When the temperature of a gas is raised from | vedclass

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

Question

(A) The root mean square (rms) speed of a gas molecule is given by the formula $v = \sqrt{\frac{3RT}{M}}$.Since $R$ and $M$ are constants,we have $v \

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) The root mean square (rms) speed of a gas molecule is given by the formula $v = \sqrt{\frac{3RT}{M}}$.Since $R$ and $M$ are constants,we have $v \propto \sqrt{T}$.Initial temperature $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$.Final temperature $T_2 = 90^{\circ} C = 90 + 273 = 363 \ K$.The ratio of the rms speeds is $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{363}{300}} = \sqrt{1.21} = 1.1$.The percentage increase in rms speed is given by $\frac{v_2 - v_1}{v_1} 	imes 100 = (\frac{v_2}{v_1} - 1) 	imes 100$.Substituting the value,we get $(1.1 - 1) 	imes 100 = 0.1 	imes 100 = 10 \%$.

When the temperature of a gas is raised from $27^{\circ} C$ to $90^{\circ} C$,the increase in the rms velocity of the gas molecules is: (in $\%$)

Similar Questions

Equal volumes of two gases,having their densities in the ratio of $1: 16$,exert equal pressures on the walls of two containers. The ratio of their rms speeds $(c_1: c_2)$ is

At a given temperature,the ratio of $r.m.s.$ velocities of a hydrogen molecule and a helium atom will be:

Define absolute temperature.

The r.m.s. speed of the molecules of a gas at $100^{\circ} C$ is $v$. The temperature at which the r.m.s. speed will be $\sqrt{3} v$ is: (in $^{\circ} C$)

If $V_H, V_N$,and $V_O$ represent the $rms$ speeds of hydrogen,nitrogen,and oxygen molecules at a given temperature,respectively,then which of the following is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module