At temperature T,N molecules of gas A each ha | vedclass

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

Question

(D) For an ideal gas at temperature $T$,the average kinetic energy per molecule is given by $\frac{1}{2} m v_{rms}^2 = \frac{3}{2} k_B T$.For gas $A$

Vedclass · Accepted Answer

$2/3$

Vedclass · Answer

(D) For an ideal gas at temperature $T$,the average kinetic energy per molecule is given by $\frac{1}{2} m v_{rms}^2 = \frac{3}{2} k_B T$.For gas $A$ with mass $m$ at temperature $T$:The mean square velocity is $\langle v_A^2 angle = \frac{3 k_B T}{m}$.Since the motion is isotropic,the mean square of the $x$-component is $\langle v_{Ax}^2 angle = w^2 = \frac{1}{3} \langle v_A^2 angle = \frac{k_B T}{m}$.For gas $B$ with mass $2m$ at temperature $T$:The mean square velocity is $v^2 = \langle v_B^2 angle = \frac{3 k_B T}{2m}$.Now,calculating the ratio $w^2 / v^2$:$\frac{w^2}{v^2} = \frac{k_B T / m}{3 k_B T / 2m} = \frac{1}{m} 	imes \frac{2m}{3} = \frac{2}{3}$.Thus,the ratio is $2/3$.

Similar Questions

If the temperature of a gas is increased from $27^{\circ} C$ to $159^{\circ} C$, then the percentage increase in the $rms$ speed of the gas molecules is

Let $\overline{V}$, $V_{\text{rms}}$, and $V_{p}$ denote the mean speed, root mean square speed, and most probable speed of the molecules each of mass $m$ in an ideal monoatomic gas at absolute temperature $T$ Kelvin. Which statement$(s)$ is/are correct?

Consider an ideal gas in a closed container at $300 \ K$. The container is then heated so that the average velocity of particles of the gas increases by a factor of $4$. What would be the final temperature (in $^{\circ} C$)?

The temperature of a gas is $-80^{\circ} C$. To what temperature should the gas be heated so that the r.m.s. speed is increased by $2$ times (in $^{\circ} C$)?

If the rms speed of the molecules of a gas at a temperature of $77^{\circ} C$ is $50 \,ms^{-1}$, then the rms speed of the same gas molecules at a temperature of $150.5^{\circ} C$ is (in $\,ms^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module