MCQ
The temperature of a gas is $ -68^\circ C$. At ...... $^oC$ temperature will the average kinetic energy of its molecules be twice that of at $ -68^\circ C$
- ✓$ 137$
- B$ 127$
- C$ 100$
- D$ 105$
$\Rightarrow 2 = \frac{{{T_2}}}{{{T_1}}} $
$\Rightarrow {T_2} = 2{T_1}$
$ \Rightarrow {T_2} = 2(273 - 68) = 410K = 137^\circ C$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.