A monatomic ideal gas initially at a certain pressure has a mean free path ₁ . It undergoes a reversible adiabatic expansion until its pressure drops to 1 32 of its initial value. The new mean free path of the gas molecules becomes ₂ . The ratio ₂ ₁ is equal to:

Step-by-step solution

The mean free path is given by: = 1 2 d^2 n_V Since the number density n_V = N V , for a fixed amount of gas, the mean free path is directly proportional to the volume: V For a reversible adiabatic process, the relationship between pressure and volume is: P V^ = constant V P^ -1/ For a monatomic ideal gas, the ratio of specific heats is = 5 3 . Therefore, V P^ -3/5 , which implies P^ -3/5 . The ratio of the final mean free path to the initial mean free path is: ₂ ₁ = ( P₂ P₁ )^ -3/5 Given that P₂ = P₁ 32 , we have