If the pressure in a closed vessel is reduced | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The mean free path $\lambda$ of a gas molecule is given by the formula $\lambda = \frac{1}{\sqrt{2} \pi d^2 n}$,where $d$ is the diameter of the m

Vedclass · Accepted Answer

Is increased

Vedclass · Answer

(B) The mean free path $\lambda$ of a gas molecule is given by the formula $\lambda = \frac{1}{\sqrt{2} \pi d^2 n}$,where $d$ is the diameter of the molecule and $n$ is the number density (number of molecules per unit volume).Since the pressure $P$ of an ideal gas is related to the number density by $P = nkT$,reducing the pressure at a constant temperature implies that the number density $n$ decreases.As $n$ is in the denominator of the expression for $\lambda$,a decrease in $n$ leads to an increase in the mean free path $\lambda$.Therefore,when gas is drawn out of a closed vessel,the mean free path of the molecules increases.

If the pressure in a closed vessel is reduced by drawing out some gas,the mean free path of the molecules

Similar Questions

The mean free path $\ell$ for a gas molecule depends upon the diameter $d$ of the molecule as:

The vapour of a substance behaves as a gas:

The change in volume $V$ with respect to an increase in pressure $P$ has been shown in the figure for a non-ideal gas at four different temperatures ${T_1}, {T_2}, {T_3}$ and ${T_4}$. The critical temperature of the gas is

The mean free path of a molecule of $He$ gas is $\alpha$. Its mean free path along any arbitrary coordinate axis will be ...........

Vedclass Test Series

Exam Paper Generator

Online Exam Module