The average translational kinetic energy and | vedclass

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

Question

(A) The average translational kinetic energy $E$ of a gas molecule is given by $E = \frac{3}{2} k_B T$. Thus,$E \propto T$.Given $T_1 = 300 \, K$ and

Vedclass · Accepted Answer

$12.42 	imes 10^{-21} \, J, \, 684 \, m/s$

Vedclass · Answer

(A) The average translational kinetic energy $E$ of a gas molecule is given by $E = \frac{3}{2} k_B T$. Thus,$E \propto T$.Given $T_1 = 300 \, K$ and $T_2 = 600 \, K$,the ratio is $\frac{E_2}{E_1} = \frac{T_2}{T_1} = \frac{600}{300} = 2$.Therefore,$E_2 = 2 	imes E_1 = 2 	imes 6.21 	imes 10^{-21} \, J = 12.42 	imes 10^{-21} \, J$.The $r.m.s.$ speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$. Thus,$v_{rms} \propto \sqrt{T}$.Therefore,$\frac{(v_{rms})_2}{(v_{rms})_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{600}{300}} = \sqrt{2} \approx 1.414$.$(v_{rms})_2 = 1.414 	imes 484 \, m/s \approx 684 \, m/s$.Thus,the values are $12.42 	imes 10^{-21} \, J$ and $684 \, m/s$.

The average translational kinetic energy and the $r.m.s.$ speed of molecules in a sample of oxygen gas at $300 \, K$ are $6.21 \times 10^{-21} \, J$ and $484 \, m/s$ respectively. The corresponding values at $600 \, K$ are nearly (assuming ideal gas behaviour):

Similar Questions

Fill in the blanks:
$(i)$ At absolute zero,the volume of an ideal gas is ...... .
$(ii)$ With an increase in temperature,the pressure of a gas ...... .
$(iii)$ All molecular motion will stop at ...... .
$(iv)$ Due to ...... at higher altitudes from the surface of the Earth,the air becomes cooler.

$A$ flat plate is moving normal to its plane through a gas under the action of a constant force $F$. The gas is kept at a very low pressure. The speed of the plate $v$ is much less than the average speed $u$ of the gas molecules. Which of the following options is/are true?

Moon has no atmosphere because

Vedclass Test Series

Exam Paper Generator

Online Exam Module