At what temperature will the root mean square | vedclass

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

Question

(B) The root mean square $(rms)$ speed of a gas molecule is given by $v_{rms} = \sqrt{\frac{3k_B T}{m}}$.For the molecules to escape the Earth's atmos

Vedclass · Accepted Answer

$8.360 	imes 10^4 \, K$

Vedclass · Answer

(B) The root mean square $(rms)$ speed of a gas molecule is given by $v_{rms} = \sqrt{\frac{3k_B T}{m}}$.For the molecules to escape the Earth's atmosphere,their $v_{rms}$ must be equal to the escape velocity of the Earth,$v_e = 11.2 \, km/s = 1.12 	imes 10^4 \, m/s$.Equating $v_{rms} = v_e$,we get $\sqrt{\frac{3k_B T}{m}} = v_e$.Squaring both sides: $\frac{3k_B T}{m} = v_e^2$.Solving for $T$: $T = \frac{m v_e^2}{3k_B}$.Substituting the given values: $T = \frac{(2.76 	imes 10^{-26} \, kg) 	imes (1.12 	imes 10^4 \, m/s)^2}{3 	imes 1.38 	imes 10^{-23} \, JK^{-1}}$.$T = \frac{2.76 	imes 10^{-26} 	imes 1.2544 	imes 10^8}{4.14 	imes 10^{-23}}$.$T = \frac{3.462144 	imes 10^{-18}}{4.14 	imes 10^{-23}} \approx 0.83626 	imes 10^5 \, K = 8.3626 	imes 10^4 \, K$.Rounding to the nearest option,we get $8.360 	imes 10^4 \, K$.

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 $v_e = 11.2 \, km/s$)

Similar Questions

Two containers $A$ and $B$ contain an ideal gas at the same temperature. If the masses of the molecules are $m_A$ and $m_B$ respectively,what is the relationship between their momenta $p_A$ and $p_B$?

The temperature at which the $r.m.s.$ velocity of hydrogen molecules is $4.5$ times that of an oxygen molecule at $47^{\circ} C$ is (Molecular weight of hydrogen and oxygen molecules are $2$ and $32$ respectively). (in $^{\circ} C$)

Let $\bar{v}$,$v_{rms}$ and $v_p$ respectively denote the mean speed,root mean square speed and most probable speed of the molecules in an ideal monoatomic gas at absolute temperature $T$. The mass of the molecule is $m$. Then:

The effect of temperature on Maxwell's speed distribution is correctly shown by

The temperature at which oxygen molecules will have the same r.m.s. speed as helium molecules at $57^{\circ} C$ is (molecular masses of oxygen and helium are $32$ and $4$ respectively.) (in $K$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module