The root mean square (rms) speed of X_2 gas i | vedclass

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

Question

(C) The formula for root mean square speed is $C_{rms} = \sqrt{\frac{3RT}{M}}$.Initially,for $X_2$ gas at temperature $T$,$C_1 = x = \sqrt{\frac{3RT}{

Vedclass · Accepted Answer

$2x$

Vedclass · Answer

(C) The formula for root mean square speed is $C_{rms} = \sqrt{\frac{3RT}{M}}$.Initially,for $X_2$ gas at temperature $T$,$C_1 = x = \sqrt{\frac{3RT}{M_{X_2}}}$.When temperature is doubled $(T_2 = 2T)$ and $X_2$ dissociates into $2X$,the molar mass becomes $M_2 = \frac{M_{X_2}}{2}$.The new rms speed $C_2$ is given by $C_2 = \sqrt{\frac{3R(2T)}{M_{X_2}/2}} = \sqrt{4 	imes \frac{3RT}{M_{X_2}}} = 2 	imes \sqrt{\frac{3RT}{M_{X_2}}}$.Substituting $x$ for the initial speed,we get $C_2 = 2x \ m/s$.

The root mean square (rms) speed of $X_2$ gas is $x \ m/s$ at a given temperature. When the temperature is doubled,the $X_2$ molecules dissociate completely into atoms. The root mean square speed of the sample of gas then becomes (in $m/s$):

Similar Questions

Which of the following is correct regarding molecular speeds?

Explain the effect of the mass of a molecule on the Maxwell-Boltzmann speed distribution.

The ratio between the most probable speed of $N_2$ at $400 \ K$ and $CO$ at $800 \ K$ is (molar mass of $N_2 = 28 \ g \ mol^{-1}$,molar mass of $CO = 28 \ g \ mol^{-1}$)

If the average velocity of $N_2$ molecule is $0.3 \, m/s$ at $27 \, ^\circ C$,then the velocity will be $0.6 \, m/s$ at ........... $K$.

The relationship among most probable velocity,average velocity,and root mean square velocity is respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module