JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Heat & Thermodynamics

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. If

ℓ_{1}

and

ℓ_{2}

are the lengths of the gas column, before and after the expansion respectively, then the value of

\frac{T _{1}}{T _{2}}

will be:

Answer: (D)

Solution

For an adiabatic process,

T V^{γ - 1} = constant

. Since the cylinder has constant cross-sectional area

A

, the volume is proportional to length:

V = A ℓ

. Therefore,

T_{1} (A ℓ_{1})^{γ - 1} = T_{2} (A ℓ_{2})^{γ - 1}

, which simplifies to

T_{1} ℓ_{1}^{γ - 1} = T_{2} ℓ_{2}^{γ - 1}

. Rearranging:

\frac{T _{1}}{T _{2}} = (\frac{ℓ _{2}}{ℓ _{1}})^{γ - 1}

. For a monoatomic ideal gas,

γ = \frac{5}{3}

, so

γ - 1 = \frac{2}{3}

. Therefore,

\frac{T _{1}}{T _{2}} = (\frac{ℓ _{2}}{ℓ _{1}})^{2/3}

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option