Question

4. The mean free path of molecules in a gas is $200\mathrm{nm}$.

a. What will be the mean free path if the pressure is doubled while the temperature is held constant?

b. What will be the mean free path if the absolute temperature is doubled while the pressure is held constant?

Short answer

a) The mean free path will reduce by half when the pressure is doubled, making it $100nm$in width.

b) As the temperature rises, the mean free path will double to$400nm.$

Step-by-step solution

Given Information (Part a) 

The mean free path of molecules in a gas is$200nm.$

Explanation (Part a) 

Mean free path $\lambda$is given as,

$\lambda =\frac{1}{4\sqrt{2}\pi N/V{r}^2}$

The number density is, however, inversely proportional to it. The number density in an ideal gas, as you have already seen in exercise 1, is inversely proportional to the number density.

$\frac{N}{V}=\frac{p{N}_A}{RT}$

The number density of a system is proportional to the pressure and inversely proportional to its temperature, according to the equation. Hence, the mean free path will depend on temperature and pressure in reverse proportion.

Final Answer (Part a)

Hence, the mean free path will reduce by half when the pressure is doubled, making it $100nm$in width.

Given Information (Part b)  

The mean free path of molecules in a gas is $200nm$.

Explanation (Part b)  

Mean free path $\lambda$is given as,

$\Lambda =\frac{1}{4\sqrt{2}\pi N/V{r}^2}$

The number density is, however, inversely proportional to it. The number density in an ideal gas, as you have already seen in exercise 1, is inversely proportional to the number density.

$\frac{N}{V}=\frac{p{N}_A}{RT}$

The number density of a system is proportional to the pressure and inversely proportional to its temperature, according to the equation. Hence, the mean free path will depend on temperature and pressure in reverse proportion.

Final Answer (Part b)

Hence, As the temperature rises, the mean free path will double to$400nm.$