By using which equation can the force acting | vedclass

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

Question

(A) According to the kinetic theory of gases,the pressure $P$ exerted by an ideal gas on the walls of its container is given by the relation: $P = \fr

Vedclass · Accepted Answer

$P = \frac{1}{3} \rho v_{rms}^2$

Vedclass · Answer

(A) According to the kinetic theory of gases,the pressure $P$ exerted by an ideal gas on the walls of its container is given by the relation: $P = \frac{1}{3} \frac{N m}{V} v_{rms}^2$,where $N$ is the number of molecules,$m$ is the mass of each molecule,$V$ is the volume,and $v_{rms}$ is the root mean square speed. Since pressure is defined as force per unit area $(P = F/A)$,the force $F$ exerted on a wall of area $A$ is $F = P 	imes A$. Substituting the expression for pressure,we get $F = (\frac{1}{3} \frac{N m}{V} v_{rms}^2) 	imes A$. Thus,the force is derived from the fundamental kinetic theory pressure equation $P = \frac{1}{3} ho v_{rms}^2$.

By using which equation can the force acting on the walls of a container due to gas molecules be obtained?

Similar Questions

The kinetic energy of $1$ gram-mole of Argon at $127^{\circ}C$ is .... $J$.

If the average translational kinetic energy of a molecule in a gas is equal to the kinetic energy of an electron accelerating from rest through $10 \,V$, then the temperature of the gas molecule is (Boltzmann constant $= 1.38 \times 10^{-23} \,JK^{-1}$)

On colliding in a closed container,the gas molecules:

The relationship between pressure $P$,volume $V$,and total internal energy $E$ for an ideal gas is:

$A$ gas is contained in a vessel at pressure $P_0$. If the mass of all molecules is halved and their speed is doubled,what will be the final pressure?

Vedclass Test Series

Exam Paper Generator

Online Exam Module