{P}-\mathrm{V}graph for the two process is as shown in figure.

 

In the graph, \mathrm{AB} is isothermal and \mathrm{BC}  is adiabatic.
\mathrm{\mathrm{W}_{\mathrm{AB}}=} positivie (as volume is increasing)
And \mathrm{\mathrm{W}_{\mathrm{BC}}=} negative (as volume is decreasing) plus,

\mathrm{\left|\mathrm{W}_{\mathrm{BC}}\right|>\left|\mathrm{W}_{\mathrm{AB}}\right|,} as area under \mathrm{\mathrm{P}-\mathrm{V}}graph gives the work done.
  \mathrm{Hence, \mathrm{W}_{\mathrm{AB}}+\mathrm{W}_{\mathrm{BC}}=\mathrm{W}<0}

From the graph itself, it is clear that \mathrm{\mathrm{P}_{3}>\mathrm{P}_{1}}.
Hence, the correct option is (c).

 

Posted by

Ajit Kumar Dubey

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

Sigh of relief as JEE Main 2025 session 2 day 1 turns out easier than January exams
anam-khan

JEE Main 2025 April 2 shift 2 exam moderately difficult, math remains easy
anam-khan

JEE Main April 2 Exam Analysis 2025: Shift 1 maths easier compared to session 1; chemistry 'tough'

Latest Question