Question 734: Thermodynamics & KTG

Question

A monoatomic ideal gas, initially at temperature $T_1$ is enclosed in a cylinder fitted with frictionless piston. The gas is allowed to expand adiabatically to a temperature $T_2$ by releasing the piston suddenly. $L_1$ and $L_2$ are the lengths of the gas columns before and after the expansion respectively. The ratio $T_2 / T_1$ is

Accepted Answer

For an adiabatic process involving an ideal gas, the relationship between initial and final states is given by:
$ \mathrm{T}_1 \mathrm{V}_1^{\gamma-1} = \mathrm{T}_2 \mathrm{V}_2^{\gamma-1} $
From this, the ratio of the final to initial temperatures is:
$ \frac{\mathrm{T}_2}{\mathrm{T}_1} = \left(\frac{\mathrm{V}_1}{\mathrm{V}_2}ight)^{\gamma-1} $
For a monoatomic gas, the adiabatic index $\gamma$ is $\frac{5}{3}$. Thus:
$ \gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3} $
The volume of the gas is calculated using the area of the base and the length of the gas column. Thus:
$ \mathrm{V}_1 = \mathrm{A}L_1 \quad 	ext{and} \quad \mathrm{V}_2 = \mathrm{A}L_2 $
By substituting these into the expression for $\frac{\mathrm{T}_2}{\mathrm{T}_1}$:
$ \frac{\mathrm{T}_2}{\mathrm{T}_1} = \left(\frac{\mathrm{A}L_1}{\mathrm{A}L_2}ight)^{\frac{2}{3}} = \left(\frac{L_1}{L_2}ight)^{\frac{2}{3}} $
Thus, the ratio of the final to initial temperature after adiabatic expansion is:
$ \frac{\mathrm{T}_2}{\mathrm{T}_1} = \left(\frac{\mathrm{L}_1}{\mathrm{L}_2}ight)^{\frac{2}{3}} $

Answer

$\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^{2 / 3}$

Answer

$\left[\frac{\mathrm{L}_2}{\mathrm{~L}_1}\right]^{2 / 3}$

Answer

$\left[\frac{L_2}{L_1}\right]^{1 / 2}$

Answer

$\left[\frac{L_1}{L_2}\right]^{1 / 2}$

Thermodynamics & KTG

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