Q175 / 200 · MHT CET - Medical 2026 (24 April Shift 1) · Physics

A mono-atomic 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 gas columns before and after the expansion, respectively. The value of $\dfrac{T_1}{T_2}$ is $\left[\gamma=\dfrac{5}{3}\right]$

  • A
    $\left(\dfrac{L_1}{L_2}\right)^{\frac{2}{3}}$
  • Correct answer
    B
    $\left(\dfrac{L_2}{L_1}\right)^{\frac{2}{3}}$
  • C
    $\left(\dfrac{L_1}{L_2}\right)^{\frac{5}{3}}$
  • D
    $\left(\dfrac{L_2}{L_1}\right)^{\frac{5}{3}}$

Solution

For a reversible adiabatic process, the relationship between temperature and volume is given by $TV^{\gamma - 1} = \text{constant}$.

Applying this to the initial and final states of the gas:

$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$

$\dfrac{T_1}{T_2} = \left(\dfrac{V_2}{V_1}\right)^{\gamma - 1}$

Let $A$ be the cross-sectional area of the cylinder. The volume of the gas column of length $L$ is $V = AL$.

Substituting $V_1 = AL_1$ and $V_2 = AL_2$:

$\dfrac{T_1}{T_2} = \left(\dfrac{AL_2}{AL_1}\right)^{\gamma - 1} = \left(\dfrac{L_2}{L_1}\right)^{\gamma - 1}$

Given that the gas is mono-atomic, $\gamma = \dfrac{5}{3}$.

$\gamma - 1 = \dfrac{5}{3} - 1 = \dfrac{2}{3}$

Substituting this value into the equation:

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

Answer: $\left(\dfrac{L_2}{L_1}\right)^{\frac{2}{3}}$

Hence, the correct answer is option: 'B'.

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