JEE Main & Advanced Physics Kinetic Theory of Gases Sample Paper Topic Test - Kinetic Theory of Gases

The graph which represents the variation of mean kinetic energy of molecules with temperature \[{{t}^{o}}C\] is

A)

B)

C)

D)

Correct Answer: C

Solution :

[c] Mean KE of gas molecule \[E=\frac{3}{2}kT=\frac{3}{2}k(t+273)\]

where \[T=\] temperature is in kelvin and \[t=\] is in centigrade.

\[\therefore \] \[\,E=\frac{3}{2}kt+\frac{3}{2}\,\times 273\,k\]

[k = Boltzmann's constant]

By comparing this equation with standard equation of straight line

\[y=mx+c,\] we get \[m=\frac{3}{2}k\] and \[c=\frac{3}{2}\,273\,k\].

So, the graph between E and t will be straight line with positive intercept on y-axis and positive slope with r-axis.