At T(K), the root mean square (rms) velocity | vedclass

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

Question

(A) The root mean square velocity $(u_{rms})$ is given by the formula: $u_{rms} = \sqrt{\frac{3RT}{M}}$.Given $u_{rms} = 20 \ ms^{-1}$ and $M = 40 \ g

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) The root mean square velocity $(u_{rms})$ is given by the formula: $u_{rms} = \sqrt{\frac{3RT}{M}}$.Given $u_{rms} = 20 \ ms^{-1}$ and $M = 40 \ g \ mol^{-1} = 0.04 \ kg \ mol^{-1}$.Squaring both sides: $u_{rms}^2 = \frac{3RT}{M} \implies 20^2 = \frac{3RT}{0.04}$.$400 = \frac{3RT}{0.04} \implies 3RT = 400 	imes 0.04 = 16$.$RT = \frac{16}{3} \ J \ mol^{-1}$.The average kinetic energy $(KE_{avg})$ for one mole of an ideal gas is given by: $KE_{avg} = \frac{3}{2}RT$.Substituting the value of $RT$: $KE_{avg} = \frac{3}{2} 	imes \frac{16}{3} = 8 \ J \ mol^{-1}$.Thus, the correct option is $A$.

At $T(K)$, the root mean square (rms) velocity of argon (molar mass $40 \ g \ mol^{-1}$) is $20 \ ms^{-1}$. The average kinetic energy of the same gas at $T(K)$ (in $J \ mol^{-1}$) is:

Similar Questions

At $300 \ K$ the average velocity of a gas is $3 \times 10^2 \ cm / s$. The average velocity (in $cm / s$) of it at $1200 \ K$ is

Explain the speed of molecules of a gas and the average speed $(u_{av})$.

At $400 \ K$,the root mean square $(rms)$ speed of a gas $X$ (molecular weight $= 40$) is equal to the most probable speed of gas $Y$ at $60 \ K$. The molecular weight of the gas $Y$ is:

Explain the relationship between different types of molecular speeds.

For $1 \ mol$ of a gas,the average kinetic energy is $E$. Then the root mean square velocity $U_{rms}$ of the gas is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module