Question

The temperature at which the r.m.s. velocity of carbon dioxide becomes the same as that of nitrogen at 21 $^\circ$C is

• A. 462$^\circ$C
• B. 273 K
• C. 189$^\circ$C
• D. 546 K

Answer 1

Answer: (C)

189$^\circ$C

Answer 2

$u_{CO_2} = \left( \frac{3RT_1}{44} \right)^{1/2}$ $u_{N_2} = \left( \frac{3R \times 294}{28} \right) ^{1/2}$ $\left(\frac{3RT_1}{44} \right) ^{1/2} = \left(\frac{3R \times 294}{28} \right)^{1/2}$ or $T_1$ = 462 K or 189$^\circ$C