The pressure exerted by a gas is P_0. If the | vedclass

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

Question

(C) The pressure exerted by an ideal gas is given by the formula: $P = \frac{1}{3} \rho v_{rms}^2 = \frac{1}{3} \frac{M}{V} v_{rms}^2$.Since $P \propt

Vedclass · Accepted Answer

$2P_0$

Vedclass · Answer

(C) The pressure exerted by an ideal gas is given by the formula: $P = \frac{1}{3} ho v_{rms}^2 = \frac{1}{3} \frac{M}{V} v_{rms}^2$.Since $P \propto M \cdot v^2$,where $M$ is the total mass of the gas and $v$ is the root mean square velocity.Given: $M_2 = \frac{M_1}{2}$ and $v_2 = 2v_1$.Therefore,the ratio of pressures is: $\frac{P_2}{P_1} = \frac{M_2}{M_1} \left( \frac{v_2}{v_1} ight)^2$.Substituting the values: $\frac{P_2}{P_0} = \left( \frac{M_1/2}{M_1} ight) \left( \frac{2v_1}{v_1} ight)^2 = \frac{1}{2} 	imes (2)^2 = \frac{1}{2} 	imes 4 = 2$.Thus,$P_2 = 2P_0$.

The pressure exerted by a gas is $P_0$. If the mass of the molecules is halved and their velocity is doubled,the pressure will become .......

Similar Questions

If the number of molecules of ${H_2}$ is double that of ${O_2}$,then the ratio of the kinetic energy of hydrogen to that of oxygen at $300 \, K$ is:

An ideal gas is kept in a cylinder of volume $3 \,m^3$ at a pressure of $3 \times 10^5 \,Pa$. The energy of the gas is

At which of the following temperatures would the molecules of a gas have twice the average kinetic energy they have at $20^{\circ}C$ (in $^{\circ}C$)?

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

The relation between the gas pressure $P$ and average kinetic energy per unit volume $E$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module