Question

If the average velocity of $N_{2}$ molecules is 0.3 m/s at $27^{o}C$, then the velocity of 0.6 m/s will take place at

• A. 273 K
• B. 927 K
• C. 1000 K
• D. 1200 K

Answer 1

Answer: (D)

1200 K

Answer 2

$v = 0.921u$

∴ $\frac{v_{1}}{v_{2}} = \frac{u_{1}}{u_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$

∴ $\frac{0.3}{0.6} = \sqrt{\frac{300}{T_{2}}}$ or $\frac{1}{2} = \sqrt{\frac{300}{T_{2}}}$ or

$T_{2} = 300 \times 4 = 1200K$