Question 546: Thermodynamics & KTG

Question

The average translational kinetic energy of N
2
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 k
B
= 1.38 $$	imes$$ 10
$$-$$23
J/K) (Fill the nearest integer).

Accepted Answer

Given, the average translational kinetic energy of dinitrogen (N
2
) = Kinetic energy of an electron .... (i)
Translational kinetic energy of dinitrogen (N
2
)
$$KE = {3 \over 2}{K_B}T$$
Here, T = temperature of the gas,
and K
B
= Boltzmann constant.
Kinetic energy of an electron = eV
Given, the potential differential of an electron, V = 0.1 V
Substituting the values in the Eq. (i), we get
$${3 \over 2}{K_B}T = eV$$
$$ \Rightarrow {3 \over 2} 	imes 1.38 	imes {10^{ - 23}} 	imes T = 1.6 	imes {10^{ - 19}} 	imes (0.1)$$
$$T = 773K = 773 - 273^\circ C = 500^\circ C$$

Thermodynamics & KTG

The average translational kinetic energy of N 2 gas molecules at .............$^\circC becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt. (Given k B = 1.38 \times 10 -$23 J/K) (Fill the nearest integer).

Integer / Numeric Answer Type