Ask Questions, Get Answers

Want to ask us a question? Click here
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
0 votes

A thermodynamics ideal gas , initially at temperature $\;T_{1}\;$ ia enclosed in a cylinder fitted with a friction less piston . The gas is allowed to expand adiabatically to a temperature $\;T_{2}\;$ by releasing the piston suddenly . If $\;L_{1}\;$ and $\;L_{2}\;$ are the lengths of gas column before and after expansion respectively , then $\;\large\frac{T_{1}}{T_{2}}\;$ is given by


Can you answer this question?

1 Answer

0 votes
Answer : $\;(\large\frac{L_{2}}{L_{1}})^{\large\frac{2}{3}}$
Explanation :
For an adiabatic process
$PV^{r} =Constant$
$TV^{r-1} = Constant$
$T_{1}V_{1}^{r-1} = T_{2} V_{2}^{r-1}$
$\large\frac{T_{1}}{T_{2}} = (\large\frac{V_{2}}{V_{1}})^{r-1} = (\large\frac{L_{2}A}{L_{1}A})^{\large\frac{3}{5} -1}= (\large\frac{L_{2}}{L_{1}})^{\large\frac{2}{3}}$
answered Apr 9, 2014 by yamini.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App