Question

The graph which represent the variation of mean kinetic energy of molecules with temperature t⁰C is

• A.
• B.
• C.
• D.

Answer 1

Answer: (C)

Answer 2

Mean K.E. 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

∴ $E = \frac{3}{2}kt + \frac{3}{2} \times 273k$

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 $c = \frac{3}{2}273k$. So the graph between E and t will be straight line with positive intercept on E-axis and positive slope with t-axis.