Question

A monoatomic ideal gas, initially at temperature $T_1$. is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to temperature $T_2$ by releasing the piston suddenly. If $L_1$ and $L_2$ are the lengths of the gas column before and after expansion respectively, then $\frac{T_1}{T_2}$ is given by

• A. $(\frac{L_1}{L_2}{)}^{2/3}$
• B. $\frac{L_1}{L_2}$
• C. $\frac{L_2}{L_1}$
• D. $(\frac{L_2}{L_1}{)}^{2/3}$

Answer 1

The correct option is D $(\frac{L_2}{L_1}{)}^{2/3}$

Let the base area of cylinder be $n$ then

the volume of gas before expansion $=L_{1}A$

and after expansion $=L_{2}A$

Since gas is monoatomic

$\overline{r}=\frac{C_p}{C_v}=\frac{5}{3}$

We know that adiabatic process is given by

$T{V}^{r-1}=$ const

$\Rightarrow T_1{V}_1^{r-1}=T_2{V}_2^{r-1}$

$\Rightarrow \frac{T_1}{T_2}={\left(\frac{v_2}{v_1}\right)}^{r-1}={\left(\frac{A{L}_2}{A{L}_1}\right)}^{5/3-1}$

$\Rightarrow \frac{T_1}{T_2}={\left(\frac{L_2}{L_1}\right)}^{2/3}$