Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • A monatomic ideal gas, initially at temperature , is enclosed in a cylinder fitted with a friction

A monatomic ideal gas, initially at temperature \mathrm{T_1}, is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature \mathrm{T_2} by releasing the piston suddenly. If  \mathrm{\mathrm{L}_1 \: and \: \mathrm{L}_2} are the lengths of the gas column, before and after expansion, respectively, then \mathrm{T_1 / T_2} is given by
 

Option: 1

\mathrm{\left(\frac{L_1}{L_2}\right)^{2 / 3}}


 


Option: 2

\mathrm{\left(\frac{L_1}{L_2}\right)}
 


Option: 3

\mathrm{\left(\frac{L_2}{L_1}\right)}
 


Option: 4

\mathrm{\left(\frac{L_2}{L_1}\right)^{2 / 3}}


Answers (1)

best_answer

For adiabatic process,

\mathrm{ \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\mathrm{y}-1} }

Now, for gas column \mathrm{\mathrm{V}_1=\mathrm{AL}_1 \: and \: \mathrm{V}_2=\mathrm{AL}_2}

\mathrm{ \Rightarrow \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\mathrm{L}_1}{\mathrm{~L}_2} }

\mathrm{So, \quad \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{\gamma-1}=\left(\frac{\mathrm{L}_2}{\mathrm{~L}_1}\right)^{\gamma-1}}

Now, for a monatomic gas \gamma=5 / 3

\mathrm{ \therefore \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\left(\frac{\mathrm{L}_2}{\mathrm{~L}_1}\right)^{\frac{2}{3}} }
 

Posted by

Nehul

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

Karnataka: NEET candidate dies by suicide; no note recovered
anam-khan

CBSE Re-evaluation Row: Dharmendra Pradhan seeks detailed report on technical failures
anam-khan

Before NEET, CMC Vellore’s unique MBBS admissions tested aptitude along with merit; paper-leak restarts debate

Latest Question