Question 677: Thermodynamics & KTG

Question

A monoatomic ideal gas, initially at temperature $\mathrm{T}_1$ is enclosed in a cylinder fitted with massless, frictionless piston. By releasing the piston suddenly, the gas is allowed to expand adiabatically to a temperature $\mathrm{T}_2$. If $\mathrm{L}_1$ and $\mathrm{L}_2$ are the lengths of the gas columns before and after expansion respectively, then $\left(T_2 / T_1\right)$ is given by

Accepted Answer

Adiabatic process formula:
For an adiabatic process, the following formula holds:
$$ \mathrm{TV}^{\gamma-1}= 	ext{constant} $$

This means:
$$ \frac{\mathrm{T}_2}{\mathrm{T}_1} = \left(\frac{\mathrm{V}_1}{\mathrm{V}_2}ight)^{\gamma-1} $$
Value of $\gamma$ for monoatomic gas:
For a monoatomic gas, $\gamma = \frac{5}{3}$.
Calculate $\gamma-1$:
$\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$
Relating volume to length:
The volume of gas is calculated as the area ($A$) of the cylinder times the length ($L$) of the gas column.
So, $\mathrm{V}_1 = \mathrm{A} \mathrm{L}_1$ and $\mathrm{V}_2 = \mathrm{A} \mathrm{L}_2$.
Substitute in the temperature formula:
Now, substitute these volumes into the earlier formula:
$$ \frac{\mathrm{T}_2}{\mathrm{T}_1} = \left(\frac{\mathrm{A} \mathrm{L}_1}{\mathrm{A} \mathrm{L}_2}ight)^{2/3} $$

The area ($A$) cancels out:

$$ \frac{\mathrm{T}_2}{\mathrm{T}_1} = \left(\frac{\mathrm{L}_1}{\mathrm{L}_2}ight)^{2/3} $$
Final result:
So, the ratio of the final and initial temperature is:
$$ \frac{\mathrm{T}_2}{\mathrm{T}_1} = \left( \frac{\mathrm{L}_1}{\mathrm{L}_2} ight)^{2/3} $$

Answer

$\frac{\mathrm{L}_1}{\mathrm{~L}_2}$

Answer

$\frac{\mathrm{L}_2}{\mathrm{~L}_1}$

Answer

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

Answer

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

Thermodynamics & KTG

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