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

Discuss the main points to be considered while deriving the equation of pressure on the basis of the kinetic theory of gases.

The average force applied on the walls of a closed container depends as $T^{x}$ where $T$ is the temperature of an ideal gas. The value of $x$ is

Two gases are at absolute temperatures $300 \ K$ and $350 \ K$ respectively. The ratio of the average kinetic energy of their molecules is:

The graph which represents the variation of mean kinetic energy of molecules with temperature $t^{\circ} \text{C}$ is

The internal energy of air in a $4 \ m \times 4 \ m \times 3 \ m$ sized room at $1$ atmospheric pressure will be . . . . . . $\times 10^6 \ J$. (Consider air as a diatomic molecule)

Vedclass Test Series

Exam Paper Generator

Online Exam Module