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 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. Then $\frac{T_2}{T_1}$ is

• A. $\left(\frac{L_2}{L_1}\right)^{2/3}$
• B. $\left(\frac{L_1}{L_2}\right)^{2/3}$
• C. $\left(\frac{L_1}{L_2}\right)^{1/2}$
• D. $\left(\frac{L_2}{L_1}\right)^{1/2}$

Hint:

During gas expansions, volume increases ($L_2 > L_1$) which always causes temperature to drop ($T_2 < T_1$). Therefore, the ratio $\frac{T_2}{T_1}$ must evaluate to a value less than 1. This means the fractional core must put the smaller length on top, instantly filtering out options (A) and (D).

Answer 1

The Correct Option is B

Step 1: Convert lengths into volumes early.
In a uniform cylinder the cross-section $A$ is fixed, so volume scales with column length: $V = A L$, giving $\dfrac{V_1}{V_2} = \dfrac{L_1}{L_2}$. We will work the adiabatic law in terms of lengths.
Step 2: Adiabatic temperature-volume law.
For an adiabatic process, $T V^{\gamma-1} = \text{constant}$, so $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Step 3: Isolate the temperature ratio.
$\dfrac{T_2}{T_1} = \left(\dfrac{V_1}{V_2}\right)^{\gamma-1} = \left(\dfrac{L_1}{L_2}\right)^{\gamma-1}$.
Step 4: Use the monoatomic value of $\gamma$.
For a monoatomic ideal gas, $\gamma = \dfrac{5}{3}$, so $\gamma - 1 = \dfrac{2}{3}$.
Step 5: Substitute the exponent.
$\dfrac{T_2}{T_1} = \left(\dfrac{L_1}{L_2}\right)^{2/3}$.
Step 6: Sanity check and conclude.
Expansion means $L_2 > L_1$, so the ratio is less than one, i.e. the gas cools, as expected for adiabatic expansion. \[ \boxed{\frac{T_2}{T_1} = \left(\frac{L_1}{L_2}\right)^{2/3}} \]