The average translational kinetic energy of N2 gas molecules at .............$^\circ$C becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt. (Given kB = 1.38 $\times$ 10$-$23 J/K) (Fill the nearest integer).
Answer (integer)
500
Solution
Given, the average translational kinetic energy of dinitrogen (N<sub>2</sub>) = Kinetic energy of an electron .... (i)<br/><br/>Translational kinetic energy of dinitrogen (N<sub>2</sub>)<br/><br/>$KE = {3 \over 2}{K_B}T$<br/><br/>Here, T = temperature of the gas,<br/><br/>and K<sub>B</sub> = Boltzmann constant.<br/><br/>Kinetic energy of an electron = eV<br/><br/>Given, the potential differential of an electron, V = 0.1 V<br/><br/>Substituting the values in the Eq. (i), we get<br/><br/>${3 \over 2}{K_B}T = eV$<br/><br/>$$ \Rightarrow {3 \over 2} \times 1.38 \times {10^{ - 23}} \times T = 1.6 \times {10^{ - 19}} \times (0.1)$$<br/><br/>$T = 773K = 773 - 273^\circ C = 500^\circ C$
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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