A monoatomic ideal gas, initially at temperature T₁ is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T₂ by releasing the piston suddenly. If l₁ and l₂ are the lengths of the gas column, before and after the expansion respectively, then the value of T₁/T₂ will be :

Solution

Physics sir · 4 steps

Volume tracks the column length, so put V ∝ l into TV^(γ-1) = constant and you are done.

1. Adiabatic relation. For an adiabatic process on an ideal gas, TV^(γ-1) stays constant. For a monoatomic gas γ = 5/3, so γ - 1 = 2/3.
2. Volume from length. The cylinder has a fixed cross-section A, so V = Al. The constant A cancels, and V can be replaced by l.

   T₁ l₁^(2/3) = T₂ l₂^(2/3) ⇒ T₁/T₂ = (l₂/l₁)^(2/3)

3. Sense check. The gas expands, so l₂ > l₁ and the ratio is greater than 1. That means T₁ > T₂, which fits: an adiabatic expansion cools the gas.
4. So. T₁/T₂ = (l₂/l₁)^(2/3), which is option (b).

The idea to keep. In an adiabatic process, temperature and volume move in opposite directions through TV^(γ-1) = constant. Swap V for whatever is proportional to it, here the length, and keep the exponent γ - 1.

Answer from the NTA final answer key. Solution by PrepWiser.