The temperature at which the rms speed of mol | vedclass

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

Question

(C) The rms speed of molecules in a gas is given by the formula $v_{rms} = \sqrt{\frac{3 k_B T}{m}}$, where $k_B$ is the Boltzmann constant, $T$ is th

Vedclass · Accepted Answer

$927$

Vedclass · Answer

(C) The rms speed of molecules in a gas is given by the formula $v_{rms} = \sqrt{\frac{3 k_B T}{m}}$, where $k_B$ is the Boltzmann constant, $T$ is the absolute temperature in Kelvin, and $m$ is the mass of the molecule.From this relation, we can see that $v_{rms} \propto \sqrt{T}$.Given the initial temperature $T_1 = 27^{\circ} C = (27 + 273) K = 300 K$.Let the initial rms speed be $v_1$. We want the final rms speed $v_2 = 2v_1$ at temperature $T_2$.Using the proportionality $v_{rms} \propto \sqrt{T}$, we have $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.Substituting the values, we get $2 = \sqrt{\frac{T_2}{300}}$.Squaring both sides, $4 = \frac{T_2}{300}$, which gives $T_2 = 1200 K$.Converting back to Celsius, $T_2 = (1200 - 273)^{\circ} C = 927^{\circ} C$.

The temperature at which the rms speed of molecules in hydrogen gas will be double of its initial value at $27^{\circ} C$ is (in $^{\circ} C$)

Similar Questions

The gas having an average speed four times that of $SO_2$ (molecular mass $64$) is

At what temperature is the root mean square velocity of gaseous hydrogen molecules equal to that of oxygen molecules at $47^{\circ}C$ (in $; K$)?

Derive the equation of $v_{rms}$ in terms of the mass of a molecule.

At what temperature $T$ (in $K$) will the $rms$ speed of $H_2$ be equal to the $rms$ speed of $O_2$ at $47^{\circ}C$?

At a pressure of $24 \times 10^5 \, \text{dyne/cm}^2$,the volume of $O_2$ is $10 \, \text{litre}$ and its mass is $20 \, \text{gm}$. The $r.m.s.$ velocity will be ....... $m/s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module