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

All gases deviate from gas laws at

The value of critical temperature in terms of Vander Waals constants $a$ and $b$ is

Statement-$1$: Real gas approaches ideal gas behaviour for low pressures and high temperatures.
Statement-$2$: At low pressure,the density of a gas is very low.

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