Question

At what temperature the velocity of $O_2$ molecules have the same velocity as $S{O}_2$ at ${47}^{\circ }$ C?

• A. 113^oC
• B. 160^oC
• C. -113^oC
• D. -160^oC

Answer 1

The correct option is C

-113^oC

$\frac{C_1}{C_2}=\sqrt{\frac{T_1{M}_2}{T_2{M}_1}}$

$C_1=C_2$

$\frac{T_1}{T_2}=\frac{M_1}{M_2}$

$\frac{T_1}{320}=\frac{32}{64}$

$T_1=160K$

$t^{\circ }C=160-273$

$=-{113}^{\circ }C$