The mean free path of a molecule is 2.25 × 10-7 m in an ideal gas at standard temperature and pressure (0°C and 1 atm). What is the mean free path if the diameter of the molecule and the temperature of the gas are both doubled?
Question
The mean free path of a molecule is 2.25 × 10-7 m in an ideal gas at standard temperature and pressure (0°C and 1 atm). What is the mean free path if the diameter of the molecule and the temperature of the gas are both doubled?
Solution
The mean free path (λ) of a molecule in a gas is given by the formula:
λ = kT / (sqrt(2) * π * d^2 * P)
where:
- k is the Boltzmann constant,
- T is the temperature,
- d is the diameter of the molecule, and
- P is the pressure.
In this case, we are told that the diameter (d) and the temperature (T) are both doubled. Therefore, the new mean free path (λ') will be:
λ' = k(2T) / (sqrt(2) * π * (2d)^2 * P) = 2kT / (2sqrt(2) * π * d^2 * P) = λ / 2
So, the mean free path will be halved when both the diameter of the molecule and the temperature of the gas are doubled. Therefore, the new mean free path is 1.125 × 10^-7 m.
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