Assuming the expression for the pressure exer | vedclass

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

Question

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

Vedclass · Accepted Answer

$(2/3)$ of kinetic energy per unit volume of a gas.

Vedclass · Answer

(B) The pressure $P$ exerted by an ideal gas is given by the kinetic theory of gases as:$P = \frac{1}{3} \rho v_{rms}^2$where $\rho = \frac{M}{V}$ is the density of the gas and $v_{rms}$ is the root-mean-square speed.Substituting $\rho$ in the equation:$P = \frac{1}{3} \left( \frac{M}{V} \right) v_{rms}^2$Multiply and divide the right side by $2$:$P = \frac{2}{3} \left( \frac{1}{2} \frac{M}{V} v_{rms}^2 \right)$Since the total kinetic energy $K.E. = \frac{1}{2} M v_{rms}^2$,the kinetic energy per unit volume is $u = \frac{K.E.}{V} = \frac{1}{2} \rho v_{rms}^2$.Therefore,$P = \frac{2}{3} u$.

Assuming the expression for the pressure exerted by the gas,it can be shown that pressure is

Similar Questions

Estimate the average thermal energy of a helium atom at:
$(i)$ room temperature $(27^{\circ} C)$,
$(ii)$ the temperature on the surface of the Sun $(6000\; K)$,
$(iii)$ the temperature of $10$ million kelvin (the typical core temperature in the case of a star).

What is the kinetic energy of one gram mole of a gas at standard temperature and pressure? $(R = 8.31 \ J/mol \cdot K)$

In a vessel,the gas is at pressure $P_0$. If the mass of all the molecules is halved and their speed is doubled,then the resultant pressure will be

The mean kinetic energy of a gas at $300\, K$ is $100\, J.$ The mean energy of the gas at $450\, K$ is equal to ...... $J$

On the basis of the kinetic theory of gases,the gas exerts pressure because its molecules:

Vedclass Test Series

Exam Paper Generator

Online Exam Module