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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

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

$(a)\;(\large\frac{L_{1}}{L_{2}})^{\large\frac{2}{3}}\qquad(b)\;\large\frac{L_{1}}{L_{2}}\qquad(c)\;\large\frac{L_{2}}{L_{1}}\qquad(d)\;(\large\frac{L_{2}}{L_{1}})^{\large\frac{2}{3}}$

1 Answer

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
 

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