The pressure exerted by an ideal gas at a par | vedclass

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

Question

(B) According to the kinetic theory of gases,the pressure $P$ exerted by an ideal gas is given by the formula: $P = \frac{1}{3} \rho v_{rms}^2$,where

Vedclass · Accepted Answer

mean of the square of the speed of the gas molecules.

Vedclass · Answer

(B) According to the kinetic theory of gases,the pressure $P$ exerted by an ideal gas is given by the formula: $P = \frac{1}{3} \rho v_{rms}^2$,where $\rho$ is the density and $v_{rms}$ is the root mean square speed.Since $\rho = \frac{M}{V}$,we can write $P = \frac{1}{3} \frac{N m}{V} v_{rms}^2$.Also,the mean square speed is defined as $\langle v^2 \rangle = \frac{1}{N} \sum v_i^2 = v_{rms}^2$.Therefore,the pressure $P$ is directly proportional to the mean of the square of the speeds of the gas molecules,which is $\langle v^2 \rangle$.

The pressure exerted by an ideal gas at a particular temperature is directly proportional to

Similar Questions

The average kinetic energy of a monatomic molecule is $0.414 eV$ at temperature : (Use $K_{B} = 1.38 \times 10^{-23} J/K$) (in $K$)

The average kinetic energy of $H_2$ at $300 \, K$ is $E$. What is the average kinetic energy of $O_2$ at the same temperature?

The initial average kinetic energy of the molecules was $E$,when a gas sample is at $27^{\circ} C$. When the gas is heated to $327^{\circ} C$,then the final average kinetic energy will be

The ratio of the mean kinetic energy of hydrogen and nitrogen at temperatures $300 \, K$ and $450 \, K$ respectively is:

The kinetic energy of one gm-mole of a gas at normal temperature and pressure is $(R = 8.31 \text{ J/mol-K})$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module