Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The average translational kinetic energy of gas molecules at__________<img alt="{ }^{\circ} \mathrm{C}" src="https://entrancecorner.codecogs.com/gif.

The average translational kinetic energy of \mathrm{N}_{2} gas molecules at__________{ }^{\circ} \mathrm{C} becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt.(Given\: \: \mathrm{k}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} ) (Fill the nearest integer).

Answers (1)

best_answer

K_{1}= KE\: of\: N_{2}\: gas\: molecule= \frac{3}{2}K_{B}T\rightarrow \left ( 1 \right )
K_{2}= KE\: of\: electron= \left ( e \right )\left ( \Delta V \right )\rightarrow \left ( 2 \right )
K_{1}= K_{2}
T\left ( \frac{3}{2} \times 1\cdot 38\times 10^{-23}\right )= 1\cdot 6\times 10^{-19}\times 0\cdot 1
T\left ( 3\times 0\cdot 69\times 10^{-3} \right )= 1
T= \frac{1000}{3\times 0\cdot 69}
T= \frac{1000}{ 2\cdot 07}\simeq 500\, K

Posted by

vishal kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today
anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs
anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question