In a sample of chlorine gas at 300 K,the ave | vedclass

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

Question

(D) The average kinetic energy $(E)$ of a gas molecule is directly proportional to the absolute temperature $(T)$: $E \propto T$.Given $E_1 = 6.21 	i

Vedclass · Accepted Answer

$12.42 	imes 10^{-21} \ J, \ 459.6 \ m/s$

Vedclass · Answer

(D) The average kinetic energy $(E)$ of a gas molecule is directly proportional to the absolute temperature $(T)$: $E \propto T$.Given $E_1 = 6.21 	imes 10^{-21} \ J$ at $T_1 = 300 \ K$. At $T_2 = 600 \ K$,the new kinetic energy $E_2$ is:$E_2 = E_1 	imes (T_2 / T_1) = 6.21 	imes 10^{-21} 	imes (600 / 300) = 12.42 	imes 10^{-21} \ J$.The root mean square speed $(
u_{rms})$ is directly proportional to the square root of the absolute temperature: $
u_{rms} \propto \sqrt{T}$.Given $(
u_{rms})_1 = 325 \ m/s$ at $T_1 = 300 \ K$. At $T_2 = 600 \ K$,the new speed $(
u_{rms})_2$ is:$(
u_{rms})_2 = (
u_{rms})_1 	imes \sqrt{T_2 / T_1} = 325 	imes \sqrt{600 / 300} = 325 	imes \sqrt{2} \approx 325 	imes 1.414 = 459.6 \ m/s$.

In a sample of chlorine gas at $300 \ K$,the average kinetic energy per molecule is $6.21 \times 10^{-21} \ J$ and the root mean square speed $\nu_{rms}$ is $325 \ m/s$. What will be the values of these quantities at $600 \ K$?

Similar Questions

The $v_{rms}$ of gas molecules in a container is $400 \ ms^{-1}$. If half of the gas leaks out of the container at a constant temperature,the $v_{rms}$ of the remaining gas molecules will be . . . . . . $ms^{-1}$.

At what temperature will the root mean square $(rms)$ speed of oxygen molecules be sufficient to escape from the Earth's atmosphere? (Mass of an oxygen molecule $m = 2.76 \times 10^{-26} \, kg$,Boltzmann constant $k_B = 1.38 \times 10^{-23} \, JK^{-1}$,Escape velocity of Earth $v_e = 11.2 \, km/s$)

The ratio of vapour densities of two gases at the same temperature is $\frac{4}{25}$. What is the ratio of their $r.m.s.$ velocities?

The temperature of an ideal gas is increased from $120 \ K$ to $480 \ K.$ If at $120 \ K,$ the root mean square velocity of the gas molecules is $v,$ at $480 \ K$ it becomes

In thermal equilibrium,the average velocity of gas molecules is

Vedclass Test Series

Exam Paper Generator

Online Exam Module